How to find eigenvalues of following block matrix $A$ in terms of eigenvalues of matrix $B$?

$A=\begin{bmatrix} 4I-B & -B \\ -B & 2I \\ \end{bmatrix}$

Where $B$ is square matrix of order $n$ and $I$ is an identity matrix of order $n$

I have tried the following

let ,$w=\begin{bmatrix} v \\ cv \end{bmatrix}$ be an eigenvector of $A$ then with eigenvalue $\lambda_a$

then $Aw=\lambda_a w$

$\Rightarrow$ $4v-Bv-c(Bv)=\lambda_av$ and


$\Rightarrow$ $4v-\lambda_bv-c(\lambda_bv)=\lambda_av$

$-\lambda_b v+2cv=\lambda_a(cv)$

As $v$ is nonzero vector

$4-\lambda_b-c\lambda_b=\lambda_a$ and

$-\lambda_b+2c=c \lambda_a$

Now solving both the equation to find $c$.

Please verify whether my steps are correct or not?


closed as off-topic by Morgan Rodgers, Travis, zz20s, Charles, Claude Leibovici Apr 23 '16 at 4:24

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Hint: look at the vector $$ \begin{bmatrix} v\\ cv \end{bmatrix} $$ where $v$ is an eigenvector of $B$ with eigenvalue $\lambda$, and $c$ is some constant. Under what condition on $c$ is this an eigenvector for your matrix?

  • $\begingroup$ i don't get that.Please answer in brief if its possible for you $\endgroup$ – kalpeshmpopat Apr 22 '16 at 16:23
  • $\begingroup$ I did answer in brief. When you show us the work you've done so far, I might consider amplifying my remarks a little. $\endgroup$ – John Hughes Apr 22 '16 at 23:07
  • $\begingroup$ I had written what I tried.Please verify and give your valuable comment on it $\endgroup$ – kalpeshmpopat Apr 23 '16 at 3:28
  • $\begingroup$ That looks pretty good. Multiply your first equation by $c$, and then subtract to eliminate the $c\lambda_a$ term. Now you've got a quadratic in $c$ to solve. Go for it! $\endgroup$ – John Hughes Apr 23 '16 at 10:56
  • $\begingroup$ BTW: Note that by multiplying by $c$, you may introduce a false root of $c = 0$, and you'll have to check this. When you've got a solution for $c$, you get that $\lambda_a = 4 - \lambda_b - c\lambda_b$; with these values of $c$, you've therefore discovered two eigenvectors of $A$ for each eigenvector of $B$. $\endgroup$ – John Hughes Apr 23 '16 at 11:04

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