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Let $A$, $B$ be $n\times n$ lower triangular matrices.

Prove that $AB$ is also a lower triangular matrix.

How do I prove this for every $n$?

Algebraic Pavel
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FigureItOut
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  • Yes I do, and did checked if it's true in some cases, but I don't know the approach of showing it's true for every $nxn$ matrix. – FigureItOut Dec 08 '14 at 23:37
  • What conditions in $A_{i,j}$ the matrix must have to be lower triangular? – Phicar Dec 08 '14 at 23:38
  • Hint: Try writing out the definition of matrix multiplication in general, then take into account that both matrices are lower-triangular. – Daniel Hast Dec 08 '14 at 23:38
  • I know the definition of matrix multiplication, and I know that a triangular matrix is defined that $a_{ij} = 0 $ if $ i > j $ but still, I can't get how I can conclude from all this that when I multiply it will always hit the zero to get a triangular matrix – FigureItOut Dec 08 '14 at 23:42
  • As a matter of fact, you can even prove that the inverse of an upper (res. lower) triangular matrix is also an upper (res. lower) triangular matrix. As for your question, try to visualize it first (write A and B lower triangular matrices side by side and try to see what happens when you calculate $c_{1,2}$ where C=AB for instance). You'll see that the positions of the zeros in each matrix necesseraly implies that the upper part is made of zeros as well. – mvggz Dec 09 '14 at 13:29
  • @user1326293 Found anything interesting below? – mvggz Dec 09 '14 at 15:27
  • I'm still observing all the solutions in order to fully understand them, havent got to yours yet but I promise Ill take a look at it, once I find the best solution Ill give the answer :) – FigureItOut Dec 09 '14 at 15:29

4 Answers4

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Induction and block partitioning of matrices is a very useful tool for proving statements like this.

The statement is trivial for $n=1$. Assume that the products of $n\times n$ ($n\geq 1$) lower triangular matrices are lower triangular and let $A$ and $B$ be $(n+1)\times (n+1)$. Consider the paritioning $$ A:=\pmatrix{\alpha_{11} & 0 \\ a_{21} & A_{22}}, \quad B:=\pmatrix{\beta_{11} & 0 \\ b_{21} & B_{22}}, $$ where $A_{22}$ and $B_{22}$ are $n\times n$. We have $$ C:=AB= \pmatrix{\alpha_{11} & 0 \\ a_{21} & A_{22}} \pmatrix{\beta_{11} & 0 \\ b_{21} & B_{22}} = \pmatrix{\alpha_{11}\beta_{11} & 0\\\beta_{11} a_{21}+A_{22}b_{21}&A_{22}B_{22}}. $$ Hence $C$ is lower triangular if and only if $A_{22}B_{22}$ is lower triangular. But $A_{22}$ and $B_{22}$ are $n\times n$ lower triangular and by the induction assumption, their product is lower triangular. Hence $C$ is lower triangular.

Algebraic Pavel
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Hint: Suppose that $A,B$ are $n\times n$ matrices. Then if $i,j\in\{1,...,n\},$ we have that the $i^\text{th}$ row, $j^\text{th}$ column entry of $AB$ is $$\sum_{k=1}^na_{ik}b_{kj}.$$ You must show that this is $0$ when $i<j$ (note that this differs from what you said in the comments above!) so long as $A,B$ are lower triangular.

Cameron Buie
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I'd rather prove this using transformations associated to both matrices A and B. They are both assumed to be nxn square matrices.

Let u be the tranformation associated to A in the canonical basis $(e_1,..,e_n)$. Let v be the same for B.

I'll define: $ I_j = Vect(e_j,..,e_n). $ We obviously have : $I_{j+1} \subset I_j$

What does :"A and B lower triangular matrices " mean?

It means that: for every $k\in (1..n): u(e_k)\in I_k; v(e_k)\in I_k$. This is an equivalency of definition. The zero above the diagonal mean that $u(e_j)$ does not have a part expressed in terms of $e_i$, with $i\in (1..j-1)$. This implies that: $ u(I_k) \subset I_k; v(I_k) \subset I_k $

Now if I call w the tranformation associated to C=AB, we have: w=u o v (composition of tranformations)

We only have to prove that : for every $k\in (1..n): w(e_k)\in I_k$:

$ w(e_k) = u(v(e_k)) $ . But $v(e_k)\in I_k= Vect(e_k,..,e_n)$, so as it's been stated before: $u(I_k) \subset I_k \implies w(e_k) = u(v(e_k))\in I_k$

So w is a transformation associated to a lower triangular matrix

mvggz
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Hint: Write it down in a general form: $$ \left( \begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \cdots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{array} \right) \cdot \left( \begin{array}{ccc} b_{11} & \cdots & b_{1n} \\ \vdots & \cdots & \vdots \\ b_{n1} & \cdots & b_{nn} \end{array} \right) $$

Now for lower triangular matrices you have that the upper parts of the matrices are all zeros. Now just calculate using the definition of matrix multiplication. It is sufficient to show that again all upper triangular part and the diagonal is zero ...

Loreno Heer
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