[proofplan] We start from the unit-sphere formula for the [operator norm](/page/Operator%20Norm) and square it, reducing the problem to maximizing the quadratic form $x^*A^*Ax$ over Euclidean unit vectors. The matrix $A^*A$ is Hermitian positive semidefinite, so the finite-dimensional spectral theorem diagonalizes it with real nonnegative eigenvalues. In that diagonal basis, the Rayleigh quotient is a convex combination of the eigenvalues, hence its maximum on the unit sphere is the largest eigenvalue. Taking square roots gives the asserted operator norm formula. [/proofplan] [step:Rewrite the squared operator norm as a Rayleigh quotient] By the unit-sphere formula for the operator norm, [citetheorem:9320] gives \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}=\sup\{\|Ax\|_{\mathbb F^m}:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} Since the function $t\mapsto t^2$ is increasing on $[0,\infty)$, squaring this supremum gives \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}^2=\sup\{\|Ax\|_{\mathbb F^m}^2:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} For every $x\in\mathbb F^n$, the Euclidean [inner product](/page/Inner%20Product) identity gives \begin{align*} \|Ax\|_{\mathbb F^m}^2=(Ax)^*(Ax)=x^*A^*Ax. \end{align*} Therefore \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}^2=\sup\{x^*A^*Ax:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} [guided] The operator norm measures the largest Euclidean stretching factor of $A$. Because the theorem concerns the square root of an eigenvalue of $A^*A$, the natural first move is to square the norm and express $\|Ax\|_{\mathbb F^m}^2$ as a quadratic form. By the unit-sphere formula for the operator norm, [citetheorem:9320] applies to the bounded [linear map](/page/Linear%20Map) $A:\mathbb F^n\to\mathbb F^m$. In finite dimensions every linear map is bounded, and the statement already regards $A$ as a linear map between normed vector spaces with Euclidean norms. Hence \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}=\sup\{\|Ax\|_{\mathbb F^m}:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} All numbers inside this supremum are nonnegative. Since $t\mapsto t^2$ is increasing on $[0,\infty)$, the square of the supremum equals the supremum of the squares: \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}^2=\sup\{\|Ax\|_{\mathbb F^m}^2:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} Now use the defining relation of the conjugate transpose. For each $x\in\mathbb F^n$, the squared Euclidean norm of $Ax$ is \begin{align*} \|Ax\|_{\mathbb F^m}^2=(Ax)^*(Ax)=x^*A^*Ax. \end{align*} Thus the operator norm problem becomes the problem of maximizing a quadratic form over the Euclidean unit sphere: \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}^2=\sup\{x^*A^*Ax:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}. \end{align*} [/guided] [/step] [step:Verify that $A^*A$ is Hermitian positive semidefinite] Define the matrix $B\in\mathbb F^{n\times n}$ by \begin{align*} B=A^*A. \end{align*} Then \begin{align*} B^*=(A^*A)^*=A^*A=B, \end{align*} so $B$ is Hermitian. Moreover, for every $x\in\mathbb F^n$, \begin{align*} x^*Bx=x^*A^*Ax=\|Ax\|_{\mathbb F^m}^2\ge 0. \end{align*} Thus $B$ is Hermitian positive semidefinite. [/step] [step:Maximize the Rayleigh quotient of $A^*A$] We use the finite-dimensional [spectral theorem for Hermitian matrices](/theorems/925) (citing a result not yet in the wiki: Spectral theorem for Hermitian matrices). Since $B=A^*A$ is Hermitian, there exist an [orthonormal basis](/page/Orthonormal%20Basis) $v_1,\dots,v_n$ of $\mathbb F^n$ and real eigenvalues $\lambda_1,\dots,\lambda_n\in\mathbb R$ such that \begin{align*} Bv_i=\lambda_i v_i \end{align*} for every $i\in\{1,\dots,n\}$. Since $B$ is positive semidefinite, each $\lambda_i\ge 0$. Relabel the eigenvalues so that \begin{align*} \lambda_1\le \lambda_2\le \dots \le \lambda_n=\lambda_{\max}(B). \end{align*} Let $x\in\mathbb F^n$ satisfy $\|x\|_{\mathbb F^n}=1$. Write \begin{align*} x=\sum_{i=1}^n c_i v_i \end{align*} with coefficients $c_i\in\mathbb F$. Orthonormality gives \begin{align*} \sum_{i=1}^n |c_i|^2=1. \end{align*} Using the eigenvector relations and orthonormality, \begin{align*} x^*Bx=\sum_{i=1}^n \lambda_i |c_i|^2. \end{align*} Since $\lambda_i\le \lambda_{\max}(B)$ for every $i$, \begin{align*} x^*Bx\le \lambda_{\max}(B)\sum_{i=1}^n |c_i|^2=\lambda_{\max}(B). \end{align*} Taking $x=v_n$ gives \begin{align*} v_n^*Bv_n=\lambda_{\max}(B). \end{align*} Therefore \begin{align*} \sup\{x^*Bx:x\in\mathbb F^n,\ \|x\|_{\mathbb F^n}=1\}=\lambda_{\max}(B). \end{align*} [/step] [step:Take square roots to obtain the operator norm formula] Combining the first step with the Rayleigh quotient computation for $B=A^*A$, we obtain \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}^2=\lambda_{\max}(A^*A). \end{align*} The left-hand side is nonnegative, and $\lambda_{\max}(A^*A)\ge 0$ because $A^*A$ is positive semidefinite. Taking the nonnegative square root of both sides gives \begin{align*} \|A\|_{\mathcal L(\mathbb F^n,\mathbb F^m)}=\sqrt{\lambda_{\max}(A^*A)}. \end{align*} This is the desired identity. [/step]