Dimension Space Given Polynomials [on hold]

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

1

What is the dimension of the space of all polynomials $f(t)$ in $mathrmPol_2(mathbbR)$ such that $f(1) = 0$.

So we have some arbitrary equation $xt^2 + yt + z = 0$. Would the dimension of this space simply be 2 because it takes a square?

linear-algebra linear-transformations dimension-theory

share|cite|improve this question

edited Aug 3 at 12:43

zzuussee

1,101419

asked Aug 3 at 12:05

DoofusAnarchy

74

put on hold as off-topic by Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh Aug 3 at 15:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
  – Arnaud Mortier
  Aug 3 at 12:22
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

What is the dimension of the space of all polynomials $f(t)$ in $mathrmPol_2(mathbbR)$ such that $f(1) = 0$.

So we have some arbitrary equation $xt^2 + yt + z = 0$. Would the dimension of this space simply be 2 because it takes a square?

linear-algebra linear-transformations dimension-theory

share|cite|improve this question

edited Aug 3 at 12:43

zzuussee

1,101419

asked Aug 3 at 12:05

DoofusAnarchy

74

put on hold as off-topic by Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh Aug 3 at 15:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
  – Arnaud Mortier
  Aug 3 at 12:22
  

  
  
  
  

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

What is the dimension of the space of all polynomials $f(t)$ in $mathrmPol_2(mathbbR)$ such that $f(1) = 0$.

So we have some arbitrary equation $xt^2 + yt + z = 0$. Would the dimension of this space simply be 2 because it takes a square?

linear-algebra linear-transformations dimension-theory

share|cite|improve this question

edited Aug 3 at 12:43

zzuussee

1,101419

asked Aug 3 at 12:05

DoofusAnarchy

74

What is the dimension of the space of all polynomials $f(t)$ in $mathrmPol_2(mathbbR)$ such that $f(1) = 0$.

So we have some arbitrary equation $xt^2 + yt + z = 0$. Would the dimension of this space simply be 2 because it takes a square?

linear-algebra linear-transformations dimension-theory

share|cite|improve this question

edited Aug 3 at 12:43

zzuussee

1,101419

asked Aug 3 at 12:05

DoofusAnarchy

74

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 12:43

zzuussee

1,101419

edited Aug 3 at 12:43

zzuussee

1,101419

edited Aug 3 at 12:43

zzuussee

1,101419

1,101419

asked Aug 3 at 12:05

DoofusAnarchy

74

asked Aug 3 at 12:05

DoofusAnarchy

74

asked Aug 3 at 12:05

DoofusAnarchy

74

74

put on hold as off-topic by Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh Aug 3 at 15:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh Aug 3 at 15:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud Mortier, Henrik, José Carlos Santos, Strants, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
  – Arnaud Mortier
  Aug 3 at 12:22
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
  – Arnaud Mortier
  Aug 3 at 12:22
  

  
  
  
  

What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
– Arnaud Mortier
Aug 3 at 12:22

What is the definition of dimension? (Hint: it's not "the first integer that you notice when you look at an expression related to the question")
– Arnaud Mortier
Aug 3 at 12:22

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

So, I answer your question given some conditions. First, I assume you are talking about a subspace of $mathrmPol_2(mathbbR)$, that is all real polynomial functions of degree less than two.

---

The space you described is $U=finmathrmPol_2(mathbbR)mid f(1)=0$, i.e. all polynomials having a root at $1$, i.e. $U=f(t)=at^2+bt+cinmathrmPol_2(mathbbR)mid f(1)=a+b+c=0$.

Note first, that this is really a subspace, i.e. the null polynomial $mathbf0in U$, as $mathbf0(t)=0$ f.a. $tinmathbbR$. Second, if $f,gin U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+gin U$ for the canonical(pointwise addition in $mathrmPol_2(mathbbR)$). Lastly, for $fin U$, we have $(lambda f)(1)=lambdacdot f(1)=lambdacdot 0=0$.

---

Now, the canonical basis for $mathrmPol_2(mathbbR)$ stems from its correspondence with $mathbbR^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $mathrmPol_2(mathbbR)$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $mathrmPol_2(mathbbR)$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).

You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.

For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.

---

A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $mathbbPol_2(mathbbR)$ and $mathbbR^3$ as the corresponding space there, $(x,y,z)inmathbbR^3mid x+y+z=0$ has a similar basis of $((-1,1,0),(-1,0,1))$.

share|cite|improve this answer

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

add a comment | 

up vote
0
down vote

Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=,deg(f)le 2$.
Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,dots ,a_2in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.

Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $deg(g)le 1$ satisfies that for $f(T):=g(T)cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $gin P1$ to define a polynomial with the above condition.

Now we have found that any of your polynomials can be written as $(T-1)cdot g(T)$ with $gin P1$ and also for any $gin P1$ the polynomial $(T-1)cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.

This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
  – zzuussee
  Aug 3 at 12:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
  – Sellerie
  Aug 3 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
  – zzuussee
  Aug 3 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
  – Sellerie
  Aug 3 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
  – zzuussee
  Aug 3 at 13:05
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

So, I answer your question given some conditions. First, I assume you are talking about a subspace of $mathrmPol_2(mathbbR)$, that is all real polynomial functions of degree less than two.

---

The space you described is $U=finmathrmPol_2(mathbbR)mid f(1)=0$, i.e. all polynomials having a root at $1$, i.e. $U=f(t)=at^2+bt+cinmathrmPol_2(mathbbR)mid f(1)=a+b+c=0$.

Note first, that this is really a subspace, i.e. the null polynomial $mathbf0in U$, as $mathbf0(t)=0$ f.a. $tinmathbbR$. Second, if $f,gin U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+gin U$ for the canonical(pointwise addition in $mathrmPol_2(mathbbR)$). Lastly, for $fin U$, we have $(lambda f)(1)=lambdacdot f(1)=lambdacdot 0=0$.

---

Now, the canonical basis for $mathrmPol_2(mathbbR)$ stems from its correspondence with $mathbbR^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $mathrmPol_2(mathbbR)$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $mathrmPol_2(mathbbR)$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).

You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.

For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.

---

A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $mathbbPol_2(mathbbR)$ and $mathbbR^3$ as the corresponding space there, $(x,y,z)inmathbbR^3mid x+y+z=0$ has a similar basis of $((-1,1,0),(-1,0,1))$.

share|cite|improve this answer

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

add a comment | 

up vote
1
down vote



accepted

So, I answer your question given some conditions. First, I assume you are talking about a subspace of $mathrmPol_2(mathbbR)$, that is all real polynomial functions of degree less than two.

---

The space you described is $U=finmathrmPol_2(mathbbR)mid f(1)=0$, i.e. all polynomials having a root at $1$, i.e. $U=f(t)=at^2+bt+cinmathrmPol_2(mathbbR)mid f(1)=a+b+c=0$.

Note first, that this is really a subspace, i.e. the null polynomial $mathbf0in U$, as $mathbf0(t)=0$ f.a. $tinmathbbR$. Second, if $f,gin U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+gin U$ for the canonical(pointwise addition in $mathrmPol_2(mathbbR)$). Lastly, for $fin U$, we have $(lambda f)(1)=lambdacdot f(1)=lambdacdot 0=0$.

---

Now, the canonical basis for $mathrmPol_2(mathbbR)$ stems from its correspondence with $mathbbR^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $mathrmPol_2(mathbbR)$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $mathrmPol_2(mathbbR)$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).

You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.

For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.

---

A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $mathbbPol_2(mathbbR)$ and $mathbbR^3$ as the corresponding space there, $(x,y,z)inmathbbR^3mid x+y+z=0$ has a similar basis of $((-1,1,0),(-1,0,1))$.

share|cite|improve this answer

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

So, I answer your question given some conditions. First, I assume you are talking about a subspace of $mathrmPol_2(mathbbR)$, that is all real polynomial functions of degree less than two.

---

The space you described is $U=finmathrmPol_2(mathbbR)mid f(1)=0$, i.e. all polynomials having a root at $1$, i.e. $U=f(t)=at^2+bt+cinmathrmPol_2(mathbbR)mid f(1)=a+b+c=0$.

Note first, that this is really a subspace, i.e. the null polynomial $mathbf0in U$, as $mathbf0(t)=0$ f.a. $tinmathbbR$. Second, if $f,gin U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+gin U$ for the canonical(pointwise addition in $mathrmPol_2(mathbbR)$). Lastly, for $fin U$, we have $(lambda f)(1)=lambdacdot f(1)=lambdacdot 0=0$.

---

Now, the canonical basis for $mathrmPol_2(mathbbR)$ stems from its correspondence with $mathbbR^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $mathrmPol_2(mathbbR)$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $mathrmPol_2(mathbbR)$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).

You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.

For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.

