[proofplan] Write the standard matrix of $T$ as $A=(A_{ij})\in\mathbb{R}^{n\times m}$. The vector $T(e_j)$ is precisely the $j$-th column of $A$, so its Euclidean norm squared is the sum of the squares of the entries in that column. Summing these column norms over $j$ gives exactly the double sum defining $\|T\|_F^2$. [/proofplan] [step:Identify each image $T(e_j)$ with a column of the standard matrix] Let $A=(A_{ij})\in\mathbb{R}^{n\times m}$ be the standard matrix of $T$ with respect to the standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$, so that $T(x)=Ax$ for every $x\in\mathbb{R}^m$. For each $j\in\{1,\ldots,m\}$, the standard basis vector $e_j$ has $j$-th coordinate equal to $1$ and all other coordinates equal to $0$. Therefore the coordinates of $T(e_j)=Ae_j$ are exactly the entries in the $j$-th column of $A$: \begin{align*} T(e_j)=(A_{1j},A_{2j},\ldots,A_{nj})\in\mathbb{R}^n \end{align*} [guided] Let $A=(A_{ij})\in\mathbb{R}^{n\times m}$ denote the standard matrix of the [linear map](/page/Linear%20Map) $T:\mathbb{R}^m\to\mathbb{R}^n$. This means that, for every vector $x=(x_1,\ldots,x_m)\in\mathbb{R}^m$, the vector $T(x)\in\mathbb{R}^n$ is given by the matrix-vector product $Ax$. Fix an index $j\in\{1,\ldots,m\}$. The standard basis vector $e_j\in\mathbb{R}^m$ is the vector whose $j$-th coordinate is $1$ and whose other coordinates are $0$. Hence multiplying $A$ by $e_j$ selects the $j$-th column of $A$. Since $T(e_j)=Ae_j$, this gives \begin{align*} T(e_j)=(A_{1j},A_{2j},\ldots,A_{nj})\in\mathbb{R}^n \end{align*} This is the key point: the column vectors of the standard matrix are not auxiliary objects; they are exactly the images of the standard basis vectors under $T$. [/guided] [/step] [step:Compute the Euclidean norm of each column] By the definition of the Euclidean norm on $\mathbb{R}^n$, for every $j\in\{1,\ldots,m\}$ we have \begin{align*} |T(e_j)|^2=\sum_{i=1}^{n}A_{ij}^2 \end{align*} [/step] [step:Sum the column norms and recover the Frobenius norm] Summing the preceding identity over all $j\in\{1,\ldots,m\}$ gives \begin{align*} \sum_{j=1}^{m}|T(e_j)|^2=\sum_{j=1}^{m}\sum_{i=1}^{n}A_{ij}^2 \end{align*} Since the sums are finite, we may interchange their order: \begin{align*} \sum_{j=1}^{m}\sum_{i=1}^{n}A_{ij}^2=\sum_{i=1}^{n}\sum_{j=1}^{m}A_{ij}^2 \end{align*} By the definition of the Frobenius norm of $T$ from its standard matrix $A$, \begin{align*} \|T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}A_{ij}^2 \end{align*} Combining these equalities yields \begin{align*} \|T\|_F^2=\sum_{j=1}^{m}|T(e_j)|^2 \end{align*} as required. [/step]