# What is the number of non-conjugate $6 \times 6$ complex matrices having the characteristic polynomial $(x-5)^6=0$?

Let $D$ be the $6 \times 6$ diagonal matrix with diagonal entries $5$.

Then all the $6 \times 6$ complex matrices which are diagonalizable to $D$ are conjugate to $D$ and hence to each other.

So I should find those matrices which aren't diagonalizable to $D$ but have same characteristic equation.

I think of those matrices whose all diagonal elements are $5$ but Geometric multiplicity $\neq$ Algebraic multiplicity for $5$. But still not getting any concrete idea.

What is the general way to approach?

• What are the possible $6\times6$ Jordan matrices with the same value along the main diagonal? – amd May 1 '17 at 17:19
• Jordan normal form! – JJR May 1 '17 at 17:20
• @amd Is it that the number of Jordan matrices is same as number of non-conjugate matrices in this case? I am learning Jordan forms. I'll check the total number of them soon as you've pointed. – Error 404 May 1 '17 at 17:25
• Well, are matrices with different Jordan normal forms conjugate? – amd May 1 '17 at 17:26
• @amd I have understood that crux of this problem is in the concept of Jordan Forms. Thanks for these hints. – Error 404 May 1 '17 at 17:50

These are distinguished by the Jordan forms, each of which consists of blocks of sizes which add up to $6$. So the number of possible Jordan forms is the sixth partition number $p(6)$.