## Linear Algebra and Its Applications, exercise 1.4.18

Exercise 1.4.18. Given arbitrary nxn matrices A and B, show that the first column of AB is the same as A times the first column of B. Hint: Let x = (1, 0, …, 0) be a column vector with n components, and use the fact that A(Bx) = (AB)x.

$Bx = \begin{bmatrix} b_{11}&b_{12}&\cdots&b_{1n} \\ b_{21}&b_{22}&\cdots&b_{2n} \\ \vdots&\vdots&\ddots&\vdots \\ b_{n1}&b_{22}&\cdots&b_{nn} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} b_{11} \\ b_{21} \\ \vdots \\ b_{n1} \end{bmatrix}$

So Bx is equal to the first column of B. Similarly if C = AB we have

$Cx = \begin{bmatrix} c_{11}&c_{12}&\cdots&c_{1n} \\ c_{21}&c_{22}&\cdots&c_{2n} \\ \vdots&\vdots&\ddots&\vdots \\ c_{n1}&c_{22}&\cdots&c_{nn} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = \begin{bmatrix} c_{11} \\ c_{21} \\ \vdots \\ c_{n1} \end{bmatrix}$

So again Cx is equal to the first column of C or, to put it another way, (AB)x is equal to the first column of AB.

But we also have A(Bx) = (AB)x, which means that A times the first column of B is equal to the first column of AB.

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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