Determine Eigenvalues, Eigenvectors, Diagonalizable From a Partial Information of a Matrix

Johns Hopkins Linear Algebra Exam Problems and Solutions

Problem 180

Suppose the following information is known about a $3\times 3$ matrix $A$.
\[A\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}=6\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix},
\quad
A\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}, \quad
A\begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}.\]

(a) Find the eigenvalues of $A$.

(b) Find the corresponding eigenspaces.

(c) In each of the following questions, you must give a correct reason (based on the theory of eigenvalues and eigenvectors) to get full credit.
Is $A$ a diagonalizable matrix?
Is $A$ an invertible matrix?
Is $A$ an idempotent matrix?

(Johns Hopkins University Linear Algebra Exam)
 
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Solution.

(a) Find the eigenvalues of $A$.

From the first and the second equations, we see that $6$ and $3$ are eigenvalues of $A$.
From the second and the third equations, we have
\[A\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}=A\begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}\] and thus we have
\[\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}=A\left( \begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}- \begin{bmatrix}
2 \\
-1 \\
0
\end{bmatrix}\right)=A \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}.\] This yields that
\[A\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}=0\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix} \tag{*}\] and we conclude that $0$ is an eigenvalue.
Since the size of the matrix $A$ is $3\times 3$, it has at most three eigenvalues. Thus we found them all. Eigenvalues of $A$ are $0, 3, 6$.

(b) Find the corresponding eigenspaces.

From the first equation, the vector $\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $6$.
We also see from the second equation, the vector $\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $3$.
The equation (*) in the solution (a) also tells us that the vector $\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $0$.

Note that we have three distinct eigenvalues for $3\times 3$ matrix $A$. So all the algebraic multiplicities for eigenvalues are $1$, hence the geometric multiplicities must be $1$ since the geometric multiplicity is always less than or equal to the algebraic multiplicity.
Therefore, each eigenspace has dimension $1$. We already found a nonzero vector in each eigenspace, and thus the vector is a basis vector for each eigenspace.
We denote $E_{\lambda}$ for the eigenspace for eigenvector $\lambda$. Then we have
\[E_0=\Span\left(\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\right), \quad
E_3=\Span\left(\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\right), \quad
E_6=\Span\left(\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}\right).\]

(c) Diagonalizable matrix? Invertible matrix? Idempotent matrix?

Since $A$ has three distinct eigenvalues, $A$ is diagonalizable. (Or, algebraic multiplicities are the same as geometric multiplicities.)
Since the matrix $A$ has $0$ as an eigenvalue. Thus $A$ is not invertible.

If $A$ is idempotent matrix, then the eigenvalues of $A$ is either $0$ or $1$. Thus $A$ is not an idempotent matrix.
(For a proof of this fact, see the post Eigenvalues of an idempotent matrix.)


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