## Linear Algebra and Its Applications, Exercise 2.6.4

Exercise 2.6.4. Consider the matrix

$A = \begin{bmatrix} 2&0 \\ 0&1 \end{bmatrix}$

This matrix will “stretch” vectors along the x axis, transforming the vector $v = (x, y)$ into the vector $v' = (2x, y)$. Consider also the circle formed by all points for which $x^2 + y^2 = 1$. What shape is the curve created by transforming all points on the circle by $A$?

Answer: Transforming the points on the circle using $A$ has the effect of stretching the circle horizontally while keeping the vertical height the same. Points to the right of the y-axis (for which $x$ is positive) get stretched in a positive direction, while points to the left of the y-axis (for which $x$ is negative) get stretched in a negative direction. The resulting (closed) curve appears to be an ellipse symmetric about the two axes.

We can confirm that the new curve is an ellipse by showing that points on the curve satisfy the equation

$(x/a)^2 + (y/b)^2 = 1$

for some $a$ and $b$. We know that for all points $v = (x, y)$ on the original circle we have $x^2 + y^2 = 1$. After multiplying by $A$ for all points $v' = (x', y')$ on the new curve we have $x' = 2x$ and $y' = y$. Since $x' = 2x$ we have $x = x'/2$.

We then have

$1 = x^2 + y^2 = (x'/2)^2 + (y'/1)^2$

so that

$(x'/a)^2 + (y'/b)^2 = 1$

for $a = 2$ and $b = 1$. The new curve is thus an ellipse.

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.