Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible.

Problem 216

Let
\[A=\begin{bmatrix}
1 & 3 & 3 \\
-3 &-5 &-3 \\
3 & 3 & 1
\end{bmatrix} \text{ and } B=\begin{bmatrix}
2 & 4 & 3 \\
-4 &-6 &-3 \\
3 & 3 & 1
\end{bmatrix}.\] For this problem, you may use the fact that both matrices have the same characteristic polynomial:
\[p_A(\lambda)=p_B(\lambda)=-(\lambda-1)(\lambda+2)^2.\]

(a) Find all eigenvectors of $A$.

(b) Find all eigenvectors of $B$.

(c) Which matrix $A$ or $B$ is diagonalizable?

(d) Diagonalize the matrix stated in (c), i.e., find an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$ or $B=PDP^{-1}$.

(Stanford University Linear Algebra Final Exam Problem)
 
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Hint.

See the post how to diagonalize a matrix for a review of the diagonalization process.

Solution.

(a) All eigenvectors of $A$

The eigenvalues of $A$ are roots of the characteristic polynomial $p_A(\lambda)$, hence the eigenvalues of $A$ are
\[\lambda=1, -2\] with algebraic multiplicities $1$ and $2$, respectively.

Let us first find the eigenvectors corresponding to the eigenvalue $\lambda=-2$.
The eigenvectors are nonzero solutions of the equation
\[(A+2I)\mathbf{x}=\mathbf{0},\] where $I$ is the $3\times 3$ identity matrix, $\mathbf{x}\in \R^3$, and $\mathbf{0}$ is the three dimensional zero vector.

To solve the linear system, we use the Gauss-Jordan elimination method.
Applying elementary row operations we obtain
\begin{align*}
A+2I=\begin{bmatrix}
3 & 3 & 3 \\
-3 &-3 &-3 \\
3 & 3 & 3
\end{bmatrix} \to \cdots \to \begin{bmatrix}
1 & 1 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Therefore, the vector $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}$ must satisfy
\[x_1=-x_2-x_3.\] Hence the eigenvectors associated to $\lambda=-2$ are
\[\mathbf{x}=\begin{bmatrix}
-x_2-x_3 \\
x_2 \\
x_3
\end{bmatrix}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}+x_3\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\] for any pair of complex numbers $(x_1, x_2)\neq (0,0)$.

---

Now we find the eigenvectors corresponding to the eigenvalue $\lambda=1$.
As above, we solve the system
\[(A-I)\mathbf{x}=\mathbf{0}.\] By elementary row operations, we reduce the matrix $A-I$ and obtain
\[A-I=\begin{bmatrix}
0 & 3 & 3 \\
-3 &-6 &-3 \\
3 & 3 & 0
\end{bmatrix} \to \cdots \to \begin{bmatrix}
1 & 0 & -1 \\
0 &1 &1 \\
0 & 0 & 0
\end{bmatrix}.\] Hence the vector $\mathbf{x}$ should satisfy
\[x_1=x_3 \text{ and } x_2=-x_3.\] Thus the eigenvectors associated to $\lambda=1$ are
\[\mathbf{x}=x_3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\] for any nonzero complex number $x_3$.

(b) All eigenvectors of $B$

The eigenvalues of $B$ are roots of the characteristic polynomial $p_B(t)$, hence the eigenvalues of $B$ are
\[\lambda=1, -2\] with algebraic multiplicities $1$ and $2$, respectively.

---

Let us first find the eigenvectors corresponding to the eigenvalue $\lambda=-2$.
We need to solve the sysetm
\[(B+2I)\mathbf{x}=\mathbf{0}.\] By elementary row operations, we have
\begin{align*}
B+2I=\begin{bmatrix}
4 & 4 & 3 \\
-4 &-4 &-3 \\
3 & 3 & 3
\end{bmatrix} \to \cdots \to \begin{bmatrix}
1 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}.
\end{align*}
Thus the vector $\mathbf{x}$ satisfies
\[x_1=-x_2 \text{ and } x_3=0,\] hence the eigenvectors associated to $\lambda=-2$ are
\[\mathbf{x}=x_2\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}\] for any nonzero complex number $x_2$

---

Next, we find the eigenvectors corresponding to eigenvalue $\lambda=1$.
Since we have
\begin{align*}
B-I=\begin{bmatrix}
1 & 4 & 3 \\
-4 &-7 &-3 \\
3 & 3 & 0
\end{bmatrix}
\to \cdots \to
\begin{bmatrix}
1 & 0 & -1 \\
0 &1 &1 \\
0 & 0 & 0
\end{bmatrix}
\end{align*}
by elementary row operations, the system $(B-I)\mathbf{x}=\mathbf{0}$ has solutions
\[x_1=x_3 \text{ and } x_2=-x_3.\] Thus the eigenvectors associated to $\lambda=1$ are
\[\mathbf{x}=x_3\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\] for any nonzero complex number $x_3$.

(c) Which matrix $A$ or $B$ is diagonalizable?

From the result of (a), for each eigenvalue $\lambda$ of $A$ the geometric multiplicity of $\lambda$ is the same as the algebraic multiplicity of $\lambda$. Thus $A$ is not defective, and hence $A$ is diagonalizable.

From the result of (b), the geometric multiplicity of the eigenvalue $\lambda=-2$ of $B$ is $1$ although the algebraic multiplicity of $\lambda=-2$ is $2$. Thus the matrix $B$ is defective and hence $B$ is not diagonalizable.

(d) Diagonalize the matrix

We diagonalize the matrix $A$. From the computation of (a), the vectors
\[\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}\] form a basis of the eigenspace $E_{-2}$ associated to $\lambda=-2$.
Also the vector
\[\begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\] form a basis of the eigenspace $E_1$ associated to $\lambda=1$.
Therefore the vectors
\[\begin{bmatrix}
-1 \\
1 \\
0
\end{bmatrix}, \begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
-1 \\
1
\end{bmatrix}\] are linearly independent eigenvectors of $A$.
Thus we set
\[P=\begin{bmatrix}
-1 & -1 & 1 \\
1 &0 &-1 \\
0 & 1 & 1
\end{bmatrix} \text{ and } D=\begin{bmatrix}
-2 & 0 & 0 \\
0 &-2 &0 \\
0 & 0 & 1
\end{bmatrix}\] and we have
\[P^{-1}AP=D.\] Click here if solved 36

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