Answer:
A is diagonalizable because αA(λ) = γA(λ) for all eigenvalues of λ. ( C )
Step-by-step explanation:
The matrix
[tex]A = \left[\begin{array}{ccc}5&2&-4\\2&5&-4\\4&4&-5\end{array}\right][/tex]
This is a 3 x 3 matrix
Attached below is the prove that A is diagonalizable