Let $N$ be a square matrix with $N^m \neq 0$ for $m \in \mathbb{N}$, while $N^{m+1} = 0$. Show that the matrices $I_n, N, N^2, ..., N^m$ are l.i.

Question

Hey I am having problems with this exercise.

Let $N \in M(n; K)$ be a square matrix with the property that $N^m \neq 0$ for some $m \in \mathbb{N}$, while $N^{m+1} = 0$. Show that the matrices $I_n, N, N^2, ..., N^m$ are linearly independent in the vector space of matrices $M(n, K)$.

So I wanted to prove this with Induction over the power of m.

Let $m=1$.

The base case is when $m = 1$. In this case, we have $N^2 = N \cdot N = 0$, so the matrices $I_n$ and $N$ are linearly independent.

Now suppose that the matrices $I_n, N, N^2, ..., N^{m-1}$ are linearly independent. We want to show that $I_n, N, N^2, ..., N^{m}$ are also linearly independent.

Suppose that $a_0I_n + a_1N + a_2N^2 + \cdots + a_mN^m = 0$ for some scalars $a_0, a_1, \ldots, a_m$. We want to show that all the scalars $a_i$ are zero.

So we have that $a_0I_n + a_1N + a_2N^2 + \cdots + a_mN^{m-1} = 0 \Rightarrow N(a_0I_n + a_1N + a_2N^2 + \cdots + a_mN^{m-1}) = N0=0 \Rightarrow a_0N + a_1N^2 + \cdots + a_m-1N^{m}=0$

But now I am missing $a_0I_n$ and I don't know how to proceed. Can someone help me?

Answer 1

Before jumping to the full induction step, let's just try $m=2$. Suppose $I,N,N^2$ are nonzero but $N^{3} =0$. Suppose $aI + bN + cN^2 = 0$, i.e. we have a dependence relation. How can we show $a,b,c=0$?

First of all, multiplying both sides of $aI + bN + cN^2 = 0$ by $N^2$ we deduce that $aN^2 =0$, since the other terms cancel out, from which we deduce $a=0$. Now we are working with $bN + cN^2 = 0$. Multiplying both sides by $N$, we deduce $bN^2 = 0$, since the other terms cancel out, and again we deduce $b=0$. Finally, we have $cN^2 = 0$, so that $c=0$, and we are done.

Can you see how to generalize this argument?

Answer 2

Suppose that $$k_0I_n+k_1N+\dots+k_mN^m=0,$$ multiply by $N^m$, then we get $$k_0N^m+k_1N^{m+1}+\dots+k_mN^{m+m}=k_0N^m=0.$$ Because $N^m\neq 0$, so $k_0=0$. Similarily, we can get $k_0=\dots=k_m=0$.