[proofplan] We compare both sides by expanding them in the standard coordinates. The Frobenius norm of $T$ is the Euclidean norm of its standard matrix, so its square is the sum of the squares of all entries of $A$. On the other hand, the diagonal entries of $A^\top A$ are precisely the squared Euclidean lengths of the columns of $A$, and summing those diagonal entries gives the same double sum. [/proofplan]

[step:Expand the Frobenius norm through the entries of the standard matrix]Let $e_1,\ldots,e_m$ denote the standard basis of $\mathbb{R}^m$, and let $\varepsilon_1,\ldots,\varepsilon_n$ denote the standard basis of $\mathbb{R}^n$. Since $A=(A_{ij})$ is the standard matrix of $T$, for each $j \in \{1,\ldots,m\}$ the $j$-th column of $A$ is the coordinate vector of $T(e_j)$ in the basis $\varepsilon_1,\ldots,\varepsilon_n$; equivalently, \begin{align*} T(e_j)=\sum_{i=1}^n A_{ij}\varepsilon_i. \end{align*} By the definition of the Frobenius norm of a [linear map](/page/Linear%20Map) as the Euclidean norm of its standard matrix, \begin{align*} \|T\|_F^2=\sum_{j=1}^m\sum_{i=1}^n A_{ij}^2. \end{align*}[/step]

[guided]We first translate the linear map $T$ into coordinates, because the theorem compares a norm of $T$ with an expression built from the matrix $A$. Let $e_1,\ldots,e_m$ be the standard basis of $\mathbb{R}^m$, and let $\varepsilon_1,\ldots,\varepsilon_n$ be the standard basis of $\mathbb{R}^n$. The phrase "$A$ is the standard matrix of $T$" means exactly that the $j$-th column of $A$ records the standard coordinates of $T(e_j)$. Thus, for every $j \in \{1,\ldots,m\}$, \begin{align*} T(e_j)=\sum_{i=1}^n A_{ij}\varepsilon_i. \end{align*} The Frobenius norm of $T$ is defined by taking the Euclidean norm of this standard matrix. Therefore its square is the sum of the squares of all matrix entries: \begin{align*} \|T\|_F^2=\sum_{j=1}^m\sum_{i=1}^n A_{ij}^2. \end{align*} This is the quantity we must recover from the trace of $A^\top A$.[/guided]

[step:Compute the diagonal entries of $A^\top A$] For each $j \in \{1,\ldots,m\}$, the definition of matrix multiplication gives \begin{align*} (A^\top A)_{jj}=\sum_{i=1}^n (A^\top)_{ji}A_{ij}. \end{align*} By the definition of transpose, $(A^\top)_{ji}=A_{ij}$ for every $i \in \{1,\ldots,n\}$. Hence \begin{align*} (A^\top A)_{jj}=\sum_{i=1}^n A_{ij}^2. \end{align*} [/step]

[step:Sum the diagonal entries to identify the trace] By the definition of trace for an $m \times m$ matrix, \begin{align*} \operatorname{tr}(A^\top A)=\sum_{j=1}^m (A^\top A)_{jj}. \end{align*} Substituting the diagonal-entry computation from the previous step gives \begin{align*} \operatorname{tr}(A^\top A)=\sum_{j=1}^m\sum_{i=1}^n A_{ij}^2. \end{align*} The right-hand side is exactly the expansion of $\|T\|_F^2$ obtained above. Therefore \begin{align*} \|T\|_F^2=\operatorname{tr}(A^\top A). \end{align*} This proves the claimed trace formula. [/step]