[proofplan] We compare both norms directly from their definitions. The lower bound follows by estimating $|Ax|$ for an arbitrary Euclidean unit vector $x\in k^n$ using the finite [Cauchy-Schwarz inequality](/theorems/432) on each row of $A$, then taking the supremum over all such $x$. The upper bound follows by writing the Frobenius norm as the sum of the squared Euclidean norms of the columns $Ae_j$, where $e_1,\dots,e_n$ is the standard basis of $k^n$, and bounding each column norm by the operator norm. [/proofplan] [step:Estimate the image of an arbitrary unit vector by the Frobenius norm] Let $A=(A_{ij})\in M_{m\times n}(k)$. Let $x=(x_1,\dots,x_n)\in k^n$ satisfy $|x|=1$. For each $i\in\{1,\dots,m\}$, the $i$-th coordinate of $Ax$ is \begin{align*} (Ax)_i=\sum_{j=1}^{n}A_{ij}x_j. \end{align*} Applying the finite Cauchy-Schwarz inequality in $k^n$ to the vectors $(A_{i1},\dots,A_{in})$ and $(x_1,\dots,x_n)$ gives \begin{align*} |(Ax)_i|^2\leq \left(\sum_{j=1}^{n}|A_{ij}|^2\right)\left(\sum_{j=1}^{n}|x_j|^2\right). \end{align*} Since $|x|=1$, this becomes \begin{align*} |(Ax)_i|^2\leq \sum_{j=1}^{n}|A_{ij}|^2. \end{align*} Summing over $i\in\{1,\dots,m\}$ yields \begin{align*} |Ax|^2=\sum_{i=1}^{m}|(Ax)_i|^2\leq \sum_{i=1}^{m}\sum_{j=1}^{n}|A_{ij}|^2=\|A\|_F^2. \end{align*} [guided] Fix $A=(A_{ij})\in M_{m\times n}(k)$ and choose an arbitrary vector $x=(x_1,\dots,x_n)\in k^n$ with $|x|=1$. We want to bound the quantity appearing in the operator norm, namely $|Ax|$, in terms of the Frobenius norm of $A$. For each row index $i\in\{1,\dots,m\}$, the $i$-th coordinate of the product $Ax$ is \begin{align*} (Ax)_i=\sum_{j=1}^{n}A_{ij}x_j. \end{align*} This is the [inner product](/page/Inner%20Product)-type pairing of the $i$-th row of $A$ with the coordinate vector $x$. The finite Cauchy-Schwarz inequality in $k^n$ therefore gives \begin{align*} |(Ax)_i|^2\leq \left(\sum_{j=1}^{n}|A_{ij}|^2\right)\left(\sum_{j=1}^{n}|x_j|^2\right). \end{align*} The hypothesis $|x|=1$ means exactly that \begin{align*} \sum_{j=1}^{n}|x_j|^2=1. \end{align*} Substituting this into the previous estimate gives, for every $i\in\{1,\dots,m\}$, \begin{align*} |(Ax)_i|^2\leq \sum_{j=1}^{n}|A_{ij}|^2. \end{align*} Now we pass from coordinate estimates to the Euclidean norm in $k^m$. By definition, \begin{align*} |Ax|^2=\sum_{i=1}^{m}|(Ax)_i|^2. \end{align*} Using the coordinate estimate for each $i$ and summing gives \begin{align*} |Ax|^2\leq \sum_{i=1}^{m}\sum_{j=1}^{n}|A_{ij}|^2. \end{align*} The double sum on the right is precisely $\|A\|_F^2$. Hence \begin{align*} |Ax|^2\leq \|A\|_F^2. \end{align*} This proves the desired bound for every Euclidean unit vector $x\in k^n$. [/guided] [/step] [step:Take the supremum over unit vectors to obtain the lower bound] By definition of the Euclidean induced operator norm, \begin{align*} \|A\|_{\mathrm{op}}=\sup\{|Ax|:x\in k^n,\ |x|=1\}. \end{align*} The previous step proves $|Ax|\leq \|A\|_F$ for every $x\in k^n$ with $|x|=1$. Taking the supremum over all such $x$ gives \begin{align*} \|A\|_{\mathrm{op}}\leq \|A\|_F. \end{align*} [/step] [step:Rewrite the Frobenius norm as the sum of squared column norms] For each $j\in\{1,\dots,n\}$, let $e_j\in k^n$ denote the $j$-th standard basis vector, whose coordinates are $(e_j)_j=1$ and $(e_j)_\ell=0$ for $\ell\neq j$. Then $Ae_j\in k^m$ is the $j$-th column of $A$, so its $i$-th coordinate is $(Ae_j)_i=A_{ij}$. Therefore \begin{align*} |Ae_j|^2=\sum_{i=1}^{m}|A_{ij}|^2. \end{align*} Summing over $j\in\{1,\dots,n\}$ gives \begin{align*} \|A\|_F^2=\sum_{i=1}^{m}\sum_{j=1}^{n}|A_{ij}|^2=\sum_{j=1}^{n}|Ae_j|^2. \end{align*} [/step] [step:Bound every column by the operator norm and sum] For each $j\in\{1,\dots,n\}$, the standard basis vector $e_j$ satisfies $|e_j|=1$. Hence, by the definition of $\|A\|_{\mathrm{op}}$, \begin{align*} |Ae_j|\leq \|A\|_{\mathrm{op}}. \end{align*} Using the column formula from the previous step, we obtain \begin{align*} \|A\|_F^2=\sum_{j=1}^{n}|Ae_j|^2\leq \sum_{j=1}^{n}\|A\|_{\mathrm{op}}^2=n\|A\|_{\mathrm{op}}^2. \end{align*} Both $\|A\|_F$ and $\|A\|_{\mathrm{op}}$ are non-negative [real numbers](/page/Real%20Numbers), so taking square roots gives \begin{align*} \|A\|_F\leq \sqrt{n}\,\|A\|_{\mathrm{op}}. \end{align*} Together with the lower bound already proved, this establishes \begin{align*} \|A\|_{\mathrm{op}}\leq \|A\|_F\leq \sqrt{n}\,\|A\|_{\mathrm{op}}. \end{align*} [/step]