# Finding the eigenvalues of $\begin{pmatrix} a & b \\ b & a \end{pmatrix}$ matrix without the determinant

So we're reading Axler's Linear Algebra in class and haven't been taught the determinant. But we're asked to find the eigenvalues of

$$\begin{pmatrix} a & b \\ b & a \end{pmatrix}$$

How can I go about doing this? I tried finding something that would satisfy

$$\begin{pmatrix} a - \lambda & b \\ b & a-\lambda \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = 0$$ but all I get is $$y= \cfrac{\lambda - a}{b} x = \cfrac{b}{\lambda - a}x$$

I'm not sure what to do with this. Setting $$y$$ to either of those equalities doesn't give a matrix product of $$0$$. Obviously I see that if set $$y$$ to the first equality then I get a matrix product whose first (but not second) row is zero, and similarly if I set $$y$$ to the second equality. But what does this mean?

• From your equation, it follows that either $x=0$, or that $\frac{\lambda-a}{b}$ is its own reciprocal: thus $\lambda=a \pm b$ (because else $y=0$). – Mindlack Nov 25 '19 at 19:34
• One can guess the eigenvectors... first guess: $(1, 1)^T$; second guess: $(1, -1)^T$; oh, no need for the third guess! – Yuval Nov 25 '19 at 19:35

Hint:

You are so close: simplifying by $$x$$,

$$\frac{\lambda -a}b=\frac b{\lambda -a}$$ can be solved for $$\lambda$$.

Just guess the eigenvectors $$\begin{pmatrix}1\\1\end{pmatrix}$$ and $$\begin{pmatrix}1\\-1\end{pmatrix}$$.

If $$A = \pmatrix{a & b \\ b & a}$$ then notice that $$(A-aI)^2 = \pmatrix{0 & b \\ b & 0}^2 = b^2I$$

so the polynomial $$(x-a)^2-b^2 = (x-a+b)(x-a-b)$$

annihilates $$A$$. In fact, it is the minimal polynomial for $$A$$ so the eigenvalues are $$a-b$$ and $$a+b$$.