Linear Algebra

Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue

Ohio State University exam problems and solutions in mathematics

Problem 476

Let
\[A=\begin{bmatrix}
1 & 2 & 1 \\
-1 &4 &1 \\
2 & -4 & 0
\end{bmatrix}.\] The matrix $A$ has an eigenvalue $2$.
Find a basis of the eigenspace $E_2$ corresponding to the eigenvalue $2$.

(The Ohio State University, Linear Algebra Final Exam Problem)

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Solution.

By definition, the eigenspace $E_2$ corresponding to the eigenvalue $2$ is the null space of the matrix $A-2I$.
That is, we have
\[E_2=\calN(A-2I).\]

We reduce the matrix $A-2I$ by elementary row operations as follows.
\begin{align*}
&A-2I=\begin{bmatrix}
-1 & 2 & 1 \\
-1 &2 &1 \\
2 & -4 & -2
\end{bmatrix}\\[6pt] &\xrightarrow{\substack{R_2-R_1\\R_3+2R_1}}
\begin{bmatrix}
-1 & 2 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}
\xrightarrow{-R_1}
\begin{bmatrix}
1 & -2 & -1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}

Thus, the solutions $\mathbf{x}$ of $(A-2I)\mathbf{x}=\mathbf{0}$ satisfy $x_1=2x_2+x_3$.
Thus, the null space $\calN(A-2I)$ consists of vectors
\[\mathbf{x}=\begin{bmatrix}
2x_2+x_3 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}\] for any scalars $x_2, x_3$.

Hence we have
\begin{align*}
E_2=\calN(A-2I)=\Span\left(\, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \,\right).
\end{align*}

It is straightforward to see that the vectors $\begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix}$ are linearly independent, hence they form a basis of $E_2$.

Thus, a basis of $E_2$ is
\[\left\{\, \begin{bmatrix}
2 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
1
\end{bmatrix} \,\right\}.\]

Final Exam Problems and Solution. (Linear Algebra Math 2568 at the Ohio State University)

This problem is one of the final exam problems of Linear Algebra course at the Ohio State University (Math 2568).

The other problems can be found from the links below.

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