[proofplan]
We verify the four norm axioms directly from the coordinate definition. The standard matrix identifies a [linear map](/page/Linear%20Map) $T:\mathbb{R}^m\to\mathbb{R}^n$ with the finite array of its entries, and the Frobenius norm is exactly the Euclidean length of that array. Nonnegativity, definiteness, and homogeneity follow entrywise; the triangle inequality follows from the finite [Cauchy-Schwarz inequality](/theorems/432) applied to the two arrays of matrix entries.
[/proofplan]
[step:Record how standard matrices behave under linear operations]
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 $\mathbb{R}^n$. For $T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, the entries of $A(T)$ are determined by
\begin{align*}
T(e_j)=\sum_{i=1}^{n} A(T)_{ij}\varepsilon_i
\end{align*}
for each $j\in\{1,\ldots,m\}$.
Let $S,T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, and let $c\in\mathbb{R}$. For each $j\in\{1,\ldots,m\}$,
\begin{align*}
(S+T)(e_j)=S(e_j)+T(e_j)=\sum_{i=1}^{n}\bigl(A(S)_{ij}+A(T)_{ij}\bigr)\varepsilon_i.
\end{align*}
By uniqueness of coordinates in the basis $\varepsilon_1,\ldots,\varepsilon_n$,
\begin{align*}
A(S+T)_{ij}=A(S)_{ij}+A(T)_{ij}
\end{align*}
for every $i\in\{1,\ldots,n\}$ and $j\in\{1,\ldots,m\}$. Similarly,
\begin{align*}
(cT)(e_j)=cT(e_j)=\sum_{i=1}^{n} cA(T)_{ij}\varepsilon_i,
\end{align*}
so
\begin{align*}
A(cT)_{ij}=cA(T)_{ij}
\end{align*}
for every $i$ and $j$.
[/step]
[step:Verify nonnegativity and definiteness from the entries]
For every $T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, each number $A(T)_{ij}^2$ is nonnegative. Hence the finite sum $\sum_{i=1}^{n}\sum_{j=1}^{m}A(T)_{ij}^2$ is nonnegative, and therefore
\begin{align*}
\|T\|_F\ge 0.
\end{align*}
If $T=0$, then $T(e_j)=0$ for every $j$, so $A(T)_{ij}=0$ for every $i$ and $j$. Thus $\|T\|_F=0$.
Conversely, suppose $\|T\|_F=0$. Then
\begin{align*}
\sum_{i=1}^{n}\sum_{j=1}^{m} A(T)_{ij}^2=0.
\end{align*}
Every summand is nonnegative, so each summand must be zero. Hence $A(T)_{ij}=0$ for all $i,j$. Therefore $T(e_j)=0$ for every standard basis vector $e_j$. Since every $x\in\mathbb{R}^m$ has a unique representation $x=\sum_{j=1}^{m}x_j e_j$ with $x_j\in\mathbb{R}$, linearity gives
\begin{align*}
T(x)=\sum_{j=1}^{m}x_jT(e_j)=0.
\end{align*}
Thus $T=0$. This proves definiteness.
[/step]
[step:Verify absolute homogeneity by scaling the standard matrix]
Let $T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$ and let $c\in\mathbb{R}$. From the scalar multiplication rule for standard matrices proved above,
\begin{align*}
\|cT\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m} A(cT)_{ij}^2=\sum_{i=1}^{n}\sum_{j=1}^{m} \bigl(cA(T)_{ij}\bigr)^2.
\end{align*}
Factoring out $c^2$ from the finite sum gives
\begin{align*}
\|cT\|_F^2=c^2\sum_{i=1}^{n}\sum_{j=1}^{m} A(T)_{ij}^2=c^2\|T\|_F^2.
\end{align*}
Both $\|cT\|_F$ and $|c|\|T\|_F$ are nonnegative [real numbers](/page/Real%20Numbers), so taking the nonnegative square root yields
\begin{align*}
\|cT\|_F=|c|\|T\|_F.
\end{align*}
[/step]
[step:Prove the triangle inequality for the array of matrix entries]
We first record the finite Cauchy-Schwarz estimate needed for the entries. If $(a_{ij})$ and $(b_{ij})$ are two real arrays indexed by $1\le i\le n$ and $1\le j\le m$, then
\begin{align*}
\left|\sum_{i=1}^{n}\sum_{j=1}^{m} a_{ij}b_{ij}\right|\le \left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2\right)^{1/2}\left(\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2\right)^{1/2}.
\end{align*}
Indeed, if all $b_{ij}$ vanish, the inequality is immediate. Otherwise define
\begin{align*}
\lambda:=\frac{\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}}{\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2}.
\end{align*}
The nonnegativity of the finite sum of squares gives
\begin{align*}
0\le \sum_{i=1}^{n}\sum_{j=1}^{m}(a_{ij}-\lambda b_{ij})^2.
\end{align*}
Expanding the square and using the definition of $\lambda$ gives
\begin{align*}
0\le \sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2-\frac{\left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}\right)^2}{\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2}.
