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Attempt to proof following statement (kernel and image of a matrix)
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Let $overrightarrow u, overrightarrow v,overrightarrow w$ be an orthonormal basis of $mathbb R^3$ with respect to the standard inproduct. We define the matrix $A$ as $A= overrightarrow uoverrightarrow u^tr + overrightarrow voverrightarrow v^tr + overrightarrow woverrightarrow w^tr$.
Then show that $A=I_3$.
Hint: Use the preceding exercise and show that $Ker(A-I_3)= mathbb R^3$.
(i) Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$). This means the value on the diagonals will be equal to $1$ in a matrix of $3times 3$.
View it as
$A= beginbmatrix overrightarrow u_1 & overrightarrow v_1 & overrightarrow w_1 \ overrightarrow u_2 & overrightarrow v_2 & overrightarrow w_2 \ overrightarrow u_3 & overrightarrow v_3 & overrightarrow w_3 endbmatrixcdot beginbmatrix overrightarrow u_1^tr & overrightarrow v_1^tr & overrightarrow w_1^tr \ overrightarrow u_2^tr & overrightarrow v_2^tr & overrightarrow w_2^tr \ overrightarrow u_3^tr & overrightarrow v_3^tr & overrightarrow w_3^tr endbmatrix= beginbmatrix 1 & 0 & 0\ 0 & 1 & 0\ 0& 0& 1 endbmatrix= I_3$
(ii) Well if $A= I_3$ then we can conclude from the previous excercise that it is equal to $mathbb R^3$.
L.S.:
Will proofs like this (i.e. informal proofs) if completed be accepted in maths journals or if someone claims to have a solution on one or more of the remaining 6 Millenium Prize Problems?
linear-algebra proof-verification proof-writing linear-transformations alternative-proof
asked Jul 18 at 11:01
Anonymous I
804725
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Let $overrightarrow u, overrightarrow v,overrightarrow w$ be an orthonormal basis of $mathbb R^3$ with respect to the standard inproduct. We define the matrix $A$ as $A= overrightarrow uoverrightarrow u^tr + overrightarrow voverrightarrow v^tr + overrightarrow woverrightarrow w^tr$.
Then show that $A=I_3$.
Hint: Use the preceding exercise and show that $Ker(A-I_3)= mathbb R^3$.
(i) Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$). This means the value on the diagonals will be equal to $1$ in a matrix of $3times 3$.
View it as
$A= beginbmatrix overrightarrow u_1 & overrightarrow v_1 & overrightarrow w_1 \ overrightarrow u_2 & overrightarrow v_2 & overrightarrow w_2 \ overrightarrow u_3 & overrightarrow v_3 & overrightarrow w_3 endbmatrixcdot beginbmatrix overrightarrow u_1^tr & overrightarrow v_1^tr & overrightarrow w_1^tr \ overrightarrow u_2^tr & overrightarrow v_2^tr & overrightarrow w_2^tr \ overrightarrow u_3^tr & overrightarrow v_3^tr & overrightarrow w_3^tr endbmatrix= beginbmatrix 1 & 0 & 0\ 0 & 1 & 0\ 0& 0& 1 endbmatrix= I_3$
(ii) Well if $A= I_3$ then we can conclude from the previous excercise that it is equal to $mathbb R^3$.
L.S.:
Will proofs like this (i.e. informal proofs) if completed be accepted in maths journals or if someone claims to have a solution on one or more of the remaining 6 Millenium Prize Problems?
linear-algebra proof-verification proof-writing linear-transformations alternative-proof
asked Jul 18 at 11:01
Anonymous I
804725
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
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Let $overrightarrow u, overrightarrow v,overrightarrow w$ be an orthonormal basis of $mathbb R^3$ with respect to the standard inproduct. We define the matrix $A$ as $A= overrightarrow uoverrightarrow u^tr + overrightarrow voverrightarrow v^tr + overrightarrow woverrightarrow w^tr$.
Then show that $A=I_3$.
