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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?

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The roots of that polynomial are $0$, $a$, and $c$. So, you have only one eigenvalue if and only if $a=c=0$.

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  • $\begingroup$ So the answer for: "What value or values ​​should the parameters take for matrix A to have three real eigenvalues equal?" Would be: a=0 and c=0? $\endgroup$ – SWAT Mar 24 at 23:36
  • $\begingroup$ 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. $a=c=0$ $\endgroup$ – Vinayak Suresh Mar 24 at 23:37
  • $\begingroup$ Sorry, i misunderstood the question. I thought that you were after $3$ distinct eigenvalues. I shall edit my answer. $\endgroup$ – José Carlos Santos Mar 24 at 23:45
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This answer builds on the one given by Jose Carlos Santos. He is correct in saying that the roots of the polynomial are 0, a, and c. In other words, λ = 0, λ = a, and λ = c are the eigenvalues of that system. If you want all 3 eigenvalues to be equal, then a = c = 0. Furthermore, b can then take any value.

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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.

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