# Linear Algebra Proof of Mean Value Theorem

Is this a valid proof of the mean value theorem?

Suppose $f$ is a continuous differentiable function on the interval $[a,b]$ then there exists some $c\in(a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$.

# Proof:

Let $B=\{e_1,e_2\}$ be the standard basis for $\mathbb{R}^2$. Then let $\theta = \tan^{-1}\left(\frac{f(b)-f(a)}{b-a}\right)$, that is, the angle that the secant line forms with the $x$-axis. Then consider the the linear transformation, $T$, defined by rotation matrix by $\theta$ on the standard basis:

$$T = \left(\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right)$$

Then $T(B)$ is a basis one of whose component vectors is parallel with the secant line. Thus we may apply Rolle's theorem and the result follows. $\blacksquare$

My thinking here is that by rotating the basis vectors we can just consider it as $f(a)=f(b)$ and Rolle's theorem gives it to us right away then. I guess the intuition behind the proof is that you can just turn your head until it's a curve that satisfies the conditions of Rolle's theorem, but is that valid can I actually do this?

• You haven't written a proper proof, but the MVT can indeed be shown easily using Rolle's theorem. – Olivier Oct 1 '17 at 1:25
• Normally this would be done by a shear rather than by a rotation. – Michael Hardy Oct 1 '17 at 1:28

However, you can prove the mean value theorem from Rolle's theorem: define a new function to be equal to $f$ minus the line through $(a,f(a))$ and $(b,f(b))$ and then apply Rolle's Theorem.
The linear transformation whose matrix is $$\begin{bmatrix} 1 & 0 \\ -m & 1 \end{bmatrix},$$ where $m$ is the slope of the secant line, will work. And that is just what is usually done, even if it's not phrased in the language of linear algebra and matrices.