Linear Algebra and Its Applications, Exercise 2.6.5

Exercise 2.6.5. Suppose we have two points x and y and a third point z halfway between the first two points. Show that for any linear transformation represented by a matrix A the (transformed) point Az is halfway between Ax and Ay.

Answer: If z is halfway between x and y then we have z = \frac{1}{2}(x + y).  For example, if x = (1, 2) and y = (3, 8) then

z = \frac{1}{2}(\begin{bmatrix} 1 \\ 2 \end{bmatrix} + \begin{bmatrix} 3 \\ 8 \end{bmatrix}) = \frac{1}{2} \begin{bmatrix} 4 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ 5 \end{bmatrix}

We then have

Az = A(\frac{1}{2}(x+y)) = \frac{1}{2}A(x+y) = \frac{1}{2}(Ax+Ay)

But this implies that Az is halfway between Ax and Ay, which is what we were asked to show.

For example, suppose that

A = \begin{bmatrix} 1&2 \\ -1&1 \end{bmatrix}

Then using our example points above we have

Ax = \begin{bmatrix} 1&2 \\ -1&1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \end{bmatrix}

Ay = \begin{bmatrix} 1&2 \\ -1&1 \end{bmatrix} \begin{bmatrix} 3 \\ 8 \end{bmatrix} = \begin{bmatrix} 19 \\ 5 \end{bmatrix}

Az = \begin{bmatrix} 1&2 \\ -1&1 \end{bmatrix} \begin{bmatrix} 2 \\ 5 \end{bmatrix} = \begin{bmatrix} 12 \\ 3 \end{bmatrix}

\frac{1}{2}(Ax+Ay) = \frac{1}{2}(\begin{bmatrix} 5 \\ 1 \end{bmatrix} + \begin{bmatrix} 19 \\ 5 \end{bmatrix}) = \frac{1}{2} \begin{bmatrix} 24 \\ 6 \end{bmatrix} = \begin{bmatrix} 12 \\ 3 \end{bmatrix} = Az

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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