Theorem 1   (Gram-Schmidt Orthogonalization) Let $\{\mathbf{u_k}\}_{k=1}^n$ be a basis for an inner product space $V$ with inner product $\big<,\big>$ . Define $\mathbf{v_1}=\tfrac{\mathbf{u_1}}{\mid\mid\mathbf{u_1}\mid\mid}$ and $\{\mathbf{m_k}\}_{k=2}^n$ recursively by \begin{equation} \mathbf{m_k}=\mathbf{u_k}-\big<\mathbf{u_k},\mathbf{v_1}\big>\mathbf{v_1}- \big<\mathbf{u_k},\mathbf{v_2}\big>\mathbf{v_2}-\ldots -\big<\mathbf{u_k},\mathbf{v_{k-1}}\big>\mathbf{v_{k-1}}\text{,} \end{equation}where $\mathbf{v_k}=\tfrac{\mathbf{m_k}}{\mid\mid\mathbf{m_k}\mid\mid}$ for $2\leq k\leq n$ . Then $\{\mathbf{v_k}\}_{k=1}^n$ is an orthonormal basis for $V$ .

Proof. We proceed by induction on $n$ . In the case $n=1$ , we suppose $\{\mathbf{u_k}\}_{k=1}^n=\{\mathbf{u_k}\}_{k=1}^1=\mathbf{u_1}$ is a basis for the inner product space $V$ . Letting $\mathbf{v_1}=\tfrac{\mathbf{u_1}}{\mid\mid\mathbf{u_1}\mid\mid}$ , it is clear that $\mathbf{v_1}\in \spn\big(\mathbf{u_1}\big)$ , whence it follows that $\spn\big(\mathbf{v_1}\big)=\spn\big(\mathbf{u_1}\big)=V$ . Thus $\mathbf{v_1}$ is an orthonormal basis for $V$ , and the result holds for $n=1$ . Now let $n\geq 1\in\mathbb{N}$ , and suppose the result holds for arbitrary $n$ . Let $\{\mathbf{u_k}\}_{k=1}^{n+1}$ be a basis for an inner product space $V$ . By the inductive hypothesis we may use $\{\mathbf{u_k}\}_{k=1}^n$ to construct an orthonormal set of vectors $\{\mathbf{v_k}\}_{k=1}^n$ such that $\spn\big(\{\mathbf{v_k}\}_{k=1}^n\big)=\spn\big(\{\mathbf{u_k}\}_{k=1}^n\big)$ . In accordance with the procedure outlined in the statement of the theorem, let $\mathbf{m_{n+1}}$ be defined as \begin{equation*} \mathbf{u_{n+1}}- \big<\mathbf{u_{n+1}},\mathbf{v_1}\big>\mathbf{v_1} -\big<\mathbf{u_{n+1}},\mathbf{v_2}\big>\mathbf{v_2} -\ldots -\big<\mathbf{u_{n+1}},\mathbf{v_n}\big>\mathbf{v_n} =\mathbf{u_{n+1}}-\sum\limits_{i=1}^n \big<\mathbf{u_{n+1}},\mathbf{v_i}\big>\mathbf{v_i}\text{.} \end{equation*}First we show that the vectors $\mathbf{v_1},\mathbf{v_2},\ldots,\mathbf{v_n},\mathbf{m_{n+1}}$ are mutually orthogonal. Consider the inner product of $\mathbf{m_{n+1}}$ with $\mathbf{v_j}$ for $1\leq j\leq n$ . By construction, we have \begin{equation*} \big<\mathbf{m_{n+1}},\mathbf{v_j}\big>=\biggl<\mathbf{u_{n+1}} -\sum_{i=1}^n\big<\mathbf{u_{n+1}}, \mathbf{v_i}\big>\mathbf{v_i},\mathbf{v_j}\biggr> =\big<\mathbf{u_{n+1}},\mathbf{v_j}\big> -\biggl<\sum\limits_{i=1}^n\big<\mathbf{u_{n+1}},\mathbf{v_i}\big> \mathbf{v_i},\mathbf{v_j}\biggr>\text{.} \end{equation*}Now since $\{\mathbf{v_k}\}_{k=1}^n$ is an orthonormal set of vectors, whence $\big<\mathbf{v_i},\mathbf{v_j}\big>=\delta_{ij}$ , each term in the summation on the right-hand side of the preceding equation will vanish except for the term where $i=j$ . Thus by this and the preceding equation, we have
Thus $\mathbf{m_{n+1}}$ is orthogonal to $\mathbf{v_j}$ for $1\leq j\leq n$ , so we may take $\mathbf{v_{n+1}}=\tfrac{\mathbf{m_{n+1}}}{\mid\mid\mathbf{m_{n+1}}\mid\mid}$ to have $\{\mathbf{v_k}\}_{k=1}^{n+1}$ an orthonormal set of vectors. Finally we show that $\{\mathbf{v_k}\}_{k=1}^{n+1}$ is a basis for $V$ . By construction, each $\mathbf{v_k}$ is a linear combination of the vectors $\{\mathbf{u_k}\}_{k=1}^{n+1}$ , so we have $n+1$ orthogonal, hence linearly independent vectors in the $n+1$ dimensional space $V$ , from which it follows that $\{\mathbf{v_k}\}_{k=1}^{n+1}$ is a basis for $V$ . Thus the result holds for $n+1$ , and by the principle of induction, for all $n$ .