Linear Algebra - "Explain why v*w and 0 are eigenvalues of M"

[ksdhart2]

Hi all,

I'm having difficulties with a problem in my Linear Algebra course, and I think this problem is meant to be a follow-up to previous one, although I'm not quite seeing how they relate The problem is:

3) Let v and w be length n vectors (aka nx1 matrices) and let the matrix M = vw^T. Assume that \(\displaystyle v \cdot w \ne 0\).

a) Explain why \(\displaystyle v \cdot w\) and 0 are eigenvalues of M, and then explain why M is diagonalizable.

b) For \(\displaystyle v=\begin{pmatrix}1\\ 2\\ -1\\ 1\end{pmatrix}\) and \(\displaystyle w=\begin{pmatrix}3\\ 1\\ 2\\ -1\end{pmatrix}\), diagonalize the resulting M.

In the hints, he says it may be helpful to revisit a previous homework involving M=vw^T (I've posted about that question here). He then says:

• Consider Mv. What will this result in?
• Suppose z is a vector orthagonal to w (i.e. \(\displaystyle w \cdot z =0\)), then consider Mz
• What is the dimension of the subspace of all vectors orthagonal to w?

Initially I was completely stumped. I tried following his hints, but they don't seem to be leading me anywhere. Here's what I have so far:

\(\displaystyle v=\begin{pmatrix}v_1\\ v_2\\ ...\\ v_n\end{pmatrix}\) and \(\displaystyle w=\begin{pmatrix}w_1\\ w_2\\ ...\\ w_n\end{pmatrix}\)

\(\displaystyle M=vw^T\) so \(\displaystyle Mv=vw^Tv=\begin{pmatrix}\left(v\cdot w\right)v_1\\ \left(v\cdot \:w\right)v_2\\ ...\\ \left(v\cdot \:w\right)v_n\end{pmatrix}\)

I had trouble expressing z in terms of abstract terms, so I used a concrete example:

\(\displaystyle v=\begin{pmatrix}8\\ 5\\ 4\end{pmatrix}\) and \(\displaystyle w=\begin{pmatrix}2\\ 9\\ 7\end{pmatrix}\)

\(\displaystyle \begin{pmatrix}2\\ 9\\ 7\end{pmatrix}\cdot \begin{pmatrix}\alpha \\ \beta \\ \gamma \end{pmatrix}=0\) has infinitely many solutions, but one such is \(\displaystyle z=\begin{pmatrix}1\\ 6\\ -8\end{pmatrix}\)

Based on this and a few other trials, I have reasonably concluded that for any z orthogonal to w, Mz=O where O is the zero matrix. I'm not certain of this result, nor do I know how to prove it (or disprove it). Furthermore, I'm not sure where to go from here. Even if I am correct and Mz=O, what does that mean for me? How does that help me solve the problem?

Also, what does the last hint even mean? Perhaps I just wasn't playing attention in class, but I don't even know what a subspace is, nor what its dimension might be, nor how that relates. We have no textbook for this class and I've tried looking up other resources, but those often leave me more confused than before, as they use different notation and different terminology than I'm used to seeing in class. Any help sorting this mess out would be very much appreciated. These intensely abstract and purely theoretical questions are always the ones that frustrate me the most.

 

[HallsofIvy]

I think you have misunderstood the notation. If w is given by \(\displaystyle \begin{bmatrix}w_1 \\ w_2 \\ w_3\end{bmatrix}\) then \(\displaystyle w^T= \begin{bmatrix} w_1 & w_2 & w_3\end{bmatrix}\). With \(\displaystyle v= \begin{bmatrix}v_1 \\ v_2\\ v_3\end{bmatrix}\), \(\displaystyle vw^T= \begin{bmatrix}v_1 \\ v_2\\ v_3\end{bmatrix}\begin{bmatrix}w_1 & w_2 & w_3\end{bmatrix}\) is the 3 by 3 matrix \(\displaystyle \begin{bmatrix}v_1w_1 & v_1w_2 & v_1w_3 \\ v_2w_1 & v_2w_2 & v_2w_3 \\ v_3w_1 & v_3w_2 & v_3w_3\end{bmatrix}\)

 

[ksdhart2]

Okay, so I ended up leaving this question blank and turning in the rest of the homework. I now have the solution, but it only leaves me more confused than ever. Perhaps someone can explain it, or at the very least point me in the direction of some easy to follow resources that might help... Here's the full solution to part 3a. 3b was easy for me, and so I won't post the solution.

3a) \(\displaystyle M=vw^T\) Then \(\displaystyle Mv=(vw^T)v=v(w^Tv)=v(v \cdot w)=(v \cdot w)v\) So, v is an eigenvector, with eigenvalue \(\displaystyle v \cdot w\).

Now if \(\displaystyle w \cdot z = 0\), where z is orthagonal to w, then \(\displaystyle Mz=(vw^T)z=v(w^Tz)=v(w \cdot z)=0v=0.\) Since \(\displaystyle Mz=0=0z\), z is an eigenvector, with eigenvalue 0.

Now dimension of all vectors orthagonal to w will be n - 1. (Specifically to find all z with \(\displaystyle w \cdot z = 0\). This is the equation \(\displaystyle w^Tz=0\) which is a 1xn homogeneous system with matrix \(\displaystyle w^T = \begin{pmatrix}w_1&w_2&...&w_n\end{pmatrix}\). It will involved n - 1 parameters and just one pivot.)

So \(\displaystyle dim\left(\epsilon _{\left(0\right)}\right) \ge n-1\) and \(\displaystyle dim\left(\epsilon _{\left(v\cdot w\right)}\right)\ge 1\), but their sum will be \(\displaystyle \le n\). This forces \(\displaystyle dim\left(\epsilon _{\left(0\right)}\right) = n-1\) and \(\displaystyle dim\left(\epsilon _{\left(v\cdot w\right)}\right) = 1\), and M will be diagonalizable.

I think I follow all of the steps he took, but I don't understand why any of these steps are the right ones. I'm specifically confused as to why we can say that 0 is an eigenvalue because \(\displaystyle Mz=0z\). I don't recall that bit of information ever being mentioned in class. Seeing the solution come out so nice and concise really makes me believe that this problem is a trivially easy one if you really understand what eigenvectors are and how they relate to matrices, but clearly I do not. I have, at best, a mechanical understanding of the steps to take to arrive at the desired answer. Any help will be greatly appreciated.