Linear Algebra and Its Applications, Exercise 1.5.7

Posted on January 8, 2011 by hecker

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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