## Linear Algebra and Its Applications, exercise 1.4.13

Exercise 1.4.13. Provide an example of multiplying two 3×3 triangular matrices, to confirm the general case that the product of triangular matrices is itself a triangular matrix. Prove the general case based on the definition of matrix multiplication.

Answer: An example of multiplying two 3×3 upper triangular matrices:

$\begin{bmatrix} 1&1&1 \\ 0&1&1 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&2&3 \\ 0&1&2 \\ 0&0&1 \end{bmatrix} = \begin{bmatrix} 1&3&6 \\ 0&1&3 \\ 0&0&1 \end{bmatrix}$

In general, for the two upper triangular matrices A and B, where A is mxn and B is nxp, we have

$a_{ij} = 0, i > j$ and $b_{ij} = 0, i > j$

For the product C = AB we have

$c_{ij} = \sum_{k=1}^{k=n} a_{ik}b_{kj} \: \rm for \: i = 1, \ldots, m \: \rm and \: j = 1, \ldots, p$

Assume $i > j$. Then

$1 \le k \le j \Rightarrow i > k \Rightarrow a_{ik} = 0 \Rightarrow a_{ik}b_{kj} = 0 \: \rm for \: k = 1, \ldots, j$

and

$j < k \le n \Rightarrow k > j \Rightarrow b_{kj} = 0 \Rightarrow a_{ik}b_{kj} = 0 \: \rm for \: k = j+1, \ldots, n$

If i > j we thus have

$a_{ik}b_{kj} = 0 \: \rm for \: k = 1, \ldots, n$

But then if i > j we have

$c_{ij} = \sum_{k=1}^{k=n} a_{ik}b_{kj} = \sum_{k=1}^{k=n} 0 = 0$

Since $c_{ij} = 0$ for i > j, C = AB is an upper triangular matrix as well.

A similar argument shows that the product of two lower triangular matrices is also lower triangular. (For a lower triangular matrix A we would have $a_{ij} = 0$ if i < j.)

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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