## 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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