[proofplan] We compare the squares of the two Frobenius norms, since Frobenius norms are nonnegative. The [trace formula for the Frobenius norm](/theorems/9174) rewrites each square as the trace of $M^\top M$. Expanding $(QAP)^\top(QAP)$ uses the product rule for transposes and the orthogonality relations $Q^\top Q=I_n$ and $PP^\top=I_m$; the remaining trace identity is proved directly from matrix entries. [/proofplan]

[step:Rewrite both Frobenius norms by the trace formula] For a matrix $M \in \mathbb{R}^{n \times m}$, let $L_M: \mathbb{R}^m \to \mathbb{R}^n$ denote the [linear map](/page/Linear%20Map) \begin{align*} L_M(x)=Mx. \end{align*} By the trace formula for the Frobenius norm [citetheorem:9174] applied to $L_M$, whose standard matrix is $M$, we have \begin{align*} \|M\|_F^2=\operatorname{tr}(M^\top M). \end{align*} Applying this with $M=QAP$ and with $M=A$ gives \begin{align*} \|QAP\|_F^2=\operatorname{tr}((QAP)^\top(QAP)). \end{align*} and \begin{align*} \|A\|_F^2=\operatorname{tr}(A^\top A). \end{align*} [/step]

[step:Use orthogonality and trace cyclicity to identify the two traces]The transpose product rule gives \begin{align*} (QAP)^\top=P^\top A^\top Q^\top. \end{align*} Since $Q$ is orthogonal, $Q^\top Q=I_n$. Therefore \begin{align*} (QAP)^\top(QAP)=P^\top A^\top Q^\top QAP=P^\top A^\top A P. \end{align*} Define $B \in \mathbb{R}^{m \times m}$ by $B=A^\top A$. We prove directly that \begin{align*} \operatorname{tr}(P^\top B P)=\operatorname{tr}(BPP^\top). \end{align*} Indeed, using the definition of matrix multiplication and trace, \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^m P_{ji}B_{jk}P_{ki}. \end{align*} Reordering the finite sums gives \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{j=1}^m \sum_{k=1}^m B_{jk}\sum_{i=1}^m P_{ki}P_{ji}. \end{align*} The inner sum is the $(k,j)$-entry of $PP^\top$, so this equals \begin{align*} \sum_{j=1}^m \sum_{k=1}^m B_{jk}(PP^\top)_{kj}=\operatorname{tr}(BPP^\top). \end{align*} Since $P$ is orthogonal, $PP^\top=I_m$. Hence \begin{align*} \operatorname{tr}(P^\top A^\top A P)=\operatorname{tr}(A^\top A I_m)=\operatorname{tr}(A^\top A). \end{align*} Combining this with the previous computation yields \begin{align*} \operatorname{tr}((QAP)^\top(QAP))=\operatorname{tr}(A^\top A). \end{align*}[/step]

[guided]We want to show that multiplying $A$ on the left by $Q$ and on the right by $P$ does not change the trace expression for the Frobenius norm. The first task is to expand the square of the transformed matrix. The transpose product rule reverses the order of the factors, so \begin{align*} (QAP)^\top=P^\top A^\top Q^\top. \end{align*} Multiplying this by $QAP$ gives \begin{align*} (QAP)^\top(QAP)=P^\top A^\top Q^\top QAP. \end{align*} The hypothesis that $Q$ is orthogonal means $Q^\top Q=I_n$, where $I_n$ is the identity matrix in $\mathbb{R}^{n \times n}$. Substituting this relation removes the left orthogonal factor: \begin{align*} (QAP)^\top(QAP)=P^\top A^\top A P. \end{align*} It remains to show that the right orthogonal factor also disappears after taking trace. Define $B \in \mathbb{R}^{m \times m}$ by $B=A^\top A$. We compute the trace of $P^\top B P$ from entries. Since $P^\top B P$ is an $m \times m$ matrix, \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{i=1}^m (P^\top B P)_{ii}. \end{align*} Expanding the $(i,i)$-entry by matrix multiplication gives \begin{align*} (P^\top B P)_{ii}=\sum_{j=1}^m \sum_{k=1}^m (P^\top)_{ij}B_{jk}P_{ki}. \end{align*} Since $(P^\top)_{ij}=P_{ji}$, this becomes \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{i=1}^m \sum_{j=1}^m \sum_{k=1}^m P_{ji}B_{jk}P_{ki}. \end{align*} All sums are finite, so we may reorder them: \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{j=1}^m \sum_{k=1}^m B_{jk}\sum_{i=1}^m P_{ki}P_{ji}. \end{align*} The inner sum is exactly the $(k,j)$-entry of $PP^\top$: \begin{align*} (PP^\top)_{kj}=\sum_{i=1}^m P_{ki}(P^\top)_{ij}=\sum_{i=1}^m P_{ki}P_{ji}. \end{align*} Therefore \begin{align*} \operatorname{tr}(P^\top B P)=\sum_{j=1}^m \sum_{k=1}^m B_{jk}(PP^\top)_{kj}. \end{align*} The right-hand side is the trace of $BPP^\top$, because \begin{align*} \operatorname{tr}(BPP^\top)=\sum_{j=1}^m (BPP^\top)_{jj}=\sum_{j=1}^m \sum_{k=1}^m B_{jk}(PP^\top)_{kj}. \end{align*} Now use the orthogonality of $P$, namely $PP^\top=I_m$. Hence \begin{align*} \operatorname{tr}(P^\top A^\top A P)=\operatorname{tr}(A^\top A I_m)=\operatorname{tr}(A^\top A). \end{align*} Together with the earlier expansion, this proves \begin{align*} \operatorname{tr}((QAP)^\top(QAP))=\operatorname{tr}(A^\top A). \end{align*}[/guided]

[step:Take square roots to recover equality of Frobenius norms] From the trace identities in the previous steps, \begin{align*} \|QAP\|_F^2=\operatorname{tr}((QAP)^\top(QAP))=\operatorname{tr}(A^\top A)=\|A\|_F^2. \end{align*} Both $\|QAP\|_F$ and $\|A\|_F$ lie in $[0,\infty)$ by the definition of the Frobenius norm. Since the square function is injective on $[0,\infty)$, it follows that \begin{align*} \|QAP\|_F=\|A\|_F. \end{align*} This proves the claimed orthogonal invariance of the Frobenius norm. [/step]