031 Review Part 1, Problem 4

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

Solution:
Let  $A={\begin{bmatrix}1&1\\0&1\end{bmatrix}}.$
First, notice that  ${\text{det }}A=1\neq 0.$
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  $1.$  Additionally, there is only one linearly independent eigenvector.
Hence,  $A$  is not diagonalizable and the statement is false.

Final Answer:
FALSE

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