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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!

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    $\begingroup$ If we apply $q(x) = (x-1)^5$ on $I+N$ we get $q(I+N)=\dots$ $\endgroup$
    – dan_fulea
    Oct 21, 2019 at 23:03
  • $\begingroup$ @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) ? $\endgroup$
    – testguy807
    Oct 21, 2019 at 23:14
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    $\begingroup$ 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 $\lambda$. $\endgroup$
    – Joppy
    Oct 22, 2019 at 4:40
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    $\begingroup$ As $S^{-1}(I+N)S=I+S^{-1}NS$, all possible Jordan forms of $I+N$ are $I+$ all possible Jordan forms of $N$. $\endgroup$
    – A.Γ.
    Oct 22, 2019 at 4:45

2 Answers 2

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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 $N+I$.

Knowing all this we get these options of the jordan canonical form (up to rearranging etc.) (also, I'm using lower triangular matrices, I know some people use upper, these are equivalent):

$$ \left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right] $$ $$ \left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} \\ {0} & {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {0} & {1}\end{array}\right],\left[\begin{array}{ccccc}{1} & {0} & {0} & {0} & {0} \\ {1} & {1} & {0} & {0} & {0} \\ {0} & {1} & {1} & {0} & {0} \\ {0} & {0} & {1} & {1} & {0} \\ {0} & {0} & {0} & {1} & {1}\end{array}\right] $$

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If the characteristic polynomial of a matrix $A$ is $\chi_A(t) = \det(tI - A)$, then the characteristic polynomial of $A + I$ is $$\chi_{A + I}(t) = \det(tI - (A + I)) = \det((t-1)I - A) = \chi_A(t-1).$$

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