How to show linear independence of three elements connected by a linear transformation.

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Let $V$ be a three dimensional vector space over the field $mathbbQ.$ Suppose $T: V to V$ be a linear transformation and $T(x)=y,$ $T(y)=z,$ $T(z)=x+y,$ for certain $x,y,z in V,x neq0.$ Prove that $x,y$ and $z$ are linearly independent.

What I did: Consider a linear homogeneous relation $ax+by+cz=0$ for $a,b,c in mathbbQ.$ Since $xneq0,$ we also have $z-x neq 0.$ We also get $(aI + bT +cT^2)(x)=0.$ After that I cannot conclude anything. I need some help.

linear-algebra matrices vector-spaces linear-transformations minimal-polynomials

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asked Jul 30 at 18:26

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Let $V$ be a three dimensional vector space over the field $mathbbQ.$ Suppose $T: V to V$ be a linear transformation and $T(x)=y,$ $T(y)=z,$ $T(z)=x+y,$ for certain $x,y,z in V,x neq0.$ Prove that $x,y$ and $z$ are linearly independent.

What I did: Consider a linear homogeneous relation $ax+by+cz=0$ for $a,b,c in mathbbQ.$ Since $xneq0,$ we also have $z-x neq 0.$ We also get $(aI + bT +cT^2)(x)=0.$ After that I cannot conclude anything. I need some help.

linear-algebra matrices vector-spaces linear-transformations minimal-polynomials

share|cite|improve this question

asked Jul 30 at 18:26

user371231

411410

add a comment | 

up vote
2
down vote

favorite

4

up vote
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down vote

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4

Let $V$ be a three dimensional vector space over the field $mathbbQ.$ Suppose $T: V to V$ be a linear transformation and $T(x)=y,$ $T(y)=z,$ $T(z)=x+y,$ for certain $x,y,z in V,x neq0.$ Prove that $x,y$ and $z$ are linearly independent.

What I did: Consider a linear homogeneous relation $ax+by+cz=0$ for $a,b,c in mathbbQ.$ Since $xneq0,$ we also have $z-x neq 0.$ We also get $(aI + bT +cT^2)(x)=0.$ After that I cannot conclude anything. I need some help.

linear-algebra matrices vector-spaces linear-transformations minimal-polynomials

share|cite|improve this question

asked Jul 30 at 18:26

user371231

411410

Let $V$ be a three dimensional vector space over the field $mathbbQ.$ Suppose $T: V to V$ be a linear transformation and $T(x)=y,$ $T(y)=z,$ $T(z)=x+y,$ for certain $x,y,z in V,x neq0.$ Prove that $x,y$ and $z$ are linearly independent.

What I did: Consider a linear homogeneous relation $ax+by+cz=0$ for $a,b,c in mathbbQ.$ Since $xneq0,$ we also have $z-x neq 0.$ We also get $(aI + bT +cT^2)(x)=0.$ After that I cannot conclude anything. I need some help.

linear-algebra matrices vector-spaces linear-transformations minimal-polynomials

share|cite|improve this question

asked Jul 30 at 18:26

user371231

411410

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 18:26

user371231

411410

asked Jul 30 at 18:26

user371231

411410

asked Jul 30 at 18:26

user371231

411410

411410

add a comment | 

add a comment | 

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Note: I'm not sure I ever use that $V$ is 3-dimensional below. This makes me concerned; be warned. But I'm pretty convinced the argument is correct.

First let's show that $x$ and $y$ are linearly independent. Proceed by contradiction. Suppose instead $x$ and $y$ are linearly dependent, which means that $y=ax$ for some $a in mathbbQ$ (since $x$ is not zero, $x$ and $y$ linearly dependent $Rightarrow y = ax$ for some $a in mathbbQ$).

Now $z = T(y) = T(ax) = aT(x) = a^2x$. And $x+y = T(z) = T(a^2x) = a^3x$. So $a^3x = x+y = x+ax Rightarrow (a^3-a-1)x = mathbf0$. Since $x$ is not $mathbf0$, we must have that $a^3 - a + 1 = 0 $; but this is impossible for $a in mathbbQ$ (by the rational root theorem, checking that $pm 1$ are not roots).

Ok so we've shown that $x$ and $y$ must be linearly independent.

Let's now show that $x$, $y$, and $z$ are linearly independent. Proceed by contradiction again. Suppose instead that $x$, $y$, and $z$ are linearly dependent, which means that $z = ax + by$ for some $a,b in mathbbQ$ (since we know $x$ and $y$ are linearly independent, we can guarantee this is possible).