Hint: Use the preceding exercise and show that $Ker(A-I_3)= mathbb R^3$.
(i) Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$). This means the value on the diagonals will be equal to $1$ in a matrix of $3times 3$.
View it as
$A= beginbmatrix overrightarrow u_1 & overrightarrow v_1 & overrightarrow w_1 \ overrightarrow u_2 & overrightarrow v_2 & overrightarrow w_2 \ overrightarrow u_3 & overrightarrow v_3 & overrightarrow w_3 endbmatrixcdot beginbmatrix overrightarrow u_1^tr & overrightarrow v_1^tr & overrightarrow w_1^tr \ overrightarrow u_2^tr & overrightarrow v_2^tr & overrightarrow w_2^tr \ overrightarrow u_3^tr & overrightarrow v_3^tr & overrightarrow w_3^tr endbmatrix= beginbmatrix 1 & 0 & 0\ 0 & 1 & 0\ 0& 0& 1 endbmatrix= I_3$
(ii) Well if $A= I_3$ then we can conclude from the previous excercise that it is equal to $mathbb R^3$.
L.S.:
Will proofs like this (i.e. informal proofs) if completed be accepted in maths journals or if someone claims to have a solution on one or more of the remaining 6 Millenium Prize Problems?
linear-algebra proof-verification proof-writing linear-transformations alternative-proof
asked Jul 18 at 11:01
Anonymous I
804725
Let $overrightarrow u, overrightarrow v,overrightarrow w$ be an orthonormal basis of $mathbb R^3$ with respect to the standard inproduct. We define the matrix $A$ as $A= overrightarrow uoverrightarrow u^tr + overrightarrow voverrightarrow v^tr + overrightarrow woverrightarrow w^tr$.
Then show that $A=I_3$.
Hint: Use the preceding exercise and show that $Ker(A-I_3)= mathbb R^3$.
(i) Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$). This means the value on the diagonals will be equal to $1$ in a matrix of $3times 3$.
View it as
$A= beginbmatrix overrightarrow u_1 & overrightarrow v_1 & overrightarrow w_1 \ overrightarrow u_2 & overrightarrow v_2 & overrightarrow w_2 \ overrightarrow u_3 & overrightarrow v_3 & overrightarrow w_3 endbmatrixcdot beginbmatrix overrightarrow u_1^tr & overrightarrow v_1^tr & overrightarrow w_1^tr \ overrightarrow u_2^tr & overrightarrow v_2^tr & overrightarrow w_2^tr \ overrightarrow u_3^tr & overrightarrow v_3^tr & overrightarrow w_3^tr endbmatrix= beginbmatrix 1 & 0 & 0\ 0 & 1 & 0\ 0& 0& 1 endbmatrix= I_3$
(ii) Well if $A= I_3$ then we can conclude from the previous excercise that it is equal to $mathbb R^3$.
L.S.:
Will proofs like this (i.e. informal proofs) if completed be accepted in maths journals or if someone claims to have a solution on one or more of the remaining 6 Millenium Prize Problems?
linear-algebra proof-verification proof-writing linear-transformations alternative-proof
asked Jul 18 at 11:01
Anonymous I
804725
asked Jul 18 at 11:01
Anonymous I
804725
asked Jul 18 at 11:01
Anonymous I
804725
asked Jul 18 at 11:01
Anonymous I
804725
804725
add a comment |Â
add a comment |Â
2 Answers
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The fact that you are treating $u,v,w$ like $e_1,e_2,e_3$ of $mathbb R^3$,means nothing at all. I do not see any similarity between $u,v,w$ an the standard basis, apart from the fact that both are orthonormal.
The way to see $A$, which you have written, is incorrect. In fact, if say $u = (u_1,u_2,u_3)$, then $uu^T$ is a $3 times 3$ matrix, whose $(i,j)$th entry is $u_iu_j$. Similarly with $v$ and $w$. So, what you have is:
$$
A_ij = u_iu_j + v_iv_j + w_iw_j quad 1 leq i,j leq 3
$$
You can see, for example $A_11$ does not match the description I gave.
