# Eigenvalues and Eigenvectors of Matrix Whose Diagonal Entries are 3 and 9 Elsewhere

## Problem 379

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}.\]

(*Harvard University, Linear Algebra Final Exam Problem*)

Add to solve later

Sponsored Links

Contents

## Hint.

Instead of computing the characteristic polynomial $p(t)=\det(A-tI)$ of $A$, consider the matrix $B=A+6I$.

Then use the relation between eigenvalues of $A$ and $B$.

See the problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for more about the relation.

## Solution.

Let $B=A+6I$, where $I$ is the $4 \times 4$ identity matrix. Then every entry of $B$ is $9$. Consider the vector $\mathbf{v}=[1\, 1\, 1\, 1]^{\trans}$.

Then we have

\begin{align*}

B\mathbf{v}=\begin{bmatrix}

9 & 9 & 9 & 9 \\

9 &9 & 9 & 9 \\

9 & 9 & 9 & 9 \\

9 & 9 & 9 & 9

\end{bmatrix}\begin{bmatrix}

1 \\

1 \\

1 \\

1

\end{bmatrix}=36\begin{bmatrix}

1 \\

1 \\

1 \\

1

\end{bmatrix}=36\mathbf{v}.

\end{align*}

It follows that $36$ is an eigenvalue of $B$ and the vector $\mathbf{v}$ is an associated eigenvector.

We apply the elementary row operations and obtain

\[B\to \begin{bmatrix}

1 & 1 & 1 & 1 \\

0 &0 & 0 & 0 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0

\end{bmatrix}.\] Hence the rank of $B$ is $1$, and it follows from the rank-nullity theorem that the nullity is $3$.

Thus, $0$ is an eigenvalue of $B$ and its geometric multiplicity is $3$.

The algebraic multiplicity is always greater than or equal to geometric multiplicity.

But the algebraic multiplicity of $0$ cannot be $4$, since $36$ is another eigenvalue.

As a result, the matrix $B$ has eigenvalues $36$ and $0$ with algebraic multiplicities $1$ and $3$.

The eigenvectors corresponding to $36$ are

\[a\begin{bmatrix}

1 \\

1 \\

1 \\

1

\end{bmatrix},\]
where $a$ is any nonzero complex number.

The eigenvector corresponding to $0$ are obtained by solving $B\mathbf{x}=\mathbf{0}$.

The solutions satisfy

\[x_1=-x_2-x_3-x_4.\]
Hence the eigenvectors corresponding to $0$ are

\[\mathbf{x}=x_2\begin{bmatrix}

-1 \\

1 \\

0 \\

0

\end{bmatrix}+x_3\begin{bmatrix}

-1 \\

0 \\

1 \\

0

\end{bmatrix}+x_4\begin{bmatrix}

-1 \\

0 \\

0 \\

1

\end{bmatrix},\]
where $(x_2, x_3, x_4)\neq (0,0,0)$.

Since $A=B-6I$, it follows that the matrix $A$ has eigenvalues $30$ and $-6$ with algebraic multiplicity $1$ and $3$. Their associated eigenvectors are exactly the eigenvectors for $B$.

(See Problem “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$” for a proof of this fact.)

Namely, the eigenvectors corresponding to $30$ are

\[a\begin{bmatrix}

1 \\

1 \\

1 \\

1

\end{bmatrix},\]
where $a$ is any nonzero complex number.

The eigenvectors corresponding to $-6$ are

\[x_2\begin{bmatrix}

-1 \\

1 \\

0 \\

0

\end{bmatrix}+x_3\begin{bmatrix}

-1 \\

0 \\

1 \\

0

\end{bmatrix}+x_4\begin{bmatrix}

-1 \\

0 \\

0 \\

1

\end{bmatrix},\]
where $(x_2, x_3, x_4)\neq (0,0,0)$.

Add to solve later

Sponsored Links

## 1 Response

[…] 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“. […]