[proofplan] We compute directly from the standard matrix of the rank-one map. Its $(i,j)$-entry is $u_i v_j$, so the square of the Frobenius norm is the finite double sum of $(u_i v_j)^2$. This double sum factors as the product of the squared Euclidean norms of $u$ and $v$, and taking non-negative square roots gives the desired identity. [/proofplan]

[step:Write the standard matrix of the rank-one map] Let $A\in\mathbb{R}^{n\times m}$ be the standard matrix of the [linear map](/page/Linear%20Map) $u\otimes v:\mathbb{R}^m\to\mathbb{R}^n$. For $x=(x_1,\ldots,x_m)\in\mathbb{R}^m$, the definition of the Euclidean [inner product](/page/Inner%20Product) gives \begin{align*} (u\otimes v)(x)=(x\cdot v)u=\left(\sum_{j=1}^m x_j v_j\right)u. \end{align*} Therefore the $i$-th component of $(u\otimes v)(x)$ is \begin{align*} \bigl((u\otimes v)(x)\bigr)_i=\sum_{j=1}^m u_i v_j x_j. \end{align*} By the definition of the standard matrix of a linear map, this means \begin{align*} A_{ij}=u_i v_j \end{align*} for every $i\in\{1,\ldots,n\}$ and every $j\in\{1,\ldots,m\}$. [/step]

[step:Factor the Frobenius norm sum into two Euclidean norm sums]By the definition of the Frobenius norm of the matrix $A$, we have \begin{align*} \|u\otimes v\|_F^2=\sum_{i=1}^n\sum_{j=1}^m A_{ij}^2. \end{align*} Substituting $A_{ij}=u_i v_j$ gives \begin{align*} \|u\otimes v\|_F^2=\sum_{i=1}^n\sum_{j=1}^m u_i^2 v_j^2. \end{align*} Since both sums are finite, distributivity of multiplication over addition gives \begin{align*} \sum_{i=1}^n\sum_{j=1}^m u_i^2 v_j^2=\left(\sum_{i=1}^n u_i^2\right)\left(\sum_{j=1}^m v_j^2\right). \end{align*} Using the definition of the Euclidean norm on $\mathbb{R}^n$ and $\mathbb{R}^m$, this becomes \begin{align*} \|u\otimes v\|_F^2=|u|^2 |v|^2. \end{align*}[/step]

[guided]The goal is to compare two norms: the Frobenius norm of the rank-one map and the Euclidean norms of the vectors that define it. The Frobenius norm is defined by summing the squares of all matrix entries, so the relevant object is the standard matrix $A\in\mathbb{R}^{n\times m}$ of $u\otimes v$. From the previous step, each entry of $A$ is \begin{align*} A_{ij}=u_i v_j. \end{align*} Therefore the square of the Frobenius norm is \begin{align*} \|u\otimes v\|_F^2=\sum_{i=1}^n\sum_{j=1}^m A_{ij}^2. \end{align*} Substituting the entry formula gives \begin{align*} \|u\otimes v\|_F^2=\sum_{i=1}^n\sum_{j=1}^m (u_i v_j)^2. \end{align*} Since $u_i$ and $v_j$ are [real numbers](/page/Real%20Numbers), $(u_i v_j)^2=u_i^2 v_j^2$, so \begin{align*} \|u\otimes v\|_F^2=\sum_{i=1}^n\sum_{j=1}^m u_i^2 v_j^2. \end{align*} The key point is that the $i$-dependent factor and the $j$-dependent factor separate. Because the sums are finite, we may factor by the ordinary distributive law: \begin{align*} \sum_{i=1}^n\sum_{j=1}^m u_i^2 v_j^2=\left(\sum_{i=1}^n u_i^2\right)\left(\sum_{j=1}^m v_j^2\right). \end{align*} By definition of the Euclidean norm, \begin{align*} |u|^2=\sum_{i=1}^n u_i^2. \end{align*} Similarly, \begin{align*} |v|^2=\sum_{j=1}^m v_j^2. \end{align*} Substituting these two identities yields \begin{align*} \|u\otimes v\|_F^2=|u|^2 |v|^2. \end{align*}[/guided]

[step:Take the non-negative square root] The Frobenius norm and the Euclidean norms are non-negative by definition. Hence taking the non-negative square root of \begin{align*} \|u\otimes v\|_F^2=|u|^2 |v|^2 \end{align*} gives \begin{align*} \|u\otimes v\|_F=|u|\,|v|. \end{align*} This is the desired identity. [/step]