For example, $A_23 = u_2u_3 + v_2v_3 + w_2w_3$. It is not at all clear why this entry should be zero.
To show that $A = I_3$, we use the exercise given : show that $ker(A-I_3) = mathbb R^3$. I will now only say $I$ rather than $I_3$.
Naturally, we must first find some elements in the kernel. Let's try operating $A - I$ on $u$. What is $Au - Iu$?
$$
Au - Iu = uu^Tu + v(v^Tu) + w(w^Tu) - u = u +0+0 - u = 0
$$
so $(A-I)u = 0$. Note that $v^Tu = w^Tu = 0$ from orthogonality, and $u^Tu = 1$ from normality.
Show that $(A-I)v = (A-I)w = 0$ similarly.
So the kernel of $(A-I)$ contains $u,v,w$, which are three linearly independent elements of $mathbb R^3$. Can you see why the kernel must be equal to $mathbb R^3$?
Now use the exercise to complete the problem.
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
add a comment |Â
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1
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Short answer: no, informal proofs are not permitted, at least, not this informal.
There will always be steps skipped in a mathematics research paper; nobody wants to read every step. However, the skipped steps should be within easy grasp of the reader, able to be filled in with relatively little effort for somebody in the field.
But, importantly, those steps have to exist. There has to be a way to fill in those gaps, all of them, right from axioms to the theorem proven. Maybe some are in a theorem printed elsewhere, or maybe they are just skipped for expediency, but they should be accessible to the interested reader.
This is the problem with your proofs. There are a number of issues, but I think this statement is fairly representative:
Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$).
I understand that you're going for an informal, more intuitive tone here, but it's not clear what's under the intuition. What exactly does it mean to "treat" one basis like another? Do they produce the same coordinate vectors? Since these bases can be treated as the same basis, does this mean we don't have to use any change of basis formulas? In what circumstances can we replace our bases with the standard basis?
By not being properly clear, you've left the connection between your previous statement and your next statement nebulous. Basically, you've then leapt to writing down the conclusion, in full. Somewhere in that gap of similar "treatment", there needs to be substantial logic, and it's not clear that it occurs in your proof.
Of course, learning how to write proofs properly is not easy. There's a lot of really tricky little issues and conventions that you need to learn to navigate. This proof is nowhere near a research standard, but nobody writes research standard proofs out of the gate.
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The fact that you are treating $u,v,w$ like $e_1,e_2,e_3$ of $mathbb R^3$,means nothing at all. I do not see any similarity between $u,v,w$ an the standard basis, apart from the fact that both are orthonormal.
The way to see $A$, which you have written, is incorrect. In fact, if say $u = (u_1,u_2,u_3)$, then $uu^T$ is a $3 times 3$ matrix, whose $(i,j)$th entry is $u_iu_j$. Similarly with $v$ and $w$. So, what you have is:
$$
A_ij = u_iu_j + v_iv_j + w_iw_j quad 1 leq i,j leq 3
$$
You can see, for example $A_11$ does not match the description I gave.
For example, $A_23 = u_2u_3 + v_2v_3 + w_2w_3$. It is not at all clear why this entry should be zero.
To show that $A = I_3$, we use the exercise given : show that $ker(A-I_3) = mathbb R^3$. I will now only say $I$ rather than $I_3$.
Naturally, we must first find some elements in the kernel. Let's try operating $A - I$ on $u$. What is $Au - Iu$?
$$
Au - Iu = uu^Tu + v(v^Tu) + w(w^Tu) - u = u +0+0 - u = 0
$$
so $(A-I)u = 0$. Note that $v^Tu = w^Tu = 0$ from orthogonality, and $u^Tu = 1$ from normality.
Show that $(A-I)v = (A-I)w = 0$ similarly.
So the kernel of $(A-I)$ contains $u,v,w$, which are three linearly independent elements of $mathbb R^3$. Can you see why the kernel must be equal to $mathbb R^3$?