Now
$$x+y = T(z) = T(ax + by) = aT(x) + bT(y) = ay + bz = ay + b(ax+by) = (ab)x + (a+b^2)y.$$
Rearranging gives $(ab-1)x + (a+b^2-1)y = mathbf0$. Since $x$ and $y$ are linearly independent, we have then that $ab-1 = 0$ and $a+b^2-1 = 0$.

Now $ab-1 = 0 Rightarrow ab = 1 Rightarrow b = 1/a$ and $a neq 0$ and $b neq 0$.

Plugging this into the second equation, we have $a + (1/a)^2 - 1 = 0 Rightarrow a^3 - a^2 + 1 = 0$. And again, this is impossible for $a in mathbbQ$ (by the rational root theorem).

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answered Jul 31 at 4:46

xmq

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Note: I'm not sure I ever use that $V$ is 3-dimensional below. This makes me concerned; be warned. But I'm pretty convinced the argument is correct.

First let's show that $x$ and $y$ are linearly independent. Proceed by contradiction. Suppose instead $x$ and $y$ are linearly dependent, which means that $y=ax$ for some $a in mathbbQ$ (since $x$ is not zero, $x$ and $y$ linearly dependent $Rightarrow y = ax$ for some $a in mathbbQ$).

Now $z = T(y) = T(ax) = aT(x) = a^2x$. And $x+y = T(z) = T(a^2x) = a^3x$. So $a^3x = x+y = x+ax Rightarrow (a^3-a-1)x = mathbf0$. Since $x$ is not $mathbf0$, we must have that $a^3 - a + 1 = 0 $; but this is impossible for $a in mathbbQ$ (by the rational root theorem, checking that $pm 1$ are not roots).

Ok so we've shown that $x$ and $y$ must be linearly independent.

Let's now show that $x$, $y$, and $z$ are linearly independent. Proceed by contradiction again. Suppose instead that $x$, $y$, and $z$ are linearly dependent, which means that $z = ax + by$ for some $a,b in mathbbQ$ (since we know $x$ and $y$ are linearly independent, we can guarantee this is possible).

Now
$$x+y = T(z) = T(ax + by) = aT(x) + bT(y) = ay + bz = ay + b(ax+by) = (ab)x + (a+b^2)y.$$
Rearranging gives $(ab-1)x + (a+b^2-1)y = mathbf0$. Since $x$ and $y$ are linearly independent, we have then that $ab-1 = 0$ and $a+b^2-1 = 0$.

Now $ab-1 = 0 Rightarrow ab = 1 Rightarrow b = 1/a$ and $a neq 0$ and $b neq 0$.

Plugging this into the second equation, we have $a + (1/a)^2 - 1 = 0 Rightarrow a^3 - a^2 + 1 = 0$. And again, this is impossible for $a in mathbbQ$ (by the rational root theorem).

share|cite|improve this answer

answered Jul 31 at 4:46

xmq

534

add a comment | 

up vote
0
down vote



accepted

Note: I'm not sure I ever use that $V$ is 3-dimensional below. This makes me concerned; be warned. But I'm pretty convinced the argument is correct.

First let's show that $x$ and $y$ are linearly independent. Proceed by contradiction. Suppose instead $x$ and $y$ are linearly dependent, which means that $y=ax$ for some $a in mathbbQ$ (since $x$ is not zero, $x$ and $y$ linearly dependent $Rightarrow y = ax$ for some $a in mathbbQ$).

Now $z = T(y) = T(ax) = aT(x) = a^2x$. And $x+y = T(z) = T(a^2x) = a^3x$. So $a^3x = x+y = x+ax Rightarrow (a^3-a-1)x = mathbf0$. Since $x$ is not $mathbf0$, we must have that $a^3 - a + 1 = 0 $; but this is impossible for $a in mathbbQ$ (by the rational root theorem, checking that $pm 1$ are not roots).

Ok so we've shown that $x$ and $y$ must be linearly independent.

Let's now show that $x$, $y$, and $z$ are linearly independent. Proceed by contradiction again. Suppose instead that $x$, $y$, and $z$ are linearly dependent, which means that $z = ax + by$ for some $a,b in mathbbQ$ (since we know $x$ and $y$ are linearly independent, we can guarantee this is possible).

Now
$$x+y = T(z) = T(ax + by) = aT(x) + bT(y) = ay + bz = ay + b(ax+by) = (ab)x + (a+b^2)y.$$
Rearranging gives $(ab-1)x + (a+b^2-1)y = mathbf0$. Since $x$ and $y$ are linearly independent, we have then that $ab-1 = 0$ and $a+b^2-1 = 0$.

Now $ab-1 = 0 Rightarrow ab = 1 Rightarrow b = 1/a$ and $a neq 0$ and $b neq 0$.

