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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?

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6 Answers 6

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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.

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  • $\begingroup$ Oh, I see, then I can express every column of A^n as something times Ae_1, which is zero. $\endgroup$
    – Shannon
    Commented Mar 24, 2012 at 4:15
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    $\begingroup$ 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 $vec{0}$ $\endgroup$
    – YEU
    Commented Dec 5, 2018 at 6:46
  • $\begingroup$ @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. $\endgroup$ Commented Dec 5, 2018 at 7:32
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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 $T_n^n=0$).

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  • $\begingroup$ I liked your proof, but who is $v$ in the matrix $A$? $\endgroup$
    – user162343
    Commented Dec 3, 2014 at 14:23
  • $\begingroup$ $v$ is a $n\times 1$ matrix which contains the $n$ first entries of $A$ in the last column. $\endgroup$ Commented Dec 3, 2014 at 14:26
  • $\begingroup$ ok I see, thnks a lot I liked this proof because is very natural :) $\endgroup$
    – user162343
    Commented Dec 3, 2014 at 14:27
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An $n\times n$ upper triangular matrix $A$ has characteristic polynomial $X^n$. Thus by the theorem of Cayley-Hamilton you get $A^n=0$.

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  • $\begingroup$ 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! $\endgroup$ Commented Mar 23, 2012 at 17:01
  • $\begingroup$ I think you mean "with zero on main diagonal". Otherwise the statement about c/c polynomial is wrong as stated. $\endgroup$
    – user2468
    Commented Mar 23, 2012 at 17:18
  • $\begingroup$ For example, $A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ has the c/c poly $(x-1)(x-3)$ $\endgroup$
    – user2468
    Commented Mar 23, 2012 at 17:20
  • $\begingroup$ Sure I meant strictly upper triangular matrix as asked for in the question. $\endgroup$ Commented Mar 23, 2012 at 17:37
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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, $A^n=0$.

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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 $A^n = 0$.

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

Hint: What are the eigenvalues of your matrix?

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  • $\begingroup$ 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? $\endgroup$ Commented Jul 9, 2014 at 17:36
  • $\begingroup$ @user161825 $0$ is an eigenvalue with multiplicity $n$, and hence its characteristic polynomial is $x^n$... $\endgroup$
    – N. S.
    Commented Jul 9, 2014 at 18:27
  • $\begingroup$ Maybe my linear algebra is rusty, but isn't it, strictly speaking, the other way around? $\endgroup$ Commented Jul 9, 2014 at 20:21
  • $\begingroup$ @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 $n$. $\endgroup$
    – N. S.
    Commented Jul 9, 2014 at 22:11

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