# Find All the Eigenvalues and Eigenvectors of the 6 by 6 Matrix

## Problem 400

Find all the eigenvalues and eigenvectors of the matrix
$A=\begin{bmatrix} 10001 & 3 & 5 & 7 &9 & 11 \\ 1 & 10003 & 5 & 7 & 9 & 11 \\ 1 & 3 & 10005 & 7 & 9 & 11 \\ 1 & 3 & 5 & 10007 & 9 & 11 \\ 1 &3 & 5 & 7 & 10009 & 11 \\ 1 &3 & 5 & 7 & 9 & 10011 \end{bmatrix}.$

(MIT, Linear Algebra Homework Problem)

## Solution.

Let
$B=A-10000I,$ where $I$ is the $6 \times 6$ identity matrix. That is, we have
$B=\begin{bmatrix} 1 & 3 & 5 & 7 &9 & 11 \\ 1 & 3 & 5 & 7 &9 & 11 \\ 1 & 3 & 5 & 7 &9 & 11 \\ 1 & 3 & 5 & 7 &9 & 11 \\ 1 & 3 & 5 & 7 &9 & 11 \\ 1 & 3 & 5 & 7 &9 & 11 \\ \end{bmatrix}.$

Since all the rows are the same, the matrix $B$ is singular and hence $\lambda=0$ is an eigenvalue of $B$.
Let us determine the geometric multiplicity of $\lambda=0$ (namely, the dimension of the null space of $B$).

We apply elementary row operations to $B$ and obtain
\begin{align*}
B\xrightarrow{\text{elementary row operations}}
\begin{bmatrix}
1 & 3 & 5 & 7 &9 & 11 \\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}.
\end{align*}
Thus, if $B\mathbf{x}=\mathbf{0}$, then we have
$x_1=-3x_2-5x_3-7x_4-9x_5-11x_6.$

It follows from this that basis vectors of the eigenspace $E_0=\calN(B)$ are
$\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} -5 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} -7 \\ 0 \\ 0 \\ 1 \\ 0\\ 0 \end{bmatrix}, \begin{bmatrix} -9 \\ 0 \\ 0 \\ 0 \\ 1\\ 0 \end{bmatrix}, \begin{bmatrix} -11 \\ 0 \\ 0 \\ 0 \\ 0\\ 1 \end{bmatrix},$ and hence the geometric multiplicity corresponding to $\lambda=0$ is $5$.

By inspection, we see that
$B\mathbf{v}=36\mathbf{v},$ where
$\mathbf{v}=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1\\ 1 \end{bmatrix}.$ Thus, it yields that $\lambda=36$ is an eigenvalue of $B$ and $\mathbf{v}$ is a corresponding eigenvector.

Recall that the algebraic multiplicity of an eigenvalue is greater than or equal to the geometric multiplicity. Also the sum of algebraic multiplicities of all eigenvalues of $B$ is equal to $6$ since $B$ is a $6\times 6$ matrix.
It follows from this observation that we determine that the algebraic multiplicity of $\lambda=0$ is $5$ and the algebraic and geometric multiplicities of $\lambda=36$ are both $1$.
Hence the vector $\mathbf{v}$ form a basis of the eigenspace $E_{36}$.

Now that we have determined eigenvalues and eigenvectors of $B$, we can deduce those of $A$ as follows.
In general, if $A=B+cI$, then the eigenvalues of $A$ are $\lambda+c$, where $\lambda$ are eigenvalues of $B$.
The eigenvectors for $A$ corresponding 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 proof.)

In the current problem, we have $A=B+10000I$, and thus $c=10000$.
Therefore, the eigenvalues of $A$ are $10000, 10036$.
Eigenvectors corresponding to $10000$ are
$x_2\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \\ 0\\ 0 \end{bmatrix}+x_3\begin{bmatrix} -5 \\ 0 \\ 1 \\ 0 \\ 0\\ 0 \end{bmatrix}+x_4\begin{bmatrix} -7 \\ 0 \\ 0 \\ 1 \\ 0\\ 0 \end{bmatrix}+x_5\begin{bmatrix} -9 \\ 0 \\ 0 \\ 0 \\ 1\\ 0 \end{bmatrix}+x_6\begin{bmatrix} -11 \\ 0 \\ 0 \\ 0 \\ 0\\ 1 \end{bmatrix},$ where $(x_2, x_3, x_4, x_5, x_6)\neq (0, 0, 0, 0, 0, 0)$.

The eigenvector corresponding to $10036$ is
$a\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1\\ 1 \end{bmatrix},$ where $a$ is any nonzero scalar.

Suppose $A$ is a positive definite symmetric $n\times n$ matrix. (a) Prove that $A$ is invertible. (b) Prove that $A^{-1}$...