# How to prove a matrix is nilpotent?

Given an $n\times n$ upper triangular matrix $A$ with zero on main diagonal, show that $A^n = 0$.

I did some matrix operation and noticed that the diagonal moves up, ultimately all entries will be zero. Is there a nicer way to do it?

If $\mathbf{e}_1,\ldots,\mathbf{e}_n$ is the standard basis for $\mathbb{R}^n$ (or whatever your base field is), then notice that $A\mathbf{e}_1=\mathbf{0}$, and $$A\mathbf{e}_i \in \mathrm{span}(\mathbf{e}_1,\ldots,\mathbf{e}_{i-1}),\quad i=2,3,\ldots,n.$$

Now show that $A^n\mathbf{e}_i = \mathbf{0}$ for all $i$ to get the desired conclusion.

• Oh, I see, then I can express every column of A^n as something times Ae_1, which is zero. Commented Mar 24, 2012 at 4:15
• Just to further explain the answer: $Ae_i$ is the ith column of A which has non-zero value from entry 1 to entry i-1, i.e. $Ae_i \in span(e_1,..., e_{i-1})$. When you multiply by A again : $x = A(Ae_i)$, x is a combination of column 1 to column i-1 of A because $Ae_i$ has non-zero values for only first i-1 entries, and we know that combination of columns has non-zero value from entry 1 to entry i-2. So if we keep multiplying by A, at $A^ie_i$, we get $vec{0}$
– YEU
Commented Dec 5, 2018 at 6:46
• @YEU: I purposefully did not want to spell it all out; so thank you for making sure that my intention was completely undermined, even if six and a half years later (during which time nobody asked for further explanations...) Once you have enough reputation, you can post your own answers instead. Commented Dec 5, 2018 at 7:32

It can be showed by induction: for $n=2$ it's a simple computation. If the result is true for $n$, and $A=\pmatrix{T_n&v\\\ 0&0}$ where $T_n$ is a $n\times n$ triangular matrix with zeros on the main diagonal, then for each $p\geq 1$ we have $A^p=\pmatrix{T_n^p&T_n^{p-1}v\\\ 0&0}$. Therefore, for $p=n+1$ we get $A^{n+1}=\pmatrix{T_n^{n+1}&T_n^nv\\\ 0&0}=0$ (because $T_n^n=0$).

• I liked your proof, but who is $v$ in the matrix $A$? Commented Dec 3, 2014 at 14:23
• $v$ is a $n\times 1$ matrix which contains the $n$ first entries of $A$ in the last column. Commented Dec 3, 2014 at 14:26
• ok I see, thnks a lot I liked this proof because is very natural :) Commented Dec 3, 2014 at 14:27

An $n\times n$ upper triangular matrix $A$ has characteristic polynomial $X^n$. Thus by the theorem of Cayley-Hamilton you get $A^n=0$.

• True: actually my favourite proof of C-H for complex matrices is to triangularise the matrix and then prove by induction that an upper triangular matrix satisfies its characteristic polynomial, so this would be circular for me! I know this is not what a "real" ring theorist would do! Commented Mar 23, 2012 at 17:01
• I think you mean "with zero on main diagonal". Otherwise the statement about c/c polynomial is wrong as stated.
– user2468
Commented Mar 23, 2012 at 17:18
• For example, $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ has the c/c poly $(x-1)(x-3)$
– user2468
Commented Mar 23, 2012 at 17:20
• Sure I meant strictly upper triangular matrix as asked for in the question. Commented Mar 23, 2012 at 17:37

Suppose that $a_{i\,j}=(A)_{i\,j}=0$ for $i\ge j$ and $b_{i\,j}=(B)_{i\,j}=0$ for $i\ge j-m$, where $A$ and $B$ are $n\times n$ matrices. Consider when it is possible to have $a_{i\,j}b_{j\,k}\not=0$.

To have $a_{i\,j}\not=0$ we must have $i<j$ (i.e. $i\le j-1$).

To have $b_{j\,k}\not=0$ we must have $j<k-m$ (i.e. $j\le k-m-1$).

Therefore, to have $a_{i\,j}b_{j\,k}\not=0$, we must have $i<k-m-1$ (i.e. $i\le k-m-2$).

Thus, for $i\ge k-m-1$ $$(AB)_{i\,k}=\sum_{j=1}^na_{i\,j}b_{j\,k}=0\tag{1}$$ The matrix $A$ specified in the question has $a_{i\,j}=0$ for $i\ge j$. Using $(1)$ and induction, we have that $(A^m)_{i\,j}=0$ for $i\ge j-m+1$. This means that $(A^n)_{i\,j}=0$ for $i\ge j-n+1$, which means that $(A^n)_{i\,j}=0$ for all $1\le i,j\le n$.

Therefore, $A^n=0$.

Here's a nice graph theoretical proof. Given a $n \times n$ matrix $A$ consider the directed graph $D(A)$ on $n$ vertices where vertex $i$ is joined with vertex $j$ with an edge of weight $a_{ij}$ whenever $a_{ij} \neq 0$.

Define the weight of a walk in $D(A)$ as the product of all the entries on its edges. It can be easily seen that the $ij$ entry of $A^k$ is the sum of weights of all possible walks of length $k$ from vertex $i$ to vertex $j$ in $D(A)$.

Now if $A$ is a strictly upper triangular matrix then in $D(A)$ there is an edge between vertex $i$ and vertex $j$ only if $i < j$. Therefore $D(A)$ does not contain any cycles. And hence there cannot be any walks of length greater than or equal to $n$ in $D(A)$, proving that $A^n = 0$.

Yes, if you already covered this topic....

Hint: What are the eigenvalues of your matrix?

• How does this solve the problem? Isn't it more convenient to know that the matrix is strictly upper diagonal than knowing that it has an eigenvector with eigenvalue 0? Commented Jul 9, 2014 at 17:36
• @user161825 $0$ is an eigenvalue with multiplicity $n$, and hence its characteristic polynomial is $x^n$... Commented Jul 9, 2014 at 18:27
• Maybe my linear algebra is rusty, but isn't it, strictly speaking, the other way around? Commented Jul 9, 2014 at 20:21
• @user161825 If you are familiar with minimal polynomials it is easy to prove that over an algebraically closed field a matrix is nilpotent if and only if $0$ is the only eigenvalue.... From here it follows that over any field a matrix is nilpotent if and only if $0$ is an eigenvalue of multiplicity $n$. Commented Jul 9, 2014 at 22:11