# Possible Jordan canonical forms of identity matrix plus a nilpotent matrix

I am working on the following Linear algebra problem:

Suppose that $$N$$ is a nilpotent $$5 \times 5$$ real matrix (so $$N^5$$ is the zero matrix). List all possible Jordan canonical forms of $$I + N$$.

Here is where my thinking is at: I know how to list all possible Jordan canonical forms of a matrix, given its characteristic polynomial. I also know that, in this case, the characteristic polynomial of $$N$$ will be given by $$p_N(x) = x^5$$. However, I'm struggling with this problem because I don't know how to deduce from this what the characteristic polynomial of $$I + N$$ is. Is there a nice way to see what this is?

I searched for if there is a nice relationship between a matrix's characteristic polynomial and the characteristic polynomial of the matrix plus the identity, but I couldn't find one. Does the fact that $$N$$ is nilpotent help at all to see what this is?

Thanks!

• If we apply $q(x) = (x-1)^5$ on $I+N$ we get $q(I+N)=\dots$ Oct 21, 2019 at 23:03
• @dan_fulea I see. It's true that q(I + N) = 0. How does one see this must be the characteristic polynomial of I + N, though, rather than just some other annihilating polynomial? Is it because it has degree 5, and the unique annihilating polynomial of degree n for an n x n matrix is the characteristic polynomial (up to multiplication by a scalar) ? Oct 21, 2019 at 23:14
• It's not true that the unique annihilating polynomial of an $n \times n$ matrix is the characteristic polynomial. The characteristic polynomial of the $2 \times 2$ identity matrix is $(t - 1)^2$, but the identity matrix is also annihilated by $(t - 1)(t - \lambda)$ for any $\lambda$. Oct 22, 2019 at 4:40
• As $S^{-1}(I+N)S=I+S^{-1}NS$, all possible Jordan forms of $I+N$ are $I+$ all possible Jordan forms of $N$.
– A.Γ.
Oct 22, 2019 at 4:45

As you mentioned correctly- The characteristic polynomial of $$N$$ is $$\chi_N(x)=x^5$$, we know that not by computing, but by the nilpotent property and the dimension of $$N$$. To understand the characteristic polynomial of $$N+I$$ we need to go back to the definition: To any matrix $$M$$ it's characteristic polynomial is given by $$\det(xI-M)$$. Let's examine what happens when we add the identity matrix:
$$\det(xI-(N+I))=\det((x-1)I+N)$$
This is exactly the definition of the characteristic polynomial but instead of being a polynomial over of the variable $$x$$, it's shifted by $$1$$. It's a composition of the original characteristic polynomial with $$x-1$$, so if we had $$\chi_N(x)=x^5$$ then $$\chi_{N+I}(x)=(x-1)^5$$, and thus 1 is an eigenvalue of $$N+I$$.
$$\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right]$$ $$\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right]$$
If the characteristic polynomial of a matrix $$A$$ is $$\chi_A(t) = \det(tI - A)$$, then the characteristic polynomial of $$A + I$$ is $$\chi_{A + I}(t) = \det(tI - (A + I)) = \det((t-1)I - A) = \chi_A(t-1).$$