## 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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