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The question goes like this: There is a symmetric matrix:$$A=\begin{bmatrix}3 & 0 & 0\\ 0 & x & 2\\0 & 2 & x\end{bmatrix}$$

Find the value(s) of $x$ for which $A$ has at most two distinct eigenvalues. (Eigenvalues like $3,2,2$)

In my attempts to solve this problem, I got the characteristic equation as: $$\lambda^3-(2x+3)\lambda^2+(x^2+3x-2)\lambda-3(x^2-4)=0$$ I am unable to proceed any further than this. Should I try to solve for $\lambda$ by putting appropriate values in the equation, then find $x$?

Is there any property that I seem to be missing?

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    $\begingroup$ You can decompose the matrix into one 1x1 block (containing $3$) and a 2x2 block. So one eigenvalue is always $3$. The char. polynomial is $(z-3)((z-x)^2-4)$. $\endgroup$
    – amsmath
    Sep 1 '19 at 17:51
  • $\begingroup$ I found a similar question,math.stackexchange.com/q/1916681/592530 to mine, but I did not understand the explanation $\endgroup$
    – evamPUNdit
    Sep 1 '19 at 17:53
  • $\begingroup$ I tried out what @amsmath has mentioned, but what do I do afterwards? Please elaborate $\endgroup$
    – evamPUNdit
    Sep 1 '19 at 17:57
  • $\begingroup$ So, you can't figure out the two other eigenvalues? $\endgroup$
    – amsmath
    Sep 1 '19 at 18:04
  • $\begingroup$ @amsmath I'm getting $\lambda=\mp 2 + x$ . But the question mentions atleast 2 values of $\lambda$ to be equal. In the case of all the eigenvalues being equal I get $x=5,1$ So how to I find out the remaining values of x? $\endgroup$
    – evamPUNdit
    Sep 1 '19 at 18:28
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Observe that $$A\begin{bmatrix}1\\0\\0\end{bmatrix}=3\begin{bmatrix}1\\0\\0\end{bmatrix}.$$ Thus $\lambda=3$ is an eigenvalue of this matrix.

Also observe $$A\begin{bmatrix}0\\1\\1\end{bmatrix}=(x+2)\begin{bmatrix}0\\1\\1\end{bmatrix}.$$ Thus $\lambda=x+2$ is an eigenvalue of this matrix as well.

Now the sum of the eigenvalues is the trace of the matrix. Let the other eigenvalue be $\lambda_3$, then $$3+(x+2)+\lambda_3=2x+3 \implies \color{red}{\lambda_3=x-2}.$$

So the three eigenvalues are $\boxed{3,x+2}$ and $\boxed{x-2}$. We want at most two distinct eigenvalues. Observe that when $x=1,5$ then two of them are equal, hence only two distinct eigenvalues.

When $\color{red}{x=1}$, the eigenvalues are $\color{blue}{3,3,-1}$.

When $\color{red}{x=5}$, the eigenvalues are $\color{blue}{3,7,3}$.

When $\color{red}{x \neq 1,5}$, the eigenvalues are all $\color{blue}{\text{distinct}}$.

For no value of $x$ can all the eigenvalues be the same.


Further addition to the solution:

In case you are not aware of the trace result, you can still get the third eigenvalue by observing that $$A\begin{bmatrix}0\\1\\-1\end{bmatrix}=(x-2)\begin{bmatrix}0\\1\\-1\end{bmatrix}.$$ Thus $\lambda=x-2$ is an eigenvalue of this matrix as well.

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you should not have solved all you parentheses you had $$(3-\lambda)*[(x-\lambda)^2-4)]$$ so $$(3-\lambda)=0$$ is one solution and $$[(x-\lambda)^2-4)]=0$$ the next.

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