---

A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $mathbbPol_2(mathbbR)$ and $mathbbR^3$ as the corresponding space there, $(x,y,z)inmathbbR^3mid x+y+z=0$ has a similar basis of $((-1,1,0),(-1,0,1))$.

share|cite|improve this answer

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

So, I answer your question given some conditions. First, I assume you are talking about a subspace of $mathrmPol_2(mathbbR)$, that is all real polynomial functions of degree less than two.

---

The space you described is $U=finmathrmPol_2(mathbbR)mid f(1)=0$, i.e. all polynomials having a root at $1$, i.e. $U=f(t)=at^2+bt+cinmathrmPol_2(mathbbR)mid f(1)=a+b+c=0$.

Note first, that this is really a subspace, i.e. the null polynomial $mathbf0in U$, as $mathbf0(t)=0$ f.a. $tinmathbbR$. Second, if $f,gin U$, then $(f+g)(1)=f(1)+g(1)=0$, i.e. $f+gin U$ for the canonical(pointwise addition in $mathrmPol_2(mathbbR)$). Lastly, for $fin U$, we have $(lambda f)(1)=lambdacdot f(1)=lambdacdot 0=0$.

---

Now, the canonical basis for $mathrmPol_2(mathbbR)$ stems from its correspondence with $mathbbR^3$ and is given as $S=(1,x,x^2)$. Followingly, the dimension of $mathrmPol_2(mathbbR)$ is $3$. Note that the vectors in this basis are functions. Check maybe yourself that this basis really is linearly independent and spans $mathrmPol_2(mathbbR)$(try to express some $f(t)=at^2+bt+c$ as a linear combination of $S$).

You can now solve the equation $a+b+c=0$, defining $U$ over a condition on the coefficients of a corresponding polynomial $f(t)=at^2+bt+c$.

For this, let $c$ and $b$ just be constants, then we can express $a=-b-c$. This gives us an arbitrary choice of the constant $c$ and $b$, with $a$ resulting from these two. Followingly, the dimension of your space is $2$.

---

A corresponding basis would be $(-x^2+x,-x^2+1)$. This stems again from the correspondence of $mathbbPol_2(mathbbR)$ and $mathbbR^3$ as the corresponding space there, $(x,y,z)inmathbbR^3mid x+y+z=0$ has a similar basis of $((-1,1,0),(-1,0,1))$.

share|cite|improve this answer

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

share|cite|improve this answer

share|cite|improve this answer

edited Aug 3 at 12:50

edited Aug 3 at 12:50

edited Aug 3 at 12:50

answered Aug 3 at 12:32

zzuussee

1,101419

answered Aug 3 at 12:32

zzuussee

1,101419

answered Aug 3 at 12:32

zzuussee

1,101419

1,101419

add a comment | 

add a comment | 

up vote
0
down vote

Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=,deg(f)le 2$.
Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,dots ,a_2in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.

Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $deg(g)le 1$ satisfies that for $f(T):=g(T)cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $gin P1$ to define a polynomial with the above condition.

Now we have found that any of your polynomials can be written as $(T-1)cdot g(T)$ with $gin P1$ and also for any $gin P1$ the polynomial $(T-1)cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.

This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
  – zzuussee
  Aug 3 at 12:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
  – Sellerie
  Aug 3 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
  – zzuussee
  Aug 3 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
  – Sellerie
  Aug 3 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
  – zzuussee
  Aug 3 at 13:05
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=,deg(f)le 2$.
Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,dots ,a_2in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.

Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $deg(g)le 1$ satisfies that for $f(T):=g(T)cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $gin P1$ to define a polynomial with the above condition.

Now we have found that any of your polynomials can be written as $(T-1)cdot g(T)$ with $gin P1$ and also for any $gin P1$ the polynomial $(T-1)cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.

This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
  – zzuussee
  Aug 3 at 12:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
  – Sellerie
  Aug 3 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
  – zzuussee
  Aug 3 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
  – Sellerie
  Aug 3 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
  – zzuussee
  Aug 3 at 13:05
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=,deg(f)le 2$.
Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,dots ,a_2in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.

Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $deg(g)le 1$ satisfies that for $f(T):=g(T)cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $gin P1$ to define a polynomial with the above condition.

Now we have found that any of your polynomials can be written as $(T-1)cdot g(T)$ with $gin P1$ and also for any $gin P1$ the polynomial $(T-1)cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.

