Linear Algebra and Its Applications, Exercise 2.6.11

Exercise 2.6.11. Consider the vector space $W$ of functions $u$ in $t$ for which $d^2u/dt^2 = u$ . From the previous exercise we can express any such function as $u = c_1u_1 + c_2u_2$ where $u_1 = e^t$ and $u_2(t) = e^{-t}$ are basis vectors for W.

Suppose at $t = 0$ we have $u = x$ and $du/dt = y$ . Find $c_1$ and $c_2$ such that $u = c_1u_1 + c_2u_2$ fulfills these initial conditions. This can be considered a linear transformation from the vector space $V$ of vectors $\begin{bmatrix} x&y \end{bmatrix}^T$ into the vector space $W$ . Find the matrix corresponding to this transformation assuming the basic vectors $\begin{bmatrix} 1&0 \end{bmatrix}^T$ and $\begin{bmatrix} 0&1 \end{bmatrix}^T$ for $V$ and $u_1$ and $u_2$ for $W$ .

Answer: At $t = 0$ we have

$u(0) = c_1u_1(0) + c_2u_2(0)$

$= c_1 e^0 + c_2 e^{-0} = c_1 + c_2$

We also have

$du/dt = (d/dt) (c_1u_1 + c_2u_2)$

$= (d/dt) (c_1u_1) + (d/dt) (c_2u_2)$

$= c_1 \cdot du_1/dt + c_2 \cdot du_2/dt$

$= c_1 \cdot (d/dt) e^t + c_2 \cdot (d/dt) e^{-t}$

$= c_1 e^t + c_2(-e^{-t}) = c_1e^t - c_2e^{-t}$

so that at $t = 0$ we have

$du/dt = c_1e^0 - c_2e^{-0} = c_1 - c_2$

From our initial conditions $u = x$ and $du/dt = y$ at $t = 0$ we then have

$c_1+c_2 = x$

$c_1-c_2 = y$

Adding the two equations we have $2c_1 = x+y$ or

$c_1 = (x+y)/2 = \frac{1}{2}x + \frac{1}{2}y$

Subtracting the second equation from the first we have $2c_2 = x-y$ or

$c_2 = (x-y)/2 = \frac{1}{2}x - \frac{1}{2}y$

The linear transformation from $V$ into $W$ can be represented by the following matrix:

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

so that if $v = \begin{bmatrix} x&y \end{bmatrix}^T$ is in $V$ we have

$Av = \begin{bmatrix} \frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{2}x+\frac{1}{2}y \\ \frac{1}{2}x-\frac{1}{2}y \end{bmatrix}$

$= (\frac{1}{2}x+\frac{1}{2}y) \begin{bmatrix} 1 \\ 0 \end{bmatrix} + (\frac{1}{2}x-\frac{1}{2}y) \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

$= (\frac{1}{2}x+\frac{1}{2}y) u_1 + (\frac{1}{2}x-\frac{1}{2}y) u_2 = u$

where $u$ is in $W$ .

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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Linear Algebra and Its Applications, Exercise 2.6.11

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