Let $a$ and $b$ be $p$-dimensional non zero columns. Find the number of distinct eigenvalues of $ab^t+ba^t$?
(a)1
(b)2
(c)1 or 2
(d)1 or 2 or 3
This is a recent exam question. Couldn't figure it out so tried some examples which gave two distinct eigenvalues.
On
To make the notation a little easier, assume that $a$ and $b$ are the same length, that is, that $a^t a = b^t b$. This can be arranged without changing the matrix $ab^t+ba^t$. It has the nice result that we get an orthogonal basis of the nonzero eigenspaces.
Notice that $ab^t + ba^t = \tfrac12\left( (a+b) (a+b)^t - (a-b)(a-b)^t \right)$ so $ab^t + ba^t$ has spectral decomposition:
$$ ab^t + ba^t = \lambda_1 u_1 u_1^t + \lambda_2 u_2 u_2^t$$ where $$\begin{align} u_1 &= \frac{a+b}{ \|a+b\|_2} & \qquad u_2 &= \frac{a-b}{\|a-b\|_2}\\ \lambda_1 &=\tfrac12(a+b)^t(a+b) & \lambda_2&=-\tfrac{1}{2}(a-b)^t(a-b) \end{align}$$
So typically the matrix has three distinct eigenvalues, $\lambda_1$, $\lambda_2$, and $0$. However, taking $a=b$ or $a=-b$ nonzero gives two distinct eigenvalues as either $\lambda_2=0$ or $\lambda_1=0$ in those cases. And of course, taking $a=b=0$ to be the zero vector gives the zero matrix with only one eigenvalue, $\lambda_1=\lambda_2=0$. This is the only way for $\lambda_1=\lambda_2$ since one is positive and one is negative.
If $p=2$, then we don't get the extra $0$ eigenvalue. If $p=1$ then $a=\pm b$ and we don't get extra $0$ eigenvalues.
$ab^t + ba^t$ has rank at most two, since it is the sum of two rank one matrices. So in $p$-dim space, you know that at least $(p-2)$ of the eigenvalues are $0$.
Next, by change of orthogonal basis, you can assume without loss of generality that $a = [a_1, 0, \dots, 0]^t$ and $b = [b_1, b_2, 0 ,\dots , 0]^t$. So the top left $2\times2$ submatrix will be of the form $$ \begin{bmatrix} 2 a_1 b_1 & a_1 b_2 \\ a_1 b_2 & 0 \end{bmatrix}.$$ By looking at simple examples, you can find cases where it could take two distinct non-zero values.
Thus the answer is (d).