031 Review Part 3, Problem 7

Let  $A={\begin{bmatrix}3&0&-1\\0&1&-3\\1&0&0\end{bmatrix}}{\begin{bmatrix}3&0&0\\0&4&0\\0&0&3\end{bmatrix}}{\begin{bmatrix}0&0&1\\-3&1&9\\-1&0&3\end{bmatrix}}.$

Use the Diagonalization Theorem to find the eigenvalues of  $A$  and a basis for each eigenspace.

Foundations:
Diagonalization Theorem
An  $n\times n$  matrix  $A$  is diagonalizable if and only if  $A$  has  $n$  linearly independent eigenvectors.
In fact,  $A=PDP^{-1},$  with  $D$  a diagonal matrix, if and only if the columns of  $P$  are  $n$  linearly
independent eigenvectors of  $A.$  In this case, the diagonal entries of  $D$  are eigenvalues of  $A$  that
correspond, respectively , to the eigenvectors in  $P.$

Solution:

Step 1:
Since
${\begin{bmatrix}3&0&0\\0&4&0\\0&0&3\end{bmatrix}}$
is a diagonal matrix, the eigenvalues of  $A$  are  $3$  and  $4$  by the Diagonalization Theorem.

Step 2:

Final Answer:

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