Linear Algebra and Its Applications, Exercise 3.3.21

Exercise 3.3.21. Given three vectors $a_1$ , $a_2$ , and $b$ , and the two lines $L_1$ through the origin and $a_1$ and $L_2$ through $b$ in the direction of $a_2$ , we want to find scalar values $x_1$ and $x_2$ such that the distance $\| x_1a_1 - x_2a_2 - b \|$ between the points $x_1a_1$ and $b + x_2a_2$ is at a minimum. Write down equations for $x_1$ and $x_2$ . Solve the equations for $x = (x_1, x_2)$ when $a_1 = (1, 1, 0)$ , $a_2 = (0, 1, 0)$ , and $b = (2, 1, 4)$ .

Answer: We approach the problem by taking partial derivatives of $\| x_1a_1 - x_2a_2 - b \|^2$ with respect to $x_1$ and $x_2$ and then setting those partial derivatives to zero. We have

$\| x_1a_1 - x_2a_2 - b \|^2 = (x_1a_1 - x_2a_2 - b)^T(x_1a_1 - x_2a_2 - b)$

We take advantage of the facts that for two vectors $v$ and $w$ we have $v^Tw = w^Tv$ (the inner product is the same no matter in which order we calculate it) and $(v-w)^T = v^T - w^T$ (the transpose of a difference is the same as the difference of the transposes). We then have

$\| x_1a_1 - x_2a_2 - b \|^2 = (x_1a_1^T - x_2a_2^T - b^T)(x_1a_1 - x_2a_2 - b)$

$= x_1^2a_1^Ta_1 - x_1x_2a_1^Ta_2 - x_1a_1^Tb - x_2x_1a_2^Ta_1 + x_2^2a_2^Ta_2$

$+ x_2a_2^Tb - x_1b^Ta_1 + x_2b^Ta_2 + b^Tb$

$= x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb$

Next we take the partial derivatives with respect to $x_1$ :

$\frac{\partial}{\partial x_1} \| x_1a_1 - x_2a_2 - b \|^2$

$= \frac{\partial}{\partial x_1} (x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb)$

$= 2x_1a_1^Ta_1 - 2x_2a_1^Ta_2 - 2a_1^Tb$

and with respect to $x_2$ :

$\frac{\partial}{\partial x_2} \| x_1a_1 - x_2a_2 - b \|^2$

$= \frac{\partial}{\partial x_2} (x_1^2a_1^Ta_1 + x_2^2a_2^Ta_2 - 2x_1x_2a_1^Ta_2 - 2x_1a_1^Tb + 2x_2a_2^Tb + b^Tb)$

$= 2x_2a_2^Ta_2 - 2x_1a_1^Ta_2 + 2a_2^Tb$

Equating the partial derivatives to zero gives us the following system of equations (after simplifying by dividing by 2 and rearranging terms):

$\begin{bmatrix} a_1^Ta_1&-a_1^Ta_2 \\ -a_1^Ta_2&a_2^Ta_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} a_1^Tb \\ -a_2^Tb \end{bmatrix}$

When $a_1 = (1, 1, 0)$ , $a_2 = (0, 1, 0)$ , and $b = (2, 1, 4)$ we have $a_1^Ta_1 = 2$ , $a_2^Ta_2 = 1$ , $a_1^Ta_2 = 1$ , $a_1^Tb = 3$ , and $a_2^Tb = 1$ . The system of equations is then

$\begin{bmatrix} 2&-1 \\ -1&1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$

We can solve this by multiplying both sides of the equation on the left by

$\begin{bmatrix} 2&-1 \\ -1&1 \end{bmatrix}^{-1} = \frac{1}{2 - (-1)(-1)} \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix}$

to get

$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 1&1 \\ 1&2 \end{bmatrix} \begin{bmatrix} 3 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$

The solution is thus $x = (2, 1)$ .

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

Linear Algebra and Its Applications, Exercise 3.3.21

Leave a Reply Cancel reply

Archives

Meta

Linear Algebra and Its Applications, Exercise 3.3.21

Share this:

Like this:

Related

Leave a Reply Cancel reply

Archives

Meta