[proofplan] Write the standard matrix of $T$ and expand both sides from the definitions. The Frobenius [inner product](/page/Inner%20Product) of $T$ with itself is the sum of products of corresponding entries, hence the sum of squares of all entries. The Frobenius norm is defined from the Euclidean norm of the standard matrix, so its square is the same sum. [/proofplan] [step:Represent $T$ by its standard matrix] Let $A \in \mathbb{R}^{n \times m}$ denote the standard matrix of the [linear map](/page/Linear%20Map) $T: \mathbb{R}^m \to \mathbb{R}^n$. For $1 \leq i \leq n$ and $1 \leq j \leq m$, let $A_{ij} \in \mathbb{R}$ denote the entry of $A$ in row $i$ and column $j$. By the definition of the Frobenius norm on $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, \begin{align*} \|T\|_F=\left(\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}^2\right)^{1/2}. \end{align*} [/step] [step:Expand the Frobenius inner product of $T$ with itself] By the definition of the Frobenius inner product on $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, \begin{align*} (T,T)_F=\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}A_{ij}. \end{align*} Since $A_{ij}A_{ij}=A_{ij}^2$ for every pair of indices $(i,j)$, it follows that \begin{align*} (T,T)_F=\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}^2. \end{align*} [guided] Let $A \in \mathbb{R}^{n \times m}$ denote the standard matrix of the linear map $T: \mathbb{R}^m \to \mathbb{R}^n$, and let $A_{ij} \in \mathbb{R}$ denote its entry in row $i$ and column $j$. The Frobenius inner product is defined by multiplying corresponding entries of the two standard matrices and summing over all rows and columns. In the present case both arguments are the same linear map $T$, so both standard matrices are the same matrix $A$. Therefore the definition gives \begin{align*} (T,T)_F=\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}A_{ij}. \end{align*} For each fixed row index $i \in \{1,\ldots,n\}$ and column index $j \in \{1,\ldots,m\}$, the scalar $A_{ij}$ is a real number. Hence its product with itself is its square: \begin{align*} A_{ij}A_{ij}=A_{ij}^2. \end{align*} Substituting this equality into every summand gives \begin{align*} (T,T)_F=\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}^2. \end{align*} This is the key point: the inner product of an object with itself becomes a sum of squares of its coordinates. [/guided] [/step] [step:Identify the common sum with $\|T\|_F^2$] From the definition of $\|T\|_F$ obtained above, \begin{align*} \|T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m} A_{ij}^2. \end{align*} The preceding step gives the same expression for $(T,T)_F$. Therefore \begin{align*} \|T\|_F^2=(T,T)_F. \end{align*} This proves the theorem. [/step]