# Quiz 12. Find Eigenvalues and their Algebraic and Geometric Multiplicities

## Problem 376

(a) Let
$A=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 &1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}.$ Find the eigenvalues of the matrix $A$. Also give the algebraic multiplicity of each eigenvalue.

(b) Let
$A=\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 &1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \end{bmatrix}.$ One of the eigenvalues of the matrix $A$ is $\lambda=0$. Find the geometric multiplicity of the eigenvalue $\lambda=0$.

## Solution (a). Eigenvalues of $A$ and algebraic multiplies

Eigenvalues and their algebraic multiplicities are determined by the characteristic polynomial $p(t)$ of $A$.
By definition, the characteristic polynomial of $A$ is $p(t)=\det(A-tI)$.
We have
\begin{align*}
&p(t)=\det(A-tI)\\
&=\begin{vmatrix}
-t & 0 & 0 & 0 \\
1 &1-t & 1 & 1 \\
0 & 0 & -t & 0 \\
1 & 1 & 1 & 1-t
\end{vmatrix}\6pt] &=-t\begin{vmatrix} 1-t & 1 & 1 \\ 0 &-t &0 \\ 1 & 1 & 1-t \end{vmatrix} && \text{by the first row cofactor expansion}\\[6pt] &=-t\left(\, -t\begin{vmatrix} 1-t & 1\\ 1& 1-t \end{vmatrix} \,\right)&& \text{by the second row cofactor expansion}\\[6pt] &=t^2\left(\, (1-t)^2-1 \,\right)\\ &=t^2(t^2-2t)\\ &=t^3(t-2). \end{align*} Thus the characteristic polynomial is \[p(t)=t^3(t-2). From this, the eigenvalues of $A$ are $0$ and $2$ with algebraic multiplicities $3$ and $1$, respectively.

## Solution (b). Geometric multiplicity

We give two solutions for part (b).

### First Solution (b). (Finding the rank first)

Recall that the geometric multiplicity of $\lambda$ is the dimension of the eigenspace $E_{\lambda}=\calN(A-\lambda I)$. That is, the geometric multiplicity of $\lambda$ is the nullity of the matrix $A-\lambda I$.

Let us now consider the case $\lambda=0$.
We first find the rank of $A-0 I=A$ as follows.
\begin{align*}
A-0 I= A=\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}
\xrightarrow{R_4-R_2}
\begin{bmatrix}
0 & 0 & 0 & 0 \\
1 &1 & 1 & 1 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_1 \leftrightarrow R_2}
\begin{bmatrix}
1 &1 & 1 & 1\\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}.
\end{align*}
The last matrix is in reduced row echelon form.
Hence the rank of $A$ is $1$.
The rank-nullity theorem says that
$\text{rank of A } + \text{ nullity of A}=4.$ Thus, the nullity of $A=A-0I$ is $3$, and hence the geometric multiplicity of $\lambda=0$ is $3$.

### Second Solution (b). (Finding a basis of the eigenspace)

In this solution, we find a basis of the eigenspace $E_0$.
By definition $E_0=\calN(A-0I)=\calN(A)$.
Thus, the eigenspace $E_0$ is the null space of the matrix $A$.
We solve the equation $A\mathbf{x}=\mathbf{0}$ as follows.
The augmented matrix of this equation is
\begin{align*}
[A\mid \mathbf{0}]= \left[\begin{array}{rrrr|r}
0 & 0 & 0 & 0 &0 \\
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0\\
1 & 1 & 1 & 1 &0
\end{array} \right] \xrightarrow{R_4-R_2}
\left[\begin{array}{rrrr|r}
0 & 0 & 0 & 0 &0 \\
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0\\
0 & 0 & 0 & 0 &0
\end{array} \right] \xrightarrow{R_1 \leftrightarrow R_2}
\left[\begin{array}{rrrr|r}
1 &1 & 1 & 1 &0 \\
0 & 0 & 0 & 0 &0 \\
0 & 0 & 0 & 0 &0\\
0 & 0 & 0 & 0 &0
\end{array} \right].
\end{align*}
Hence the solution satisfies
$x_1=-x_2-x_3-x_4$ and the general solution is
\begin{align*}
\mathbf{x}=\begin{bmatrix}
-x_2-x_3-x_4 \\
x_2 \\
x_3 \\
x_4
\end{bmatrix}
=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}.
\end{align*}
Therefore, the eigenspace is
\begin{align*}
&E_0=\calN(A)\\
-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}, \text{ for any } x_2, x_3, x_4\in \C \,\right\}\10pt] &=\Span\left\{\, \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \,\right\}. \end{align*} Thus the set \[\left\{\, \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 0 \\ 1 \end{bmatrix} \,\right\} is a spanning set of $E_0$, and it is straightforward to check that the set is linearly independent.
Hence this set is a basis of $E_0$, and the dimension of $E_0$ is $3$.
The geometric multiplicity of $\lambda=0$ is the dimension of $E_0$ by definition.
Thus, the geometric multiplicity of $\lambda$ is $3$.

## Comment.

These are Quiz 12 problems for Math 2568 (Introduction to Linear Algebra) at OSU in Spring 2017.

I could have combined these two problems into one problem, and just asked to find eigenvalues and algebraic/ geometric multiplicities for each eigenvalue.
The reason I didn’t do is that I wanted to rescue students who didn’t get a correct answer in part (a) for some reasons.
Also, since one of the eigenvalue is given in (b), students could use this information to double check their solutions in (a).
(At least if you didn’t get the eigenvalue $0$, you made a mistake somewhere.)

### List of Quiz Problems of Linear Algebra (Math 2568) at OSU in Spring 2017

There were 13 weekly quizzes. Here is the list of links to the quiz problems and solutions.

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