orthonormal — Jemoka Knowledge Base

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}. Proof: 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. Proof:

\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