## Linear Algebra and Its Applications, Exercise 2.6.10

Exercise 2.6.10. Consider the vector space of functions $u$ in $t$ for which $d^2u/dt^2 = u$. Find two functions that can serve as basic vectors for the space.

Answer: For a function to be in the space its second derivative with respect to $t$ must be the original function. We know that the functions in the space cannot be polynomial functions in $t$ (e.g., $u(t) = t^n$ for some $n$) because differentiating a polynomial of degree $n$ produces a polynomial of degree $n-1$ ($(d/dt) t^n = nt^{n-1}$) and taking the second derivative produces a polynomial of degree $n-2$ ($(d^2/dt^2) t^n = (d/dt) nt^{n-1} = n(n-1)t^{n-2}$).

What about exponential functions in $t$? We know that $(d/dt) e^t = e^t$ (see for example the Wikipedia article on the exponential function) so if we choose $u_1(t) = e^t$ we then have

$(d^2/dt^2) u_1(t) = (d/dt) (d/dt) u_1(t)$

$(d/dt) (d/dt) e^t = (d/dt) e^t$

$= e^t = u_1(t)$

so that $d^2u_1/dt^2 = u_1$.

Can we find a second function in the space? In general we have $(d/dt) e^{kt} = k e^{kt}$ (this follows from the chain rule) so that

$(d^2/dt^2) e^{kt} = (d/dt) k e^{kt}$

$= k (d/dt) e^{kt} = k \cdot k e^{kt} = k^2 e^{kt}$

The function being differentiated on the left hand side of this equation will equal the function on the right hand side if $k^2 = 1$ which is true if $k = 1$ or $k = -1$. The case $k = 1$ corresponds to the function $u_1(t) = e^t$ we discussed above. The case $k = -1$ corresponds to the function $u_2(t) = e^{-t}$ for which we have

$(d^2/dt^2) u_2(t) = (d/dt) (d/dt) u_2(t)$

$= (d/dt) (d/dt) e^{-t} = (d/dt) (-1 \cdot e^{-t})$

$= -(d/dt) e^{-t} = -(-1 \cdot e^{-t})$

$= -(-e^{-t}) = e^{-t} = u_2(t)$

so that $d^2u_2/dt^2 = u_2$.

The functions $u_1$ and $u_2$ are linearly dependent (since one is not a scalar multiple of the other) and form a basis for the space. The vectors in the space consist of all functions $u$ of the form $c_1u_1 + c_2u_2$. For all such functions we have

$d^2u/dt^2 = (d^2/dt^2) (c_1u_1 + c_2u_2)$

$= (d^2/dt^2) (c_1u_1) + (d^2/dt^2) (c_2u_2)$

$= c_1 \cdot d^2u_1/dt^2 + c_2 \cdot d^2u_2/dt^2$

$= c_1u_1 + c_2u_2 = u$

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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