For what values the real parameter t is a matrix diagonalisable?

Question

What to do with number 25 in the matrix?

Answer 1

A matrix is diagonalizable if the sum of $\dim \ker (A - \lambda I)$ over all eigenvalues $\lambda$ of $A$ is $n$.

When $t = 1$, the only eigenvalue of $A$ is $1$. However, $\dim \ker(A - 1 I) < 3$, so $A$ is not diagonalizable.

However, if $t \neq 1$, then the eigenvalues of $A$ are $1$ and $t$. $\dim \ker (A - tI)$ is $1$. We then note that $\dim \ker (A - tI)$ is $1$ whenever $t \neq 24$, and is $2$ when $t = 24$.

Thus, we find that $A$ is diagonalizable if and only if $t = 24$.