# Find $x_0$ when a 3x3 symmetric matrix has equal eigenvalues

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?

• 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 $(z-3)((z-x)^2-4)$. Sep 1 '19 at 17:51
• I found a similar question,math.stackexchange.com/q/1916681/592530 to mine, but I did not understand the explanation Sep 1 '19 at 17:53
• I tried out what @amsmath has mentioned, but what do I do afterwards? Please elaborate Sep 1 '19 at 17:57
• So, you can't figure out the two other eigenvalues? Sep 1 '19 at 18:04
• @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? Sep 1 '19 at 18:28

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.

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.
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.