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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?

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  • $\begingroup$ 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 $y=0$). $\endgroup$ – Mindlack Nov 25 '19 at 19:34
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    $\begingroup$ One can guess the eigenvectors... first guess: $(1, 1)^T$; second guess: $(1, -1)^T$; oh, no need for the third guess! $\endgroup$ – Yuval Nov 25 '19 at 19:35
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Hint:

You are so close: simplifying by $x$,

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

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

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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 $a+b$.

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