Now use the exercise to complete the problem.
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
add a comment |Â
up vote
1
down vote
accepted
The fact that you are treating $u,v,w$ like $e_1,e_2,e_3$ of $mathbb R^3$,means nothing at all. I do not see any similarity between $u,v,w$ an the standard basis, apart from the fact that both are orthonormal.
The way to see $A$, which you have written, is incorrect. In fact, if say $u = (u_1,u_2,u_3)$, then $uu^T$ is a $3 times 3$ matrix, whose $(i,j)$th entry is $u_iu_j$. Similarly with $v$ and $w$. So, what you have is:
$$
A_ij = u_iu_j + v_iv_j + w_iw_j quad 1 leq i,j leq 3
$$
You can see, for example $A_11$ does not match the description I gave.
For example, $A_23 = u_2u_3 + v_2v_3 + w_2w_3$. It is not at all clear why this entry should be zero.
To show that $A = I_3$, we use the exercise given : show that $ker(A-I_3) = mathbb R^3$. I will now only say $I$ rather than $I_3$.
Naturally, we must first find some elements in the kernel. Let's try operating $A - I$ on $u$. What is $Au - Iu$?
$$
Au - Iu = uu^Tu + v(v^Tu) + w(w^Tu) - u = u +0+0 - u = 0
$$
so $(A-I)u = 0$. Note that $v^Tu = w^Tu = 0$ from orthogonality, and $u^Tu = 1$ from normality.
Show that $(A-I)v = (A-I)w = 0$ similarly.
So the kernel of $(A-I)$ contains $u,v,w$, which are three linearly independent elements of $mathbb R^3$. Can you see why the kernel must be equal to $mathbb R^3$?
Now use the exercise to complete the problem.
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The fact that you are treating $u,v,w$ like $e_1,e_2,e_3$ of $mathbb R^3$,means nothing at all. I do not see any similarity between $u,v,w$ an the standard basis, apart from the fact that both are orthonormal.
The way to see $A$, which you have written, is incorrect. In fact, if say $u = (u_1,u_2,u_3)$, then $uu^T$ is a $3 times 3$ matrix, whose $(i,j)$th entry is $u_iu_j$. Similarly with $v$ and $w$. So, what you have is:
$$
A_ij = u_iu_j + v_iv_j + w_iw_j quad 1 leq i,j leq 3
$$
You can see, for example $A_11$ does not match the description I gave.
For example, $A_23 = u_2u_3 + v_2v_3 + w_2w_3$. It is not at all clear why this entry should be zero.
To show that $A = I_3$, we use the exercise given : show that $ker(A-I_3) = mathbb R^3$. I will now only say $I$ rather than $I_3$.
Naturally, we must first find some elements in the kernel. Let's try operating $A - I$ on $u$. What is $Au - Iu$?
$$
Au - Iu = uu^Tu + v(v^Tu) + w(w^Tu) - u = u +0+0 - u = 0
$$
so $(A-I)u = 0$. Note that $v^Tu = w^Tu = 0$ from orthogonality, and $u^Tu = 1$ from normality.
Show that $(A-I)v = (A-I)w = 0$ similarly.
So the kernel of $(A-I)$ contains $u,v,w$, which are three linearly independent elements of $mathbb R^3$. Can you see why the kernel must be equal to $mathbb R^3$?
Now use the exercise to complete the problem.
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
The fact that you are treating $u,v,w$ like $e_1,e_2,e_3$ of $mathbb R^3$,means nothing at all. I do not see any similarity between $u,v,w$ an the standard basis, apart from the fact that both are orthonormal.
The way to see $A$, which you have written, is incorrect. In fact, if say $u = (u_1,u_2,u_3)$, then $uu^T$ is a $3 times 3$ matrix, whose $(i,j)$th entry is $u_iu_j$. Similarly with $v$ and $w$. So, what you have is:
$$
A_ij = u_iu_j + v_iv_j + w_iw_j quad 1 leq i,j leq 3
$$
You can see, for example $A_11$ does not match the description I gave.
