eigenvalues of a matrix $A$ plus $cI$ for some constant $c$

Question

If $A$ is a $n \times n$ real matrix with eigenvalues $\lambda_1,\lambda_2,...\lambda_n$, how does one get the eigenvalues of the matrix $A$ + c$I$, where $I$ is the identity matrix and $c$ is a non-zero real constant?

I tried to work out the characteristic polynomials, but I am wondering if there is a way to quickly get the eigenvalues.

Try to apply $(A+cI)$ to one of the eigenvectors of $A$ to and see what happens. — muaddib, May 25 '15 at 00:21

score 14 · Answer 1 · answered Mar 09 '18 at 14:18

14

I'd order the proof correctly: $Ax=\lambda x$ and $cIx = cIx$, so $Ax + cIx =\lambda x +cIx$. Then rearrange to get $(A+cI)x=(\lambda+c)x$.

answered Mar 09 '18 at 14:18

Adam

141

score 8 · Accepted Answer · answered May 25 '15 at 00:21

8

Note that if $Av = \lambda v$, then $(A+cI)v = (\lambda+c)v$.

Also, $\chi_{A+cI}(x) = \det (xI-A-cI) = \det( (x-c)I -A) = \chi_A(x-c)$.

answered May 25 '15 at 00:21

copper.hat

172,524

eigenvalues of a matrix $A$ plus $cI$ for some constant $c$

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