\end{align*}
Multiplying by the positive denominator and taking square roots proves the estimate.
Now let $S,T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$. Define real arrays $a_{ij}:=A(S)_{ij}$ and $b_{ij}:=A(T)_{ij}$. Since $A(S+T)_{ij}=a_{ij}+b_{ij}$, we compute
\begin{align*}
\|S+T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}(a_{ij}+b_{ij})^2.
\end{align*}
Expanding the square gives
\begin{align*}
\|S+T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2+2\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}+\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2.
\end{align*}
Using the finite Cauchy-Schwarz estimate on the middle sum,
\begin{align*}
\|S+T\|_F^2\le \|S\|_F^2+2\|S\|_F\|T\|_F+\|T\|_F^2.
\end{align*}
The right-hand side is $(\|S\|_F+\|T\|_F)^2$. Since both sides are nonnegative, taking nonnegative square roots gives
\begin{align*}
\|S+T\|_F\le \|S\|_F+\|T\|_F.
\end{align*}
[guided]
The only norm axiom not completely entrywise is the triangle inequality, because it compares the length of the sum of two arrays with the sum of their lengths. We first prove the finite Cauchy-Schwarz inequality for real arrays because that is the precise estimate controlling the cross term.
Let $(a_{ij})$ and $(b_{ij})$ be two real arrays indexed by $1\le i\le n$ and $1\le j\le m$. If $b_{ij}=0$ for every $i,j$, then $\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}=0$, so the desired inequality holds. Suppose instead that at least one $b_{ij}$ is nonzero. Then
\begin{align*}
\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2>0.
\end{align*}
Define the real number
\begin{align*}
\lambda:=\frac{\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}}{\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2}.
\end{align*}
This choice of $\lambda$ is the projection coefficient of the array $a$ onto the array $b$; algebraically, it is chosen so that the quadratic expansion below has the strongest possible consequence. Since every square is nonnegative,
\begin{align*}
0\le \sum_{i=1}^{n}\sum_{j=1}^{m}(a_{ij}-\lambda b_{ij})^2.
\end{align*}
Expanding the finite sum gives
\begin{align*}
0\le \sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2-2\lambda\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}+\lambda^2\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2.
\end{align*}
Substituting the definition of $\lambda$ into the last two terms gives
\begin{align*}
0\le \sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2-\frac{\left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}\right)^2}{\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2}.
\end{align*}
Multiplying by the positive number $\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2$ yields
\begin{align*}
\left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}\right)^2\le \left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2\right)\left(\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2\right).
\end{align*}
Taking nonnegative square roots proves
\begin{align*}
\left|\sum_{i=1}^{n}\sum_{j=1}^{m} a_{ij}b_{ij}\right|\le \left(\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2\right)^{1/2}\left(\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2\right)^{1/2}.
\end{align*}
Now apply this estimate to the standard matrices of two linear maps. Let $S,T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$, and define $a_{ij}:=A(S)_{ij}$ and $b_{ij}:=A(T)_{ij}$. The first step of the proof showed that the standard matrix of $S+T$ is obtained by entrywise addition, so
\begin{align*}
A(S+T)_{ij}=a_{ij}+b_{ij}.
\end{align*}
Therefore
\begin{align*}
\|S+T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}(a_{ij}+b_{ij})^2.
\end{align*}
Expanding the square separates the two lengths and the cross term:
\begin{align*}
\|S+T\|_F^2=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}^2+2\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}+\sum_{i=1}^{n}\sum_{j=1}^{m}b_{ij}^2.
\end{align*}
The first sum is $\|S\|_F^2$, and the third sum is $\|T\|_F^2$. For the middle sum, Cauchy-Schwarz gives
\begin{align*}
\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}\le \left|\sum_{i=1}^{n}\sum_{j=1}^{m}a_{ij}b_{ij}\right|\le \|S\|_F\|T\|_F.
\end{align*}
Hence
\begin{align*}
\|S+T\|_F^2\le \|S\|_F^2+2\|S\|_F\|T\|_F+\|T\|_F^2=(\|S\|_F+\|T\|_F)^2.
\end{align*}
Both $\|S+T\|_F$ and $\|S\|_F+\|T\|_F$ are nonnegative, so taking square roots preserves the inequality:
\begin{align*}
\|S+T\|_F\le \|S\|_F+\|T\|_F.
\end{align*}
[/guided]
[/step]
[step:Conclude that the Frobenius assignment satisfies the norm axioms]
We have shown that $\|T\|_F\ge 0$ for every $T$, that $\|T\|_F=0$ holds exactly when $T=0$, that $\|cT\|_F=|c|\|T\|_F$ for every $c\in\mathbb{R}$, and that $\|S+T\|_F\le \|S\|_F+\|T\|_F$ for every $S,T\in\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$. These are precisely the norm axioms on the real [vector space](/page/Vector%20Space) $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$. Therefore $\|\cdot\|_F$ is a norm on $\mathcal{L}(\mathbb{R}^m,\mathbb{R}^n)$.
[/step]
