ed84c Posted March 18, 2010 Share Posted March 18, 2010 Question: Show that [math] {Ax = v_{0}}[/math] has no solution. I know [math] v_{0}[/math] is an eigenvector of A with eigenvalue 0, and the other eigenvectors do not have 0 eigenvalues. So, [math] {Av_0= \lambda_{0} v_{0}}[/math] [math] {Av_0= 0 v_{0}}[/math] [math] {Av_0= 0}[/math] So [math] {v_0}[/math] "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" Link to comment Share on other sites More sharing options...

timo Posted March 18, 2010 Share Posted March 18, 2010 In [math]\mathbb R^N[/math] multiplying with [math]v_0^t[/math] (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 [math]v_0[/math] component from a vector will not have [math]v_0[/math] as a result when operated on any vector. Link to comment Share on other sites More sharing options...

ed84c Posted March 19, 2010 Author Share Posted March 19, 2010 Thanks Timo Link to comment Share on other sites More sharing options...

Amr Morsi Posted March 29, 2010 Share Posted March 29, 2010 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. Link to comment Share on other sites More sharing options...

timo Posted March 29, 2010 Share Posted March 29, 2010 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. Link to comment Share on other sites More sharing options...

Amr Morsi Posted March 29, 2010 Share Posted March 29, 2010 Why timo? Multiply both sides by [AA]^-1, which gives x=[AA]^-1 . 0=0. Why not? Link to comment Share on other sites More sharing options...

timo Posted March 29, 2010 Share Posted March 29, 2010 (edited) The matrix A in question has an eigenvalue of zero and hence is not invertible. Since [math] Av_0 = 0 [/math] (as given in the text) you immediately get [math] AAv_0 = A0 = 0 [/math] for [math]v_0 \neq 0 [/math]. Edited March 29, 2010 by timo tyop 1 Link to comment Share on other sites More sharing options...

Amr Morsi Posted April 2, 2010 Share Posted April 2, 2010 You are right, timo. A is not invetible and so A^2. Sorry for that error. 1 Link to comment Share on other sites More sharing options...

joigus Posted May 23, 2020 Share Posted May 23, 2020 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 x_{n}. Link to comment Share on other sites More sharing options...

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