Given a matrix $A_x$, which of the following properties is true?

Question

Let $A_x \in \mathbb{R^{n \times n}}$ be defined as $A_y = \vec x \cdot \vec x^T$, where $\vec x \in \mathbb{R^n}$, $||\vec x||>1$ and $n > 2$, then are $- \vec x $ and $\vec x $ eigen vectors of $A_x$ and if soil $\vec x$ an eigen value of $A_x$. I wasn't sure how to solve this, so I used a simple example to try to understand this. I chose $n = 3$, which gives us $A_x = \begin{bmatrix}1&2&3\\4&5&6\\7&8&9\\\end{bmatrix}$ and choosing $\vec x = \begin{bmatrix}1\\2\\3\\\end{bmatrix}$, I get $||\vec x||= \sqrt {14}>1$. But I'm not sure how $\vec x \vec x^T$ fits into this. Based on my chosen matrix $A_x$, I get eigen values to be $λ=0, λ=\frac{15+3\sqrt{33}}{2},\:λ=\frac{15-3\sqrt{33}}{2}$ and the eigen vectors to be $\begin{pmatrix}1\\ -2\\ 1\end{pmatrix},\:\begin{pmatrix}2\left(3\sqrt{33}-11\right)\\ 11+3\sqrt{33}\\ 44\end{pmatrix},\:\begin{pmatrix}-2\left(11+3\sqrt{33}\right)\\ -3\sqrt{33}+11\\ 44\end{pmatrix}$. This shows that $- \vec x$ and $\vec x$ are not eigen vectors of $A_v$ and $||\vec x ||$ is also not an eigen value of $A_x$ but I'm pretty sure that that's wrong and that I need to use the fact that $A_x = \vec x \vec x^T$ and work with that to answer my question.

Answer 1

Use the fundamental definition of an eigenvector: it’s a vector $\vec x$ such that $A\vec x=\lambda\vec x$ for some scalar $\lambda$. Here, $$A\vec x = (\vec x \vec x^T) \vec x = \vec x (\vec x^T\vec x).$$ Now $\vec x^T\vec x$ is just the dot product of $\vec x$ with itself, and that’s a scalar, so $A\vec x$ is indeed equal to a scalar times $\vec x$, hence it’s an eigenvector of $A$. You could do the same thing for $-\vec x$, but there’s really no need to since any nonzero multiple of an eigenvector is also an eigenvector.

As to why this didn’t work for your example, that’s because your $A_x\ne\vec x\vec x^T$. In fact, it’s not equal to either $\vec v\vec x^T$ or $\vec x\vec v^T$ for any vector $\vec v$. Try again with $$A=\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix} = \begin{bmatrix}1&2&3\\2&4&6\\3&6&9\end{bmatrix}$$ instead.

Answer 2

Well lets try to explicitly solve for an eigenvector $v$. We have $$A_xv = xx^T (v) = x(x^T v) = x \langle v, x \rangle = \langle v, x \rangle x.$$

Case 1, $\langle v, x \rangle = 0$: We have $$Av = 0 = 0v.$$ Therefore if $\langle v, x \rangle = 0$, then $v$ is an eigenvector of $A_x$ with an eigenvalue of $0$.

Case 2, $\langle v, x \rangle \neq 0$: Suppose $$Av = \langle v, x \rangle x = \lambda v$$ for some eigenvalue $\lambda$. Then $\lambda$ cannot be zero because $\langle v, x \rangle x \neq 0$. Thus, dividing through by $\lambda$ we get $$v = \frac{\langle v, x \rangle}{\lambda}x = NONZERO \cdot x.$$ Therefore $x$ is an eigenvector of $A_x$ and we can calculate its eigenvalue: $$A_x x = \langle x, x \rangle x = ||x||^2x,$$ so the eigenvalue associated with $x$ is $||x||^2$.

We have explicitly solved for the eigenvalues and their associated eigenvectors by covering all the cases, so these must be the only eigenvectors and eigenvalues.