## 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.

This entry was posted in linear algebra. Bookmark the permalink.