[proofplan] We express the Frobenius norm through the Euclidean norms of the columns of the matrix. Orthogonality of $Q$ means $Q^\top Q=I_m$, so each diagonal entry of $Q^\top Q$ says that the corresponding column of $Q$ has Euclidean norm $1$. Summing these $m$ squared column norms gives $\|Q\|_F^2=m$, and non-negativity of the Frobenius norm gives the stated square-root formula. [/proofplan] [step:Write the Frobenius norm as a sum over the columns of $Q$] Let $e_1,\ldots,e_m$ denote the standard basis of $\mathbb{R}^m$. For each $j \in \{1,\ldots,m\}$, define the column vector $q_j \in \mathbb{R}^m$ by \begin{align*} q_j=Q(e_j). \end{align*} By the [column formula for the Frobenius norm](/theorems/9173), [citetheorem:9173] applied to the [linear map](/page/Linear%20Map) $Q: \mathbb{R}^m \to \mathbb{R}^m$ gives \begin{align*} \|Q\|_F^2=\sum_{j=1}^m |Q(e_j)|^2. \end{align*} Using the definition of $q_j$, this becomes \begin{align*} \|Q\|_F^2=\sum_{j=1}^m |q_j|^2. \end{align*} [/step] [step:Use orthogonality to compute the length of each column] Since $Q$ is orthogonal, it satisfies \begin{align*} Q^\top Q=I_m. \end{align*} For each $j \in \{1,\ldots,m\}$, the $j$-th diagonal entry of $Q^\top Q$ is the Euclidean [inner product](/page/Inner%20Product) of the $j$-th column of $Q$ with itself. Hence \begin{align*} |q_j|^2=q_j \cdot q_j=(Q^\top Q)_{jj}=(I_m)_{jj}=1. \end{align*} [guided] The purpose of orthogonality is to convert a matrix equation into a statement about column lengths. Let $e_1,\ldots,e_m$ denote the standard basis of $\mathbb{R}^m$. For the fixed index $j \in \{1,\ldots,m\}$, define the column vector $q_j \in \mathbb{R}^m$ by \begin{align*} q_j=Q(e_j). \end{align*} Thus $q_j$ is exactly the $j$-th column of the standard matrix $Q$. Because $Q$ is orthogonal, by definition \begin{align*} Q^\top Q=I_m. \end{align*} Here $I_m$ is the identity matrix from the theorem statement. The $j$-th diagonal entry of $Q^\top Q$ is obtained by taking the dot product of column $j$ of $Q$ with column $j$ of $Q$. Therefore \begin{align*} (Q^\top Q)_{jj}=q_j \cdot q_j. \end{align*} The Euclidean norm satisfies $|q_j|^2=q_j \cdot q_j$, while the identity matrix satisfies $(I_m)_{jj}=1$. Combining these equalities with $Q^\top Q=I_m$ gives \begin{align*} |q_j|^2=q_j \cdot q_j=(Q^\top Q)_{jj}=(I_m)_{jj}=1. \end{align*} Thus every column of an [orthogonal matrix](/page/Orthogonal%20Matrix) has Euclidean length $1$. [/guided] [/step] [step:Sum the column lengths and take the nonnegative square root] Substituting $|q_j|^2=1$ into the Frobenius norm formula gives \begin{align*} \|Q\|_F^2=\sum_{j=1}^m 1=m. \end{align*} The Frobenius norm is nonnegative, and $\sqrt{m}$ is the unique nonnegative real number whose square is $m$. Therefore \begin{align*} \|Q\|_F=\sqrt{m}. \end{align*} This proves the claim. [/step]