031 Review Part 3, Problem 2

Find the eigenvalues and eigenvectors of the matrix  $A={\begin{bmatrix}1&1&1\\0&-1&1\\0&0&2\end{bmatrix}}.$

Foundations:
An eigenvector of a matrix  $A$  is a nonzero vector  ${\vec {x}}$  such that  $A{\vec {x}}=\lambda {\vec {x}}$  for some scalar  $\lambda .$
In this case, we say that  $\lambda$  is an eigenvalue of  $A.$

Solution:

Step 1:
Since  $A$  is a triangular matrix, the eigenvalues of  $A$  are the entries on the diagonal.
So, the eigenvalues of  $A$  are  $1,-1,$  and  $2.$

Step 2:
Since the matrix is triangular and all the eigenvalues are distinct, the eigenvectors of  $A$  are
${\begin{bmatrix}1\\0\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\0\end{bmatrix}},{\begin{bmatrix}0\\0\\1\end{bmatrix}}$
where each eigenvector has eigenvalue  $1,-1$  and  $2,$  respectively.

Final Answer:
The eigenvalues of  $A$  are  $1,-1$  and  $2,$  and the corresponding eigenvectors are
${\begin{bmatrix}1\\0\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\0\end{bmatrix}},{\begin{bmatrix}0\\0\\1\end{bmatrix}}.$

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