Plugging this into the second equation, we have $a + (1/a)^2 - 1 = 0 Rightarrow a^3 - a^2 + 1 = 0$. And again, this is impossible for $a in mathbbQ$ (by the rational root theorem).

share|cite|improve this answer

answered Jul 31 at 4:46

xmq

534

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Note: I'm not sure I ever use that $V$ is 3-dimensional below. This makes me concerned; be warned. But I'm pretty convinced the argument is correct.

First let's show that $x$ and $y$ are linearly independent. Proceed by contradiction. Suppose instead $x$ and $y$ are linearly dependent, which means that $y=ax$ for some $a in mathbbQ$ (since $x$ is not zero, $x$ and $y$ linearly dependent $Rightarrow y = ax$ for some $a in mathbbQ$).

Now $z = T(y) = T(ax) = aT(x) = a^2x$. And $x+y = T(z) = T(a^2x) = a^3x$. So $a^3x = x+y = x+ax Rightarrow (a^3-a-1)x = mathbf0$. Since $x$ is not $mathbf0$, we must have that $a^3 - a + 1 = 0 $; but this is impossible for $a in mathbbQ$ (by the rational root theorem, checking that $pm 1$ are not roots).

Ok so we've shown that $x$ and $y$ must be linearly independent.

Let's now show that $x$, $y$, and $z$ are linearly independent. Proceed by contradiction again. Suppose instead that $x$, $y$, and $z$ are linearly dependent, which means that $z = ax + by$ for some $a,b in mathbbQ$ (since we know $x$ and $y$ are linearly independent, we can guarantee this is possible).

Now
$$x+y = T(z) = T(ax + by) = aT(x) + bT(y) = ay + bz = ay + b(ax+by) = (ab)x + (a+b^2)y.$$
Rearranging gives $(ab-1)x + (a+b^2-1)y = mathbf0$. Since $x$ and $y$ are linearly independent, we have then that $ab-1 = 0$ and $a+b^2-1 = 0$.

Now $ab-1 = 0 Rightarrow ab = 1 Rightarrow b = 1/a$ and $a neq 0$ and $b neq 0$.

Plugging this into the second equation, we have $a + (1/a)^2 - 1 = 0 Rightarrow a^3 - a^2 + 1 = 0$. And again, this is impossible for $a in mathbbQ$ (by the rational root theorem).

share|cite|improve this answer

answered Jul 31 at 4:46

xmq

534

Note: I'm not sure I ever use that $V$ is 3-dimensional below. This makes me concerned; be warned. But I'm pretty convinced the argument is correct.

First let's show that $x$ and $y$ are linearly independent. Proceed by contradiction. Suppose instead $x$ and $y$ are linearly dependent, which means that $y=ax$ for some $a in mathbbQ$ (since $x$ is not zero, $x$ and $y$ linearly dependent $Rightarrow y = ax$ for some $a in mathbbQ$).

Now $z = T(y) = T(ax) = aT(x) = a^2x$. And $x+y = T(z) = T(a^2x) = a^3x$. So $a^3x = x+y = x+ax Rightarrow (a^3-a-1)x = mathbf0$. Since $x$ is not $mathbf0$, we must have that $a^3 - a + 1 = 0 $; but this is impossible for $a in mathbbQ$ (by the rational root theorem, checking that $pm 1$ are not roots).

Ok so we've shown that $x$ and $y$ must be linearly independent.

Let's now show that $x$, $y$, and $z$ are linearly independent. Proceed by contradiction again. Suppose instead that $x$, $y$, and $z$ are linearly dependent, which means that $z = ax + by$ for some $a,b in mathbbQ$ (since we know $x$ and $y$ are linearly independent, we can guarantee this is possible).

Now
$$x+y = T(z) = T(ax + by) = aT(x) + bT(y) = ay + bz = ay + b(ax+by) = (ab)x + (a+b^2)y.$$
Rearranging gives $(ab-1)x + (a+b^2-1)y = mathbf0$. Since $x$ and $y$ are linearly independent, we have then that $ab-1 = 0$ and $a+b^2-1 = 0$.

Now $ab-1 = 0 Rightarrow ab = 1 Rightarrow b = 1/a$ and $a neq 0$ and $b neq 0$.

Plugging this into the second equation, we have $a + (1/a)^2 - 1 = 0 Rightarrow a^3 - a^2 + 1 = 0$. And again, this is impossible for $a in mathbbQ$ (by the rational root theorem).

share|cite|improve this answer

answered Jul 31 at 4:46

xmq

534

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 4:46

xmq

534

answered Jul 31 at 4:46

xmq

534

answered Jul 31 at 4:46

xmq

534

534

add a comment | 

add a comment | 

 

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