# Eigenvalues and Algebraic/Geometric Multiplicities of Matrix $A+cI$

## Problem 378

Let $A$ be an $n \times n$ matrix and let $c$ be a complex number.

(a) For each eigenvalue $\lambda$ of $A$, prove that $\lambda+c$ is an eigenvalue of the matrix $A+cI$, where $I$ is the identity matrix. What can you say about the eigenvectors corresponding to $\lambda+c$?

(b) Prove that the algebraic multiplicity of the eigenvalue $\lambda$ of $A$ is the same as the algebraic multiplicity of the eigenvalue $\lambda+c$ of $A+cI$ are equal.

## Proof.

### (a) $\lambda+c$ is an eigenvalue of $A+cI$.

Let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. Then we have
$A\mathbf{x}=\lambda \mathbf{x}.$ It follows that we have
\begin{align*}
(A+cI)\mathbf{x}&=A\mathbf{x}+c\mathbf{x}\\
&=\lambda \mathbf{x}+c\mathbf{x}\\
&=(\lambda+c)\mathbf{x}.
\end{align*}

Thus, we obtain
$(A+cI)\mathbf{x}=(\lambda+c)\mathbf{x},$ where $\mathbf{x}$ is a nonzero vector.
Hence $\lambda+c$ is an eigenvalue of the matrix $A+cI$, and $\mathbf{x}$ is an eigenvector corresponding to $\lambda-c$.

In summary, if $\lambda$ is an eigenvalue of $A$ and $\mathbf{x}$ is an associated eigenvector, then $\lambda+c$ is an eigenvalue of $A+cI$ and $\mathbf{x}$ is an associated eigenvector corresponding to $\lambda+c$.

### (b) Algebraic multiplicities are the same

Let $p(t)=\det(A-tI)$ be the characteristic polynomial of $A$.
Let $q(t)$ be the characteristic polynomial of the matrix $A+cI$.
Then we have
\begin{align*}
q(t)&=\det\left(\, (A+cI)-t \,\right)=\det\left(\, A-(t-c) \,\right)\\
&=p(t-c).
\end{align*}

Let $\lambda_1, \dots, \lambda_k$ be distinct eigenvalues of $A$ with algebraic multiplicities $n_1, \dots, n_k$, respectively.
Then we have
$p(t)=\pm \prod_{i=1}^k (t-\lambda_i)^{n_i}.$ It follows that we have
\begin{align*}
q(t)&=p(t-c)\\
&=\pm \prod_{i=1}^k (t-c-\lambda_i)^{n_i}\\
&=\pm \prod_{i=1}^k \left(t-(\lambda_i+c)\right)^{n_i}.\\
\end{align*}
From the last equation, we read that the eigenvalues of the matrix $A+cI$ are $\lambda_i+c$ with algebraic multiplicity $n_i$ for $i=1,\dots, k$.
Thus, geometric multiplicities of $\lambda$ and $\lambda-c$ are the same.

### (c) How about geometric multiplicities?

From part (a), we know that eigenvectors of $\lambda$ are eigenvectors of $\lambda-c$.
Reversing the argument, the eigenvectors of $\lambda+c$ are eigenvectors of $\lambda$.

Thus, the eigenvectors correspond one to one, and the eigenspace of $\lambda$ is the same as the eigenspace of $\lambda+c$.
Hence their geometric multiplicities are the same.

## Applications

As applications of this problem, consider the following problems.

Problem. Find all the eigenvalues and eigenvectors of the matrix
$A=\begin{bmatrix} 3 & 9 & 9 & 9 \\ 9 &3 & 9 & 9 \\ 9 & 9 & 3 & 9 \\ 9 & 9 & 9 & 3 \end{bmatrix}.$

This is a problem of a Linear Algebra final exam at Harvard University.
For a solution of this problem, see the post “Eigenvalues and eigenvectors of matrix whose diagonal entries are 3 and 9 elsewhere“.

Problem. Find the determinant of the following matrix
$A=\begin{bmatrix} 6 & 2 & 2 & 2 &2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\ 2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6 \end{bmatrix}.$

This is also a problem of a Linear Algebra final exam at Harvard University.
See the post “Determinant of matrix whose diagonal entries are 6 and 2 elsewhere” for a solution.

### 2 Responses

1. 04/14/2017

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2. 05/05/2017

[…] to $lambda+c$ are exactly the eigenvectors for $B$ corresponding $lambda$. (See the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a […]

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##### Idempotent (Projective) Matrices are Diagonalizable

Let $A$ be an $n\times n$ idempotent complex matrix. Then prove that $A$ is diagonalizable.

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