# How to show that a $4 \times 4$ strictly upper triangular matrix is nilpotent?

If $\displaystyle A$ is a $\displaystyle 4 \times 4$-strictly upper triangular matrix, how do you show that $\displaystyle A$ is nilpotent? Would it be sufficient to write out an example of a $\displaystyle 4 \times 4$-strictly upper triangular matrix and multiply by itself till I reach $\displaystyle 0$?

• That is not sufficient but it can give you an idea for why it is true and how you can prove it. – Cameron Williams Feb 12 '15 at 14:36
• If $a,b$ are nilponent of degrees $m,n$ then $(a+b)^{m+n}=0$, i.e. $a+b$ is nilpotent. Decompose your matrix as a sum of matrices that only have one non-zero element in the strict upper triangle. Each is nilpotent of degree at most $4$. – Tom Feb 12 '15 at 14:40
• @Tom This is false. Take, for instance, $a = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$ and $b = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$, which are both nilpotent of degree 2, but $a + b = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right]$ which is not nilpotent as it squares to the identity. If $ab = ba$ and $a,b$ are nilpotent, then $a+b$ is nilpotent. – walkar Feb 12 '15 at 15:04
• That would be OK if the entries of the matrix were "unknowns", ad it would actually work. So if you can prove that $A^{4} = $ for $A = \left(\begin{array}{clcr} 0 & x_{12}&x_{13}&x_{14}\\0&0& x_{23}&x_{24}\\0&0&0& x_{34}\\0&0&0&0 \end{array} \right)$ you have a valid proof. It is a bit messy to write out though. – Geoff Robinson Feb 12 '15 at 15:06

define the symbols: $$d^{ij}_k:\{1,\dots,n\}^3 \to \{0,1\}$$ by the rule $$d^{ij}_k=\begin{cases} 1& \text{if j-i \ge k},\\ 0& \text{otherwise}. \end{cases}$$ clearly an $n \times n$ upper triangular matrix may be written as $[a_{ij}d_0^{ij}]$ and a strictly upper triangular matrix may be written as $U=[u^{ij}d_1^{ij}]$

you may verify that: $$d^{ij}_a d^{jk}_b= d^{ik}_{a+b} \tag{1}$$ now $$U^2_{ik} = \sum_{j=1}^n u_{ij}d_1^{ij}u_{jk}d_1^{jk} \\ = d^{ik}_2 \sum_{j=1}^n u_{ij}u_{jk}$$ (applying (1) )

it follows that the $n^{th}$ power of $U$ is a matrix of the form $[d_n^{ij}x_{ij}]$, in which all entries are zero

You can prove by induction that a strictly upper triangular $n \times n$ matrix $A$ satisfies $A^{n} = $ by showing that it left multiplies each memeber of the standard basis of column vectors $e_{i}$ (with $1$ in position $i$ and $0$ elsewhere) to $0$. It is clear that $Ae_{1}$ is the zero vector, and if $n >1,$ you can assume by induction that $A^{n-1}e_{j}$ is the zero vector for $2 \leq j \leq n$ ( just consider the way $A$ acts on the span of $e_{2},\ldots,e_{n}$ - it just acts like a strictly upper triangular $(n-1) \times (n-1)$ matrix on that space which is why you can use induction. Hence $A^{n}$ sends each basis vector to zero, so is the zero linear transformation.

You can easily see that for any strictly upper triangular matrix the diagonal elements of $A^n$ are all $0$ for any $n\ge 1$. And, since a matrix $A$ is nilpotent iff tr$(A^n)=0\quad \forall n>0$ you have the statement.

• Though the asker may have to prove trace 0 implies nilpotent, but that isn't so bad. Especially if you have a lot of the characterizations of trace and determinant using eigenvalues. – walkar Feb 12 '15 at 14:56
• The last statement of this answer is not necessarily true over fields of prime characteristic, whereas the statement about $4 \times 4$ strictly upper triangular matrices being nilpotent works over any field. – Geoff Robinson Feb 12 '15 at 15:00
• @Geoff: Yes, good point! I was thinking only to fields of caracteristic $0$. – Emilio Novati Feb 12 '15 at 15:43

Remember that the entries on the first minor diagonal are $$(A^2)_{k,k+1}=\sum_{i=1}^n A_{k,i}A_{i,k+1}.$$ Because $A$ is strictly upper triangular for $i\le k$ you have $A_{k,i}=0$, and for $i>k$ you have $A_{i,k+1}=0$. So when squaring your matrix, you have zeros on the first minor diagonal.

If you square the result again, you can replace all "$+1$"s above by "$+2$" and see that now $(A^2)_{k,i}=0$ for $i\le k+1$, because the first minor diagonal has zeros too, and that $(A^2)_{i,k+2}=0$ for $i\ge k+1$. So now you have two minor diagonals with zeros.

Continuing the process, you end up with a 0-matrix.

More visually speaking, assume that you have $k$ minor diagonals with zeros and square that matrix. To get the $(i,j)$-th entry of the squared matrix, you multiple the $i$-th row with the $j$-th column using the scalar product. The $i$-th row has zeros as its first $i+k$ elements, while the $j$-th column has zeros as its last $n-j+k+1$ elements. So when $$n-j+k+1+i+k\ge n\Leftrightarrow j\le i+2k+1$$ they overlap and you get 0 in that spot, giving you even $2k+1$ minor diagonals full of zeros.