## 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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