# 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.

**(c)** How about geometric multiplicities?

Contents

## 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.

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