# Ax = b question

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Question:

Show that ${Ax = v_{0}}$ has no solution.

I know $v_{0}$ is an eigenvector of A with eigenvalue 0, and the other eigenvectors do not have 0 eigenvalues.

So,

${Av_0= \lambda_{0} v_{0}}$

${Av_0= 0 v_{0}}$

${Av_0= 0}$

So ${v_0}$ "is" the null space of A (since no other eigenvectors have eigenvalues of 0).

So the question is asking me to prove there is no vector that when operated on by A gets to the null space.

I can't think of how to prove this though, apart from saying "A operating on x can only give a vector that is 0 or in the column space"

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In $\mathbb R^N$ multiplying with $v_0^t$ (the transpose) from the left should constitute a proof. I currently can't think of a more general solution, despite that intuitively it is clear that a matrix that kills the $v_0$ component from a vector will not have $v_0$ as a result when operated on any vector.

Thanks Timo

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• 2 weeks later...

Multiply both sides by A from the left and then use the notation you introduced, which is A . vo=0. Then you will have A^2 . x = 0 which gives x=0.

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AAx = 0 does not imply x=0. x=0 is not the same as "no solution". So I don't think that will help here.

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Why timo? Multiply both sides by [AA]^-1, which gives x=[AA]^-1 . 0=0.

Why not?

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The matrix A in question has an eigenvalue of zero and hence is not invertible. Since $Av_0 = 0$ (as given in the text) you immediately get $AAv_0 = A0 = 0$ for $v_0 \neq 0$.

Edited by timo
tyop
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You are right, timo. A is not invetible and so A^2. Sorry for that error.

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• 10 years later...

OK, it's been ages since you posted this, but I couldn't resist. I'm just refreshing my linear algebra.

On 3/18/2010 at 8:29 PM, ed84c said:

I know v0 is an eigenvector of A with eigenvalue 0, and the other eigenvectors do not have 0 eigenvalues.

Then I can pick a basis in which,

$A=\left(\begin{array}{ccccc} a_{1} & 0 & \cdots & 0 & 0\\ 0 & a_{2} & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & a_{n-1} & 0\\ 0 & 0 & \cdots & 0 & 0 \end{array}\right)$

And, without loss of generality,

$v_{0}=\left(\begin{array}{c} 0\\ 0\\ \vdots\\ 0\\ 1 \end{array}\right)$

Generic n-vector:

$x=\left(\begin{array}{c} x_{1}\\ x_{2}\\ \vdots\\ x_{n-1}\\ x_{n} \end{array}\right)$

Eqs. render as,

$a_{1}x_{1}=0$

$a_{2}x_{2}=0$

$\vdots$

$a_{n-1}x_{n-1}=0$

$0x_{n}=1$

So no solution for xn.

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