# Eigenvalues in a Matrix with 3 unknown variables

I'm in a Linear Algebra class, where we are currently covering eigenvalues and eigenvectors. My question is how should I solve this exercise with three variables?

Consider the matrix $$\begin{equation*} A := \begin{bmatrix} 0 & 0 & a \\ b & c & 10 \\ 0 & 0 & a \end{bmatrix} \end{equation*}$$ Question: What value or values ​​should the parameters take for matrix A to have three real eigenvalues equal?

I'm will solve this using: $$\det(A- \lambda I)$$: $$\begin{equation*} \det\begin{bmatrix} 0 & 0 & a \\ b & c & 10 \\ 0 & 0 & a \end{bmatrix} -\begin{bmatrix}\lambda&0&0\\ 0&\lambda&0\\ 0&0&\lambda\end{bmatrix} =\det\begin{pmatrix}-\lambda&0&a\\ b&c-\lambda&10\\ 0&0&a-\lambda\end{pmatrix} \end{equation*}$$ After we got the determinant we have: \begin{align*}(-\lambda)(c-\lambda)(a-\lambda)=0\end{align*}

How should I proceed to answer the question?

The roots of that polynomial are $$0$$, $$a$$, and $$c$$. So, you have only one eigenvalue if and only if $$a=c=0$$.
• Yeah, as you have shown, $0$ is an eigen value clearly no matter what $a,c$ are. So the other two must also be $0$, i.e. $a=c=0$ – Vinayak Suresh Mar 24 at 23:37
• Sorry, i misunderstood the question. I thought that you were after $3$ distinct eigenvalues. I shall edit my answer. – José Carlos Santos Mar 24 at 23:45
By inspection, we know that both $$a$$ and $$c$$ are eigenvalues: $$A(0,1,0)^T=(0,c,0)^T$$ and $$(0,0,1)A=(0,0,a)$$. Since all of the eigenvalues must be equal, $$a=c$$. The trace of $$A$$ is equal to $$a+c=2a$$, so the third eigenvalue must be $$0$$, therefore $$a=c=0$$. The value of $$b$$ is unconstrained.