031 Review Part 1, Problem 4

True or false: If $A$ is invertible, then $A$ is diagonalizable.

Solution:
Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A={\begin{bmatrix}1&1\\0&1\end{bmatrix}}.}
First, notice that Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\text{det }}A=1\neq 0.}
Therefore, $A$ is invertible.
Since $A$ is a triangular matrix, the eigenvalues of $A$ are the entries on the diagonal.
Therefore, the only eigenvalue of $A$ is 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 1.} Additionally, there is only one linearly independent eigenvector.
Hence, 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 not diagonalizable and the statement is false.

Final Answer:
FALSE

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