## 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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