This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

Let's first ask the question what dimension the vektorspace $P2$ has over your given field $K$. Here I'd assume that $P2=,deg(f)le 2$.
Within $P2$ you can find the linearly independent elements $T^0,T^1,T^2$, which is also a generator set, seeing that any polynomial from degree $le 2$ can be written as $a_2 T^2+a_1 T^1+a_0 T^0$ with $a_0,dots ,a_2in K$. Now this implies that your set $P2$ has dimension $3$ over the arbitrary field $K$.

Now we gotta think about the second condition forced upon us, which is $f(1)=0$. Let's first think about what polynomials are actually given by this equation: Since $f(1)=0$, we know that the polynomial $T-1$ divides into $f$, i.e. $f(T)=(T-1)g(T)$, where $g$ needs to be a polynomial of degree $<2$. Then again, any polynomial $g$ with degree $deg(g)le 1$ satisfies that for $f(T):=g(T)cdot(T-1)$ the above condition holds, i.e. you can use any polynomial $gin P1$ to define a polynomial with the above condition.

Now we have found that any of your polynomials can be written as $(T-1)cdot g(T)$ with $gin P1$ and also for any $gin P1$ the polynomial $(T-1)cdot g(T)$ satisfies your proposed condition. It is only natural to assume that the dimension must be the same as $P1$, which is $2$ with an analogous argument to the first paragraph.

This assumption must be proven first, which requires you to show that for any two linearly independent $g_1,g_2in P1$ the elements $(T-1)g_1(T)$ and $(T-1)g_2(T)$ are also linearly independent. Can you spot why this would suffice without the need to show that those $2$ linearly independent elements also generate your entire set?

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 12:44

Sellerie

368

answered Aug 3 at 12:44

Sellerie

368

answered Aug 3 at 12:44

Sellerie

368

368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
  – zzuussee
  Aug 3 at 12:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
  – Sellerie
  Aug 3 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
  – zzuussee
  Aug 3 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
  – Sellerie
  Aug 3 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
  – zzuussee
  Aug 3 at 13:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
  – zzuussee
  Aug 3 at 12:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
  – Sellerie
  Aug 3 at 12:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
  – zzuussee
  Aug 3 at 12:57
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
  – Sellerie
  Aug 3 at 13:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
  – zzuussee
  Aug 3 at 13:05
  

  
  
  
  

You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
– zzuussee
Aug 3 at 12:48

You are not really technically fine, as you are working over any filed $K$ in correspondence with the formal polynomials $K[T]$. There might not be a $1-1$-correspondence of $K_2[T]$ and $mathrmPol_2(K)$ as e.g. for $K$ a finite field. Thus, you simple exchange of a formal polynomial $f(T)$ and its realization in $mathrmPol_2(K)$ as used in $f(1)$ is a little tricky.
– zzuussee
Aug 3 at 12:48

I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
– Sellerie
Aug 3 at 12:55

I do agree, there is no $1-1$-correspondence between polynomials and polynomial functions, but which part of my argument uses this correspondence? Maybe my argument only works for $K_2[T]$, if that is the space of polynomials, but I have worked under the assumption that we are only interested in polynomials, not polynomial functions. Is there a problem with my argument over any field if I replace $f(1)=0$ by $1$ is a root of $f$?
– Sellerie
Aug 3 at 12:55

May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
– zzuussee
Aug 3 at 12:57

May impression was just that OP was concerned about the polynomial functions not the formal polynomials, note that I followingly assumed him working over $mathbbR$ to present my argument very similar to yours. I think this adjustment should be fine. As I said it is just a (even minor) technicality but since I'm working over finite fields a lot, I just am careful as how much the deviate in some cases. I think OP should just confirm some more surroundings of his/her question. I think that your approach, if appropriate, is always nicer from an algebraic point of view.
– zzuussee
Aug 3 at 12:57

I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
– Sellerie
Aug 3 at 13:01

I have read your answer just now and do agree, it might be more fitting to the level OP is thinking about the question. While my argument would work quite similarly to yours and therefore just as fine if I simply replaced every occurence of $K$ with $mathbbR$, I think yours is better to understand, so I'll leave mine as it is and hope that OP eventually gets to the point where reading this sparks her/his interest about general polynomials and possbily finite fields.
– Sellerie
Aug 3 at 13:01

I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
– zzuussee
Aug 3 at 13:05

I totally agree that formal polynomials over arbitrary field(or even rings) are far more interesting than the space of polynomial functions. Lets hope OP gets something out of this conversation as well.
– zzuussee
Aug 3 at 13:05

add a comment |