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Orthonormal

A set of vectors $\{v_k\}$ is orthonormal if its elements $v_k$ satisfy \[v_i \cdot v_j = \begin{cases} 1 & i = j, \\ 0 & i \neq j. \end{cases}\]

The "ortho" part of "orthonormal" refers to the fact that distinct vectors in the set are orthogonal (perpendicular): their dot product is zero. The "normal" part refers to the fact that the norm (magnitude) $||v_k||$ of every vector in the set is one, since $||v_k||^2 = v_k \cdot v_k = 1$.

If the cardinality of an orthonormal set $\{v_k\}$ equals the number of entries in each vector $v_k$ (the dimension of the vector space), then $\{v_k\}$ is an orthonormal basis of the space.

Orthonormal matrix

An orthonormal matrix is a matrix whose rows (or equivalently columns; see below) form an orthonormal basis of vectors. Note that since the rows form a basis, an orthonormal matrix must be square.

Inverse and transpose

If $M$ is an orthonormal matrix with rows $v_k$, then we have $M^TM = MM^T = I$, where $M^T$ is the transpose of $M$. The proof is as follows.

Let $P = MM^T$. By the matrix multiplication formula, $P_{ij}$ is the dot product of row $i$ of $M$ and column $j$ of $M^T$ (the latter is equivalent to row $j$ of $M$). Thus, \[P_{ij} = v_i \cdot v_j = \begin{cases} 1 & i = j, \\ 0 & i \neq j, \end{cases}\] so the entries of $P$ match those of the identity matrix $I$ exactly; hence, $MM^T = P = I$.

Since the right inverse of a matrix is also its left inverse, we can also write $M^TM = I$ and, in summary, $M^T = M^{-1}$.

Conversely, letting $M$ be a matrix with rows $v_k$, if $MM^T = I$ (or equivalently $M^TM = I$), then \[v_i \cdot v_j = I_{ij} = \begin{cases} 1 & i = j, \\ 0 & i \neq j, \end{cases}\] so $M$ must be orthonormal.

Orthonormal basis of columns

Let $M$ be an orthonormal matrix. As shown above, $(M^T)^TM^T = MM^T = I$, so $M^T$ is orthonormal. Thus, the columns of $M$, being the rows of $M^T$, also form an orthonormal basis of vectors.

Conversely, let $M$ be a matrix whose columns form an orthonormal basis of vectors. The rows of $M^T$ are orthonormal; as shown above, $MM^T = (M^T)^TM^T = I$, so the rows of $M$ itself are orthonormal.

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