1. ## Hard question about linear algebra 2

does someone know how to solve the following?

"Find all matrices A so
I A
0 I
is Diagonalizable "

this is a picture of the matrice
http://i46.tinypic.com/258v514.jpg

2. ## Re: Hard question about linear algebra 2

Am I correct in assuming that 0 entry is the zero matrix?

yes

4. ## Re: Hard question about linear algebra 2

Block matrix - Wikipedia, the free encyclopedia

Please correct me if I'm wrong, but doesn't that imply that A = 0?

5. ## Re: Hard question about linear algebra 2

I think $A=0$ is correct, bopbop. In the following I will take $M$ to be the original matrix, i.e. $M$ is the given block matrix with $A$ in the upper right corner. For generality, I'm assuming $M$ is an $n\times n$ matrix. Here's an outline of what I think supports this:

1) An $n \times n$ matrix is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to $n$. See Diagonalizable matrix - Wikipedia, the free encyclopedia for this statement.

2) Since the original matrix is upper triangular with 1's along the main diagonal, the only eigenvalue is 1.

3) From 2) we know the only eigenspace is that corresponding to the eigenvalue 1, because there were no other eigenvalues. By definition, the eigenspace belonging to 1 is $\text{Ker}(M-1I)$, where $I$ is the $n\times n$ identity matrix. For $M$ to be diagonalizable, 1) tells us that we want the dimension of $\text{Ker}(M-1I)$ (i.e. the nullity of $M-1I$) to be $n$. The matrix $M-1I$ is the $2\times 2$ block matrix with $A$ in the upper right corner and $0$'s for the other entries. Now, if $A\neq 0$, then the rank of $M-1I$ is at least one, because there will be at least one linearly independent column of $M-1I$ when $A\neq 0$. By the rank-nullity theorem, this implies the nullity of $M-1I$ is less than or equal to $n-1$. Since we require the nullity of $M-1I$ to be $n$, we must have $A=0$.

If something is unclear or incorrect let me know. Hope this helps, MisterF!