The nullity/rank needed for a matrix to be diagonalizable

A is a 5x5 matrix with rank 3. Which two of the following conditions are required for the matrix to be diagonalizable? (There may be multiple correct answers.)

(I) nullity(A-2I) = 2

(II) nullity(A-3I) = 3

(III) rank(A-2I) = 2

(IV) rank(A-3I) = 3

I believe that nullity is just the number of free variables and rank is just the number of basic variables. I know that a matrix of size n x n needs n linearly independent eigenvectors to be diagonalizable (or n distinct eigenvalues which guarantees n l.i. eigenvectors). I also know that nullity(A-yI) = multiplicity of y for a diagonalizable matrix. I just don't get how to find the multiplicity, or how else to approach the question (if multiplicity is the wrong approach).

You are almost entirely there. There's one simple fact that I think will help tie this together - distinct eigenvalues have distinct eigenspaces. Remember that a vector $\vec{v}$ is in eigenspace $U$ iff is scaled by its corresponding eigenvalue $\lambda_1$ when applying the linear transformation $A$. A given vector cannot be in two eigenspaces (with distinct eigenvectors) at once, since a vector can't be scaled by both $\lambda_1$ and $\lambda_2$, where $\lambda_1 \neq \lambda_2$.

So since we have shown there cannot be any vector $\vec{v}$ that's in both the eigenspace $U$ corresponding to $\lambda_1$ and the eigenspace $V$ corresponding to $\lambda_2$, we know the bases of $U$ and $V$ are necessarily linearly independent.

In this example, the dimension of the eigenspace corresponding to eigenvalue $2$ is $2$. This is because $\dim(Null(A-2I)) = 2$. The dimension of the eigenspace with eigenvalue $3$ is $3$. Since we know they're distinct eigenspaces, we have 5 linearly independent eigenvectors.

• Thanks for the help, I get your first 2 paragraphs but I'm a bit confused on how you arrived to the third one. How do we know the dimension of the eigenspace? We know the rank/nullity of A, but how do we know rank/nullity of A-2I and A-3I without actually knowing the entries of A? – Karan Bijani May 28 '18 at 10:48
• @KaranBijani ah, unless I'm misunderstanding, we know this from conditions I and II – rb612 May 28 '18 at 18:06

The matrix $$A=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$ is obviously diagonalizable. However $A-2I$ and $A-3I$ both have rank $5$ and nullity $0$.

It could be different if you are also told that $2$ and $3$ are eigenvalues for $A$. Let's analyze this case.

The characteristic polynomial for $A$ is $(0-X)^a(2-X)^b(3-X)^c$ for some positive integers $a$, $b$ and $c$ so that $a+b+c=5$. Note that as the rank of $A$ is $3$, the geometric multiplicity of $0$ is $2$, so $a\ge2$.

Since you want the matrix to be diagonalizable, a necessary condition is that $a=2$. Hence $b+c=3$. Therefore we necessarily have $b=1$ and $c=2$ or $b=2$ and $c=1$.

Case $b=1$, $c=2$.

The geometric multiplicity of $3$ should be $2$ as well, which means $\operatorname{nullity}(A-3I)=2$ and $\operatorname{rank}(A-3I)=1$. Moreover we know that $\operatorname{nullity}(A-2I)=1$ and $\operatorname{rank}(A-2I)=4$.

Case $b=2$, $c=1$.

The geometric multiplicity of $2$ should be $2$ as well, which means $\operatorname{nullity}(A-2I)=2$ and $\operatorname{rank}(A-2I)=1$. Moreover we know that $\operatorname{nullity}(A-3I)=1$ and $\operatorname{rank}(A-3I)=4$.

Conclusion

None of those condition is required, unless you have more restrictive assumptions on the eigenvalues.