Proof. To prove existence of the SVD, we isolate the direction of the largest action of $A \in \C^{m \times n}$ , and then proceed by induction on the dimension of $A$ . We will denote hermitian conjugation by $^T$ . Norms for vectors in $\C^n$ will be the usual euclidean 2-norm $\norm{\cdot }=\norm{\cdot}_2$ and for matrix the induced by norm of vectors.

Let $\sigma_1=\norm{A}$ . By a compactness argument, there must be vectors $v_1 \in \C^n, u^*_1 \in \C^m$ with $\norm{v_1}=1, \norm{u^*_1}=\sigma_1$ and $u^*_1=Av_1$ . Normalize $u^*_1$ by setting $u_1=u^*_1 / \norm{u^*_1}$ and consider any extensions of $v_1$ to an orthonormal basis $\{v_i\}$ of $\C^n$ and of $u_1$ to an orthonormal basis $\{u_j\}$ of $\C^m$ ; let $U_1$ and $V_1$ denote the unitary matrices with columns $\{v_i\}$ and $\{u_j\}$ respectively. Then we have

where $0$ is a column vector of dimension $m-1$ , $w^T$ is a row vector of dimension $n-1$ , and $B$ is a matrix of dimension $(m-1) \times (n-1)$ . Now,

so that $\norm{S} \geq (\sigma_1^2+w^2)^{1/2}$ . But $U_1$ and $V_1$ are unitary matrix, hence $\norm{S}=\sigma_1$ ; it therefore implies $w=0$ .

If $n=1$ or $m=1$ we are done. Otherwise the submatrix $B$ describes the action of $A$ on the subspace orthogonal to $v_1$ . By the induction hypothesis $B$ has an SVD $B=U_2 \Sigma_2 V^T_2$ . Now it is easily verified that

is an SVD of $A$ . completing the proof of existence.

For the uniqueness let $A=U \Sigma V^T$ a SVD for $A$ and let $e_i$ denote the i-th, $i=1 \cdots min(m,n)$ vector of the canonical base of $\C^n$ . As $U$ and $V$ are unitary, $\norm{Ae_i}=\sigma_i^2$ , so each $\sigma_i$ is uniquely determined.