ed84c 10 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" 0 Share this post Link to post Share on other sites

timo 563 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. 0 Share this post Link to post Share on other sites

Amr Morsi 15 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. 0 Share this post Link to post Share on other sites

timo 563 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. 0 Share this post Link to post Share on other sites

Amr Morsi 15 Posted March 29, 2010 Why timo? Multiply both sides by [AA]^-1, which gives x=[AA]^-1 . 0=0. Why not? 0 Share this post Link to post Share on other sites

timo 563 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 Share this post Link to post Share on other sites

Amr Morsi 15 Posted April 2, 2010 You are right, timo. A is not invetible and so A^2. Sorry for that error. 1 Share this post Link to post Share on other sites

joigus 107 Posted May 23 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}. 0 Share this post Link to post Share on other sites