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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

$-Bv+2cv=\lambda_a(cv)$

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

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

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?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, Travis, zz20s, Charles, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

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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?

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  • $\begingroup$ 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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