## Linear Algebra and Its Applications, Exercise 1.5.7

Exercise 1.5.7. Given the following lower triangular matrices $F = \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&0&1& \\ 0&0&0&1\end{bmatrix} \quad G = \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&2&1& \\ 0&0&0&1\end{bmatrix} \quad H = \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&0&1& \\ 0&0&2&1\end{bmatrix}$

find FGH and HGF.

Answer: We have $FGH = (FG)H$ where $FG = \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&0&1& \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&2&1& \\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&2&1& \\ 0&0&0&1 \end{bmatrix}$

so that $(FG)H = \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&2&1& \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&0&1& \\ 0&0&2&1 \end{bmatrix} = \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&2&1& \\ 0&0&2&1 \end{bmatrix}$

We also have $HGF = (HG)F$ where $HG = \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&0&1& \\ 0&0&2&1 \end{bmatrix} \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&2&1& \\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix} 1&&& \\ 0&1&&& \\ 0&2&1& \\ 0&4&2&1 \end{bmatrix}$

so that $(HG)F = \begin{bmatrix} 1&&& \\ 0&1&& \\ 0&2&1& \\ 0&4&2&1 \end{bmatrix} \begin{bmatrix} 1&&& \\ 2&1&& \\ 0&0&1& \\ 0&0&0&1 \end{bmatrix} = \begin{bmatrix} 1&&& \\ 2&1&& \\ 4&2&1& \\ 8&4&2&1 \end{bmatrix}$

Thus $FGH \ne HGF$.

UPDATE: Corrected the calculation of HGF; thanks go to Brian D. for pointing out my mistake.

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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