Houjun Liu


A list of vectors is orthonormal if each vector is orthogonal to every other vector, and they all have norm 1.

In other words:

\begin{equation} \langle e_{j}, e_{k} \rangle = \begin{cases} 1, j = k\\ 0, j \neq k \end{cases} \end{equation}

The vectors should inner-product with itself to \(1\), and be orthogonal to all others.

Additional Information

orthonormal basis

See also orthonormal basis

Norm of an Orthogonal Linear Combination

\begin{equation} \| a_1e_1 + \dots + a_{m}e_{m} \|^{2} = |a_1|^{2} + \dots + |a_{m}|^{2} \end{equation}

When \(e_1, \dots e_{m}\) are orthonormal vectors in \(V\) and \(a_1, \dots a_{m} \in \mathbb{F}\).


Recall two facts: \(e_{j}\) are orthonormal vectors, so they are 1) orthogonal to each other and have 2) norm 1. Therefore, each \(a_j e_{j}\) are also orthogonal and have norm \(a_{j}\)

And so, the orthogonal condition guarantees pythagoras, and we know that each vector being added here has norm \(a_{j}\).

And so we can just chonk out each of the vectors, apply Pythagoras to the ending bunch and the one removed.

orthonormal list is linearly independent

Its a corollary of the above is that orthonormal lists are linearly independent.


\begin{equation} a_1e_1 + \dots +a_{m}e_{m} = 0 \end{equation}

We desire that each \(a_{j}=0\) to show that this list is linearly independent.

Now, given that the linear combination of these \(e_{j}\) adds to \(0\), the summed vector is a zero-vector. So:

\begin{equation} \|a_1 e_1 + \dots +a_{m}e_{m} \| = \|0\| = 0 \end{equation}

OF course their norm squared is also \(0\).

Apply the above, then, we now have:

\begin{equation} |a_1|^{2} + \dots +|a_{m}|^{2} = \|a_1 e_1 + \dots +a_{m}e_{m} \| = 0 \end{equation}

Of course adding a list of positive numbers (\(|a_{j}|^{2}\)) together yields not a negative number, so there are no possible additive inverses that will cancel each other out. Hence, \(a_{j} = 0\), as desired. \(\blacksquare\)