Prove $\sum_{i=1}^{n}v_i^t \cdot v_i = I$ for orthonormal base

Question

Let $$ \{v_1,..,v_n\} $$ Be an orthonormal base in $R^n$ with the standard inner product.

I need to prove that: $$ \sum_{i=1}^{n}v_i^t \cdot v_i = I $$

Where $v_i$ is a row vector.

What i tried:

I tried to look at an example but still - i dont feel its getting me somewhere.

Lets take $R^2$

The base will be: $$ \{v_1,v_2\} $$

Let $v_1 = [a_1,a_2], v_2 = [b_1,b_2]$

As an orthonormal base we know that:

$$ ||v_1|| = ||v_2|| = 1 $$

And:

$$ <v_1,v_2> = 0 $$

Therefore:

$$ ||v_1||^2 = a_1^2 + a_2^2 = 1, ||v_2||^2 = b_1^2 + b_2^2 = 1 $$

$$ <v_1,v_2> = a_1b_1 + a_2b_2 = 0 $$

$$ \sum_{i = 1}^{n = 2}v_i^t \cdot v_i = \begin{bmatrix} a_1^2&a_2a_2 \\ a_2a_1&a_2^2\end{bmatrix} + \begin{bmatrix} b_1^2&b_2b_2 \\ b_2b_1&b_2^2\end{bmatrix} $$

But i dont see how i get here to $I_2$? Am i even in the right direction? what am i missing?

I would prefer a hint than a full answer - as those are my homework.

And thanks for the help.

Answer 1

Hint: Rewrite $\sum_{i=1}^n v_i^Tv_i$ as a matrix product.

Answer 2

One way would be to check that $\left(\sum_{i=1}^{n}v_i^tv_i \right) w = w$ for every column vector $w$. Make sure you believe that this is sufficient to show that $\sum_{i=1}^{n}v_i^t v_i = I$.

Answer 3

For any column vector $\vec a$,

$\left (\displaystyle \sum_1^n v_i^T v_i \right )(\vec a) = \sum_1^n v_i^T (v_i \vec a) = \sum_1^n \langle v_i, \vec a \rangle v_i^T = \vec a, \tag 1$

since the $v_i$ form an orthonormal basis of $\Bbb R^n$. But this equation simply says that

$\displaystyle \sum_1^n v_i^T v_i = I. \tag 2$

Answer 4

Since $v_i$ are mutually orthonormal row vectors, I'll assign $\mathbf q_i = v_i^T$ so $\mathbf q_i \in \mathbb R^n$ are mutually orthonormal column vectors.

What follows is not the most direct approach, but if you're in a bind in linear algebra, a useful tactic is to try squaring what you have (and possibly square-rooting).

consider $X := \sum_{i=1}^n \mathbf q_i\mathbf q_i^T $
we can check to see what happens when we square this

$ X^2 $
$= \big(\sum_{i=1}^n \mathbf q_i\mathbf q_i^T\big)\big(\sum_{i=1}^n \mathbf q_i\mathbf q_i^T\big) $
$= \big(\sum_{i=1}^n \mathbf q_i(\mathbf q_i^T\mathbf q_i)\mathbf q_i^T\big) +\big(\sum_{i=1}^n\sum_{k\neq i} \mathbf q_k(\mathbf q_k^T\mathbf q_i)\mathbf q_i^T\big)$
$= \big(\sum_{i=1}^n 1\cdot \mathbf q_i\mathbf q_i^T\big) +\big(\sum_{i=1}^n\sum_{k\neq i} 0\cdot \mathbf q_k\mathbf q_i^T\big)$
$=\big(\sum_{i=1}^n \mathbf q_i\mathbf q_i^T\big) + 0 $
$=X$

so $X^2 = X$ (i.e. it is idempotent). Thus all eigenvalues are 0 or 1. And
$\text{trace}\big(X\big) = \text{trace}\big(\sum_{i=1}^n \mathbf q_i\mathbf q_i^T\big)= \sum_{i=1}^n \text{trace}\big(\mathbf q_i\mathbf q_i^T\big)= \sum_{i=1}^n \text{trace}\big(\mathbf q_i^T\mathbf q_i\big) =n$

So $X$ has $n$ eigenvalues of 1, no eigenvalues of 0 which implies $X=I$.
(Justification: it is diagonalizable and similar to the identity matrix, or that it is invertible, and the only idempotent element that is invertible is the identity, i.e. if $X^2 = X$ and we know $X^{-1}$ exists then multiply on the right by $X^{-1}$)