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  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Foundations:
Recall:
1. The eigenvalues of a triangular matrix are the entries on the diagonal.
2. By the Diagonalization Theorem, an  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\times n}   matrix  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}   is diagonalizable
if and only if  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A}   has  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n}   linearly independent eigenvectors.

Solution:

(a)

Step 1:

Step 2:

(b)

Step 1:

Step 2:

Final Answer:
(a)
(b)

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