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If $\mathbf{A}$ and $\mathbf{B}$ are both upper square triangle matrices, show $\mathbf{AB}$ is an upper triangle matrix.

Willie Wong
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bsdshell
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    This follows immediately from the definition of matrix multiplication; have you tried writing out a typical entry of $AB$ below the main diagonal to see why it must be $0$? – Brian M. Scott Oct 18 '13 at 21:43

2 Answers2

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Let's explicate Brian M.Scott's comment: If $A=(a_{ij})$ and $B=(b_{ij})$ and $(e_1,\cdots,e_n)$ the canonical basis and since $A$ and $B$ are upper triangle matrices then

$$AB(e_i)=\sum_{k=1}^ib_{ki}\sum_{s=1}^ka_{sk}e_s=\sum_{s=1}^i\left(\sum_{k=s}^ib_{ki}a_{sk}\right)e_s=\sum_{s=1}^i\alpha_se_s$$ hence $AB$ is also an upper triangle matrix.

amWhy
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$$ \begin{bmatrix} A & B \\ 0 & C\end{bmatrix} \begin{bmatrix} D & E \\ 0 & F \end{bmatrix} = \begin{bmatrix} AD & AE + BF \\ 0 & CF \end{bmatrix}$$

We never assumed that the elements of the matrix are multiplicatively commutative, so the result generalizes to all $n$ by $n$ matrices recursively.

DanielV
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