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)
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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 $(x_2, x_3, x_4)\neq (0,0,0)$.
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