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

Find Eigenvalues and Eigenvectors. MIT Linear Algebra homework problem and solution

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)
 
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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.


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