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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).

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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.

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  • $\begingroup$ 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? $\endgroup$ – Karan Bijani May 28 '18 at 10:48
  • $\begingroup$ @KaranBijani ah, unless I'm misunderstanding, we know this from conditions I and II $\endgroup$ – rb612 May 28 '18 at 18:06
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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.

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