031 Review Part 3, Problem 1

(a) Is the matrix  $A={\begin{bmatrix}3&1\\0&3\end{bmatrix}}$  diagonalizable? If so, explain why and diagonalize it. If not, explain why not.

(b) Is the matrix  $A={\begin{bmatrix}2&0&-2\\1&3&2\\0&0&3\end{bmatrix}}$  diagonalizable? If so, explain why and diagonalize it. If not, explain why not.

ExpandFoundations:
Recall:
1. The eigenvalues of a triangular matrix are the entries on the diagonal.
2. By the Diagonalization Theorem, an  $n\times n$  matrix  $A$  is diagonalizable
if and only if  $A$  has  $n$  linearly independent eigenvectors.

Solution:

(a)

ExpandStep 1:
To answer this question, we examine the eigenvalues and eigenvectors of  $A.$
Since  $A$  is a triangular matrix, the eigenvalues are the entries on the diagonal.
Hence, the only eigenvalue of  $A$  is  $3.$

ExpandStep 2:
Now, we find a basis for the eigenspace corresponding to  $3$  by solving  $(A-3I){\vec {x}}={\vec {0}}.$
We have
${\begin{array}{rcl}\displaystyle {A-3I}&=&\displaystyle {{\begin{bmatrix}3&1\\0&3\end{bmatrix}}-{\begin{bmatrix}3&0\\0&3\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}0&1\\0&0\end{bmatrix}}.}\end{array}}$
Solving this system, we see  $x_{1}$  is a free variable and  $x_{2}=0.$
Therefore, a basis for this eigenspace is
${\bigg \{}{\begin{bmatrix}1\\0\end{bmatrix}}{\bigg \}}.$

ExpandStep 3:
Now, we know that  $A$  only has one linearly independent eigenvector.
By the Diagonalization Theorem,  $A$  must have  $2$  linearly independent eigenvectors to be diagonalizable.
Hence,  $A$  is not diagonalizable.

(b)

ExpandStep 1:
First, we find the eigenvalues of  $A$  by solving  ${\text{det }}(A-\lambda I)=0.$
Using cofactor expansion, we have
${\begin{array}{rcl}\displaystyle {{\text{det }}(A-\lambda I)}&=&\displaystyle {{\text{det }}{\Bigg (}{\begin{bmatrix}2&0&-2\\1&3&2\\0&0&3\end{bmatrix}}-{\begin{bmatrix}\lambda &0&0\\0&\lambda &0\\0&0&\lambda \end{bmatrix}}{\Bigg )}}\\&&\\&=&\displaystyle {{\text{det }}{\Bigg (}{\begin{bmatrix}2-\lambda &0&-2\\1&3-\lambda &2\\0&0&3-\lambda \end{bmatrix}}{\Bigg )}}\\&&\\&=&\displaystyle {(-1)^{(2+2)}(3-\lambda ){\text{det }}{\bigg (}{\begin{bmatrix}2-\lambda &-2\\0&3-\lambda \end{bmatrix}}{\bigg )}}\\&&\\&=&\displaystyle {(3-\lambda )(2-\lambda )(3-\lambda ).}\end{array}}$
Therefore, setting
$(3-\lambda )(2-\lambda )(3-\lambda )=0,$
we find that the eigenvalues of  $A$  are  $3$  and  $2.$

ExpandStep 2:
Now, we find a basis for each eigenspace by solving  $(A-\lambda I){\vec {x}}={\vec {0}}$  for each eigenvalue  $\lambda .$
For the eigenvalue  $\lambda =2,$  we have
${\begin{array}{rcl}\displaystyle {A-2I}&=&\displaystyle {{\begin{bmatrix}2&0&-2\\1&3&2\\0&0&3\end{bmatrix}}-{\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}}}\\&&\\&=&\displaystyle {\begin{bmatrix}0&0&-2\\1&1&2\\0&0&1\end{bmatrix}}\\&&\\&\sim &\displaystyle {{\begin{bmatrix}1&1&0\\0&0&1\\0&0&0\end{bmatrix}}.}\end{array}}$
We see that  $x_{2}$  is a free variable. So, a basis for the eigenspace corresponding to  $2$  is
${\Bigg \{}{\begin{bmatrix}-1\\1\\0\end{bmatrix}}{\Bigg \}}.$

ExpandStep 3:
For the eigenvalue  $\lambda =3,$  we have
${\begin{array}{rcl}\displaystyle {A-3I}&=&\displaystyle {{\begin{bmatrix}2&0&-2\\1&3&2\\0&0&3\end{bmatrix}}-{\begin{bmatrix}3&0&0\\0&3&0\\0&0&3\end{bmatrix}}}\\&&\\&=&\displaystyle {\begin{bmatrix}-1&0&-2\\1&0&2\\0&0&0\end{bmatrix}}\\&&\\&\sim &\displaystyle {{\begin{bmatrix}1&0&2\\0&0&0\\0&0&0\end{bmatrix}}.}\end{array}}$
We see that  $x_{2}$  and  $x_{3}$  are free variables. So, a basis for the eigenspace corresponding to  $3$  is
${\Bigg \{}{\begin{bmatrix}0\\1\\0\end{bmatrix}},{\begin{bmatrix}-2\\0\\1\end{bmatrix}}{\Bigg \}}.$

ExpandStep 4:
Since  $A$  has  $3$  linearly independent eigenvectors,  $A$  is diagonalizable by the Diagonalization Theorem.
Using the Diagonalization Theorem, we can diagonalize  $A$  using the information from the steps above.
So, we have
$D={\begin{bmatrix}2&0&0\\0&3&0\\0&0&3\end{bmatrix}},P={\begin{bmatrix}-1&0&-2\\1&1&0\\0&0&1\end{bmatrix}}.$

ExpandFinal Answer:
(a)     $A$  is not diagonalizable.
(b)     $A$  is diagonalizable and  $D={\begin{bmatrix}2&0&0\\0&3&0\\0&0&3\end{bmatrix}},P={\begin{bmatrix}-1&0&-2\\1&1&0\\0&0&1\end{bmatrix}}.$

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