For example, $A_23 = u_2u_3 + v_2v_3 + w_2w_3$. It is not at all clear why this entry should be zero.
To show that $A = I_3$, we use the exercise given : show that $ker(A-I_3) = mathbb R^3$. I will now only say $I$ rather than $I_3$.
Naturally, we must first find some elements in the kernel. Let's try operating $A - I$ on $u$. What is $Au - Iu$?
$$
Au - Iu = uu^Tu + v(v^Tu) + w(w^Tu) - u = u +0+0 - u = 0
$$
so $(A-I)u = 0$. Note that $v^Tu = w^Tu = 0$ from orthogonality, and $u^Tu = 1$ from normality.
Show that $(A-I)v = (A-I)w = 0$ similarly.
So the kernel of $(A-I)$ contains $u,v,w$, which are three linearly independent elements of $mathbb R^3$. Can you see why the kernel must be equal to $mathbb R^3$?
Now use the exercise to complete the problem.
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
answered Jul 18 at 11:40
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
32k22463
32k22463
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
add a comment |Â
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
I must admit that I do not have enough experience (of paper writing / publishing) to comment on the second question.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 11:46
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
The vectors that become zero is because of the fact that (let's say $overrightarrow u_1 times overrightarrow u_2$= 0 comes from the fact that those vectors are othogonal to eachother.
– Anonymous I
Jul 18 at 12:15
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
"So the kernel of (A−I) contains u,v,w, which are three linearly independent elements of R3. Can you see why the kernel must be equal to R3?" Well because of the proof that it is shown in the link I provided from my previous question.
– Anonymous I
Jul 18 at 12:39
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
If the kernel contains $u,v,w$ then since the kernel is a subspace, it contains the smallest subspace which contains $u,v,w$,which is the span of $u,v,w$, which is $mathbb R^3$. Of course, $mathbb R^3$ contains the kernel, whence they are both equal. This is another way to see the point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 18 at 15:22
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Short answer: no, informal proofs are not permitted, at least, not this informal.
There will always be steps skipped in a mathematics research paper; nobody wants to read every step. However, the skipped steps should be within easy grasp of the reader, able to be filled in with relatively little effort for somebody in the field.
But, importantly, those steps have to exist. There has to be a way to fill in those gaps, all of them, right from axioms to the theorem proven. Maybe some are in a theorem printed elsewhere, or maybe they are just skipped for expediency, but they should be accessible to the interested reader.
This is the problem with your proofs. There are a number of issues, but I think this statement is fairly representative:
Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$).
I understand that you're going for an informal, more intuitive tone here, but it's not clear what's under the intuition. What exactly does it mean to "treat" one basis like another? Do they produce the same coordinate vectors? Since these bases can be treated as the same basis, does this mean we don't have to use any change of basis formulas? In what circumstances can we replace our bases with the standard basis?
By not being properly clear, you've left the connection between your previous statement and your next statement nebulous. Basically, you've then leapt to writing down the conclusion, in full. Somewhere in that gap of similar "treatment", there needs to be substantial logic, and it's not clear that it occurs in your proof.
Of course, learning how to write proofs properly is not easy. There's a lot of really tricky little issues and conventions that you need to learn to navigate. This proof is nowhere near a research standard, but nobody writes research standard proofs out of the gate.
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
add a comment |Â
up vote
1
down vote
Short answer: no, informal proofs are not permitted, at least, not this informal.
There will always be steps skipped in a mathematics research paper; nobody wants to read every step. However, the skipped steps should be within easy grasp of the reader, able to be filled in with relatively little effort for somebody in the field.
But, importantly, those steps have to exist. There has to be a way to fill in those gaps, all of them, right from axioms to the theorem proven. Maybe some are in a theorem printed elsewhere, or maybe they are just skipped for expediency, but they should be accessible to the interested reader.
This is the problem with your proofs. There are a number of issues, but I think this statement is fairly representative:
Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$).
