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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