Why must $M$ be diagonalizable?

Question

I am trying to figure out why the answer to the following question is (a) $M$ must be diagonalizable based on the given information.

I know that for $M$ to be diagonalizable the sum of the dimensions of the eigenspaces must equal $n$ — but we don't know what $n$ is. How can we determine what $n$ is, or is there some other way of knowing $M$ must be diagonalizable?

Suppose that $(t-2)^2(t-1)(t+2)t$ is the characteristic polynomial of a matrix $M$, and the null space of $M−2I$ has dimension $2$.

Which of (a), (b) or (c) below is the correct statement?

(a) $M$ must be diagonalizable.

(b) $M$ might be diagonalizable.

(c) $M$ cannot be diagonalizable.

Thanks in advance!

Answer 1

Yes, we do know the value if $n$: it is equal to $5$, since it must be equal to the degree of the characteristic polynomial! So, you can conclude that the matrix is indeed diagonalizable.

Answer 2

The degree of the characteristic polynomial is $5$, hence the size of the matrix $M$ is $5 \times 5$.

Since we are given that dim$(M-2I)$ is $2$ and other than $\lambda=2$ we have three distinct eigen values. Therefore, the eigen vectors will form a basis, hence diagonalizable.

Answer 3

The characteristic polynomial is split but has a single non-simple root, namely $2$ which is a double root. In the case of a split characteristic polynomial, the matrix is diagonalisable if and only if for every root (eigenvalue), the dimension of the corresponding eigenspace equals the dimension of the corresponding generalised eigenspace, the latter being always equal to the multiplicity of the root. For simple roots there is no way this can fail, so one just needs to check any multiple roots; in the current case $M$ is diagonalisable if and only if the dimension of the eigenspace for the eigenvalue$~2$ equals (the multiplicity of that root which is) $2$. But that is given to be the case.