I understand that you're going for an informal, more intuitive tone here, but it's not clear what's under the intuition. What exactly does it mean to "treat" one basis like another? Do they produce the same coordinate vectors? Since these bases can be treated as the same basis, does this mean we don't have to use any change of basis formulas? In what circumstances can we replace our bases with the standard basis?
By not being properly clear, you've left the connection between your previous statement and your next statement nebulous. Basically, you've then leapt to writing down the conclusion, in full. Somewhere in that gap of similar "treatment", there needs to be substantial logic, and it's not clear that it occurs in your proof.
Of course, learning how to write proofs properly is not easy. There's a lot of really tricky little issues and conventions that you need to learn to navigate. This proof is nowhere near a research standard, but nobody writes research standard proofs out of the gate.
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Short answer: no, informal proofs are not permitted, at least, not this informal.
There will always be steps skipped in a mathematics research paper; nobody wants to read every step. However, the skipped steps should be within easy grasp of the reader, able to be filled in with relatively little effort for somebody in the field.
But, importantly, those steps have to exist. There has to be a way to fill in those gaps, all of them, right from axioms to the theorem proven. Maybe some are in a theorem printed elsewhere, or maybe they are just skipped for expediency, but they should be accessible to the interested reader.
This is the problem with your proofs. There are a number of issues, but I think this statement is fairly representative:
Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$).
I understand that you're going for an informal, more intuitive tone here, but it's not clear what's under the intuition. What exactly does it mean to "treat" one basis like another? Do they produce the same coordinate vectors? Since these bases can be treated as the same basis, does this mean we don't have to use any change of basis formulas? In what circumstances can we replace our bases with the standard basis?
By not being properly clear, you've left the connection between your previous statement and your next statement nebulous. Basically, you've then leapt to writing down the conclusion, in full. Somewhere in that gap of similar "treatment", there needs to be substantial logic, and it's not clear that it occurs in your proof.
Of course, learning how to write proofs properly is not easy. There's a lot of really tricky little issues and conventions that you need to learn to navigate. This proof is nowhere near a research standard, but nobody writes research standard proofs out of the gate.
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
Short answer: no, informal proofs are not permitted, at least, not this informal.
There will always be steps skipped in a mathematics research paper; nobody wants to read every step. However, the skipped steps should be within easy grasp of the reader, able to be filled in with relatively little effort for somebody in the field.
But, importantly, those steps have to exist. There has to be a way to fill in those gaps, all of them, right from axioms to the theorem proven. Maybe some are in a theorem printed elsewhere, or maybe they are just skipped for expediency, but they should be accessible to the interested reader.
This is the problem with your proofs. There are a number of issues, but I think this statement is fairly representative:
Let $overrightarrow u = overrightarrow u_1,overrightarrow u_2, overrightarrow u_3 , v=overrightarrow v_1,overrightarrow v_2,overrightarrowv_3,w=overrightarrow w_1,overrightarrow w_2,overrightarrow w_3$ be orthonormal bases (That means you can treat them like $overrightarrow e_1, overrightarrow e_2, overrightarrow e_3$ of $mathbb R^3$).
I understand that you're going for an informal, more intuitive tone here, but it's not clear what's under the intuition. What exactly does it mean to "treat" one basis like another? Do they produce the same coordinate vectors? Since these bases can be treated as the same basis, does this mean we don't have to use any change of basis formulas? In what circumstances can we replace our bases with the standard basis?
By not being properly clear, you've left the connection between your previous statement and your next statement nebulous. Basically, you've then leapt to writing down the conclusion, in full. Somewhere in that gap of similar "treatment", there needs to be substantial logic, and it's not clear that it occurs in your proof.
Of course, learning how to write proofs properly is not easy. There's a lot of really tricky little issues and conventions that you need to learn to navigate. This proof is nowhere near a research standard, but nobody writes research standard proofs out of the gate.
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
answered Jul 18 at 11:46
Theo Bendit
12.1k1844
12.1k1844
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