## 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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