Proximity to an orthonormal basis implies linear independence

Question

This is a problem from Prasolov's Geometry:

Given an orthonormal basis $e_1,\dots,e_n$ and a set of vectors $a_1,\dots,a_n$ such that the angle between the vectors $e_i$ and $a_i$ equals $\alpha_i$ for each $i$, prove that if $$\cos \alpha_1+\cdots+\cos \alpha_n>\sqrt{n(n-1)} ,$$ then the vectors $a_1,\dots,a_n$ are linearly independent.

The problems seem more like a linear algebra problem, but I failed to solve it. Any hint please?

Answer 1

The function $p:\{a_{1},\ldots a_{n}\}\to\sum_{i=1}^{n}\cos\alpha_{i}$, for $\lVert a_{i}\rVert=1$, seems to be a measure of proximity to the given standard orthonormal basis $\{e_{1},\ldots e_{n}\}$: $1\leq p\leq n$.

What is the $\max_{\{a_{1},\ldots a_{n}\}\subset H^{n-1}}p$? (Spoiler: it's $\sqrt{n(n-1)}$). Let's first try to put the hyperplane $H^{n-1}$ "flat" on the $span(e_{1},\ldots e_{n-1})$ with the $e_{n}$ being its unit normal. We're free to choose the best alignment possible for $a_{i}=e_{i}$, $1\leq i\leq n-1$. It doesn't matter in this case but let's choose $a_{n}=\frac{1}{\sqrt{n-1}}\sum_{i=1}^{n-1}e_{i}$ for symmetry reasons. We have achieved a respectable $p_{0}=n-1$.

However, $p$ can be made slightly larger if we tilt the hyperplane "pulling" the unlucky $a_{n}$ (chosen as above) towards $e_{n}$ by an angle $\beta$. For example, in 2D, we'd get a better (the maximum actually) $p_{\beta}=\sqrt{2}>1=p_{0}$ by aligning $a_{1}$and $a_{2}$ along $\beta=45{}^{\circ}$ between $e_{1}$ and $e_{2}$.

Let's do this. It's a simple but cumbersome trigonometric and vector arithmetic, followed by differentiation to find the best optimal angle. I hope someone can write this in a matrix algebraic-way or even fancier.

Again, for symmetry reasons all $\alpha_{i}=\alpha$, $1\leq i\leq n-1$, so we have $$ p_{\beta}=(n-1)\cos\alpha+\sin\beta $$

I will be skipping some details now, adding them back by request possibly with a diagram showing the triangles. The best $\cos\alpha$ we could get from orthogonally projecting now detached $e_{1},\ldots$ back onto the tilted hyperplane to find the best $a_{1},\ldots$: $\langle e_{1},N_{\beta}\rangle=cos(\tfrac{\pi}{2}+\alpha)=-\sin\alpha$. We therefore need to find the unit normal to the hyperplane, $$ N_{\beta}=\frac{e_{n}-\tan\beta\cdot a_{n}}{\sqrt{1+tan^{2}\beta}} $$

After some manipulations we'll get

$$ \cos\alpha=\sqrt{1-\frac{\sin^{2}\beta}{n-1}} $$

Now let $x=\sin\beta$ and we have

$$ p(x)=(n-1)\sqrt{1-\frac{x^{2}}{n-1}}+x $$

Differentiating by $x$ and solving $p^{\prime}(x)=0$ we find the best tilt angle $x=\sin\beta=\sqrt{\frac{n-1}{n}}=\cos\alpha$ (so it turns out all the $n$ angles are in fact equal $\alpha_1=\ldots=\alpha_n$ as they should be) and the largest possible by construction $$ p=\sqrt{n(n-1)}. $$

Any value of $p$ larger than that would have to come from a $\{a_{1},\ldots a_{n}\}$ not constrained to a hyperplane being therefore a linear independent set.

Answer 2

I can prove under the stronger assumption that $\sum \cos\alpha_i > n -\frac{1}{2}\ \ (=\sqrt{n(n-1)+\frac{1}{4}}\ $). The idea is the look at the matrix $A$ with the column unit vectors $a_i$ and notice that $A=I+\Delta$, where the Frobenius norm of $\Delta$ is $<1$ (so the spectral radius of $\Delta $ is $<1$) and that guarantees invertibility.

Answer 3

I can get somewhat close. Let's make the hypotheses that $$ \sum_{i=1}^n \cos(\alpha_i) \ge \sqrt{n(n-1)} + \frac{1}{n} $$

Observe that the quantity $$ n - \sqrt{n(n-1)} = \frac{1}{1 + \sqrt{\frac{n-1}{n}}} $$ Unfortunately, this is $>\frac{1}{2}$. (It does converge to $\frac{1}{2}$, but this won't help in what follows.) On the other hand, you can check that $$ \frac{1}{1 + \sqrt{\frac{n-1}{n}}} - \frac{1}{n} < \frac{1}{2} $$ We'll just take this as read.

We may assume that the $a_i$ are unit vectors, in which case $$ \cos \alpha_i = \left\langle e_i,a_i\right\rangle $$ The implication of the stronger hypothesis is then that $$ \sum_{i=1}^n \|e_i - a_i\|^2 = 2\left(n - \sum_{i=1}^n \left\langle e_i,a_i\right\rangle\right) < 2\left(n - \sqrt{n(n-1)} - \frac{1}{n}\right) < 1 $$

Now, assume that the $a_i$ are dependent. This means we can find some unit vector $v$ that is orthogonal to all the $a_i$; in particular, we have that $$ \left\langle v,e_i - a_i\right\rangle = \left\langle v,e_i\right\rangle - \left\langle v, a_i\right\rangle = \left\langle v,e_i\right\rangle $$

Putting things together, we have $$ 1 = \|v\|^2 = \sum_{i=1}^n \langle v,e_i\rangle^2 = \sum_{i=1}^n \left\langle v,e_i - a_i\right\rangle^2 \le \sum_{i=1}^n \|v\|^2\|e_i - a_i\|^2 = \sum_{i=1}^n \|e_i - a_i\|^2 < 1 $$ by the work we did above. The resulting contradiction tells us that, in fact, the $a_i$ were independent.