# How to prove $A^n = 0$ if $A$ is an $n\times n$ strictly upper triangular matrix [duplicate]

I would like to prove the statement on the title. I tried to prove it using the definition of multiply of matrices, however it seems like weak to me. Can't decide how to make a correct proof of the statement above.

Edit: Those ns are the same ns. So that, the dimension of the Matrix have to match the power to which it is raised.

## marked as duplicate by A.Γ., Leucippus, user99914, Dap, Gottfried HelmsOct 27 '17 at 3:39

• Might be helpful to write matrix multiplication as a bunch of vector inner products – qbert Oct 26 '17 at 21:28
• Does the dimension of the Matrix have to match the power to which it is raised? – MathematicianByMistake Oct 26 '17 at 21:31
• @MathematicianByMistake Yes, those ns are the same ns. – GLHF Oct 26 '17 at 21:31

A proof using Cayley-Hamilton's theorem if you are interested.

Looking at the form of $A$ (an upper triangular matrix), you can see that:

$$\chi_A(X)=X^n$$

where $\chi_A$ is its characteristic polynomial.

By Cayley-Hamilton's theorem:

$$A^n=\chi_A(A)=0.$$

• To be honest, my linear algebra knowledge is basic, so I don't know that theorem. If you can prove it by using basic algebra like the definition of multiplying of matrices etc. so I can understand better :) – GLHF Oct 26 '17 at 21:44
• Can you explain why the char. polynomial is $X^n$? – Ninja Jun 16 '18 at 20:35
• @Ninja Just compute it using the definition of the characteristic polynomial (since $A$ is an upper triangular matrix). – E. Joseph Jun 17 '18 at 10:13

Write $a^{(k)}_{ij}$ for the $(i,j)^{\text{th}}$ component of $A^k$. We'll prove by induction on $k \ge 1$ that $a_{ij}^{(k)}=0$ whenever $i \le j + k - 1$. Note that when $k=1$, this says precisely that $A$ is strictly upper-triangular, and so the base case is trivial.

Now for the induction step, fix $k \ge 1$ and suppose that $a^{(k)}_{ij} = 0$ when $i \le j + k - 1$. Fix $i,j$ and suppose that $i \le j + k \ (=j+(k+1)-1)$. Then $$a^{(k+1)}_{ij} = \sum_{\ell=0}^n a^{(k)}_{i\ell} a_{\ell j}$$ Now we know that $a_{\ell j} = 0$ whenever $\ell \le j$ so that $$a^{(k+1)}_{ij} = \sum_{\ell=j+1}^n a^{(k)}_{i\ell} a_{\ell j}$$ Moreover, if $j+1 \le \ell \le n$ then $$\ell+k-1 \ge (j+1)+k-1 = j+k \ge i$$ so that $a^{(k)}_{i\ell} = 0$ by the induction hypothesis. Hence $a^{(k+1)}_{ij} = 0$, as required.

Applying this when $k=n$ gives $a^{(n)}_{ij} = 0$ whenever $i \le j+n-1$. But this is always true, since $i \le n$ and $j+n-1 \ge n$ for all $1 \le i,j \le n$. So indeed $A^n=0$.

• This looks like the answer I'm looking for since you used only the definition of matrix multiplication, I just need to check it all over. Thanks for the answer, I will check this as the correct answer as soon as possible if this is the one I'm looking for. – GLHF Oct 26 '17 at 21:50

Similar to the question at: How to show that a $4 \times 4$ strictly upper triangular matrix is nilpotent?. Read through the answers and comments posted and you should be able to prove for all $n$.

It suffices to show $A^n x = 0$ for any $x \in F^n$. To see this, define $W_i := \{ ( x_1, x_2, \ldots, x_{n-i}, \underbrace{0, \ldots, 0}_{i~\mathrm{zeros}}) : x_1, \ldots, x_{n-i} \in F \}$ which are each linear subspaces of $F^n$. (Note that this only requires at least $i$ trailing zeros and allows more since it allows for $x_{n-i} = 0$.) Then the condition on $A$ implies that $A \cdot W_i \subseteq W_{i+1}$ for each $i$. Thus, for any $x \in F^n$: \begin{align*} x & \in W_0 = F^n \\ A x & \in W_1 \\ A^2 x & \in W_2 \\ & \vdots \\ A^n x & \in W_n = \{ 0 \}. \end{align*}