a and b such that A is similar to B

Question

Let $$A = \begin{bmatrix} -3 & 6 & 0 \\ -2 & 4 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} $$ I know the Eigenvalues of the matrix are $-1,0$ and $1$.

Let $a,b$ be real numbers and let

$$B(a,b) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & a & b \\ 0 & b & a \\ \end{bmatrix} $$

Are there real numbers $a$ and $b$ so that $A$ is similar to $B(a,b)$? If so, find an $a$ and $b$ that work, explaining how you know the matrices are similar. If not, explain why not.

Answer 1

There are many ways to solve this. One way is to use the fact that similar matrices share the same eigenvalues.

The eigenvalues of matrix $B$ can be found by solving its characteristic polynomial

\begin{align} \det \left|\begin{matrix} -\lambda & 0& 0\\ 0 &a-\lambda & b\\ 0 & b&a-\lambda \\\end{matrix}\right| &=0 \\[2ex] -\lambda(a-\lambda)^2+b^2\lambda &=0 \end{align}

with roots

$$ \lambda_1 =0, \quad \lambda_2=a+b \quad\text{and} \quad \lambda_3 =a-b .$$

Hence, for $A$ and $B$ to have the same eigenvalues it must be that

$$\begin{cases} a+b=1 \\ a-b=-1 \end{cases} \quad \Rightarrow \quad a=0, \quad \text{and} \quad b=1,$$

or

$$\begin{cases} a+b=-1 \\ a-b=1 \end{cases} \quad \Rightarrow \quad a=0, \quad \text{and} \quad b=-1.$$

Answer 2

Let's find the eigenvalues of $B(a, b)$. We need the eigenvalues to be $-1, 0, 1$, as similarity is a equivalence relation, both matrices would be similar to:

$$\left(\begin{matrix} -1&0&0\\ 0&0&0\\ 0&0&1 \end{matrix}\right)$$

$\det(\lambda I-B(a, b))= \left|\begin{matrix} \lambda&0&0\\ 0&\lambda-a&-b\\ 0&-b&\lambda-a \end{matrix}\right|\\= \lambda\left|\begin{matrix} \lambda-a&-b\\ -b&\lambda-a \end{matrix}\right|= \lambda((\lambda-a)^2-b^2)=\lambda(\lambda-a+b)(\lambda-a-b)$

As such, the eigenvalues are $0, a-b, a+b$. Setting $a=0$, $b=1$ suffices.