Can I Find the Eigenvalues of a Matrix this Way?

Question

I have a matrix $A = xx^T - yy^T,$ where both $x$ and $y$ are linearly independent $n$-column vectors, $n\geq 2$. To find the eigenvalues, I reasoned this way:

Since $x$ and $y$ are linearly independent vectors (given), then $rank(A) = 2.$ So, we have two non-zero eigenvalues and $(n-2)$ eigenvalues, each with a value of zero.

Since both $x$ and $y$ are linearly independent, they form a basis for $V = span(x,y).$ Therefore,

$Ax = (x.x)x - (x.y)y$ and

$Ay = (x.y)y - (y.y)y$

The matrix $A$ relative to $V$ is:

$A = \begin{bmatrix}x.x&x.y\\-x.y&-y.y\end{bmatrix}.$

Now,

$Av =$ $\lambda$$v$

Therefore, $(A - \lambda I_{2})v = 0$, where $A$ has been restricted to a $2 \times 2$ matrix.

When we solve this, we get the $\lambda$'s.

Is this correct so far?

Thanks.

Answer 1

Yes, your solution is correct. The restriction of $A$ to $V = \operatorname{span}(x,y)$ is valid because $V$ is an $A$-invariant subspace of $\Bbb R^n$.

Here's an alternative solution that I like. The matrix $A$ can be expressed as the product $A = PQ$ where $$ P = \pmatrix{x & y}, \quad Q = \pmatrix{x^T \\ -y^T}. $$ Because $PQ$ and $QP$ have the same non-zero eiegenvalues, it suffices to find the eigenvalues of the matrix $$ QP = \pmatrix{x^Tx & x^Ty\\ -y^Tx & -y^Ty}, $$ which confirms the validity of your approach.

Answer 2

You're right, this is an excellent solution. But probably we should add a little more. Like this.

Since $x,y$ are linearly independent, there exists a basis $V$ of the form $x,y,v_3,\ldots,v_n$. In this basis the matrix of the linear operator $z\to Az,\ z\in V$ has the form $$ B= \left( \begin{array}{rrrrr} x.x & x.y & x.v_3 & \ldots & x.v_n \\ -y.x & -y.y & -y.v_3 & \ldots & -y.v_n \\ 0 & 0 & 0 & \ldots & 0\\ \ldots &\ldots&\ldots &\ldots &\ldots\\ 0 & 0 & 0 &\ldots & 0 \\ \end{array} \right) $$ and matrices $A$ and $B$ are similar.