[proofplan] We use the hypothesis $Q^\top Q = I_n$, where $I_n$ is the identity matrix on $\mathbb{R}^n$. For every vector $x \in \mathbb{R}^n$, this identity gives $|Qx|^2 = |x|^2$, so $Q$ preserves Euclidean lengths. The induced operator norm is the supremum of the ratios $|Qx|/|x|$ over nonzero vectors $x$, and every such ratio is therefore equal to $1$. [/proofplan] [step:Compute the Euclidean length of $Qx$ from the orthogonality identity] Let $x \in \mathbb{R}^n$. By hypothesis, \begin{align*}Q^\top Q = I_n,\end{align*} where $I_n \in \mathbb{R}^{n \times n}$ is the identity matrix. Using the Euclidean [inner product](/page/Inner%20Product) written in matrix form, we obtain \begin{align*}|Qx|^2 = (Qx)^\top(Qx).\end{align*} By associativity of matrix multiplication and the transpose identity $(Qx)^\top = x^\top Q^\top$, we have \begin{align*}(Qx)^\top(Qx) = x^\top Q^\top Qx.\end{align*} Substituting $Q^\top Q = I_n$ gives \begin{align*}x^\top Q^\top Qx = x^\top I_n x.\end{align*} Since $I_nx = x$, we have \begin{align*}x^\top I_n x = x^\top x = |x|^2.\end{align*} Thus $|Qx|^2 = |x|^2$. Because Euclidean norms are nonnegative, taking square roots gives \begin{align*}|Qx| = |x|.\end{align*} [guided] Let $x \in \mathbb{R}^n$ be arbitrary. The purpose of this step is to show that applying $Q$ does not change the Euclidean length of $x$. By hypothesis, we have the matrix identity \begin{align*}Q^\top Q = I_n,\end{align*} where $I_n \in \mathbb{R}^{n \times n}$ is the identity matrix. We compute the square of the Euclidean norm because it can be written directly using the transpose: \begin{align*}|Qx|^2 = (Qx)^\top(Qx).\end{align*} The transpose of the product $Qx$ is $x^\top Q^\top$, so associativity of matrix multiplication gives \begin{align*}(Qx)^\top(Qx) = x^\top Q^\top Qx.\end{align*} Now the orthogonality identity is exactly what is needed. Replacing $Q^\top Q$ by $I_n$ yields \begin{align*}x^\top Q^\top Qx = x^\top I_n x.\end{align*} Since $I_n$ is the identity matrix, $I_nx = x$, and therefore \begin{align*}x^\top I_n x = x^\top x = |x|^2.\end{align*} Combining these equalities gives $|Qx|^2 = |x|^2$. Both $|Qx|$ and $|x|$ are nonnegative [real numbers](/page/Real%20Numbers), so taking square roots preserves equality: \begin{align*}|Qx| = |x|.\end{align*} Thus $Q$ preserves the Euclidean norm of every vector in $\mathbb{R}^n$. [/guided] [/step] [step:Evaluate the supremum defining the induced operator norm] The induced Euclidean operator norm of the [linear map](/page/Linear%20Map) $Q: \mathbb{R}^n \to \mathbb{R}^n$ is \begin{align*} \|Q\|_{\mathrm{op}} = \sup\left\{\frac{|Qx|}{|x|} : x \in \mathbb{R}^n,\ x \neq 0\right\}. \end{align*} Let $x \in \mathbb{R}^n$ with $x \neq 0$. From the previous step, $|Qx| = |x|$, and since $|x| > 0$, division by $|x|$ gives \begin{align*} \frac{|Qx|}{|x|} = 1. \end{align*} Because $n \in \mathbb{N}$ and Androma's convention is that $\mathbb{N}$ starts at $1$, the vector $e_1 = (1,0,\dots,0) \in \mathbb{R}^n$ exists and is nonzero. Hence the set whose supremum defines $\|Q\|_{\mathrm{op}}$ is nonempty, and every element of it is equal to $1$. Therefore its supremum is $1$, so \begin{align*} \|Q\|_{\mathrm{op}} = 1. \end{align*} This proves the theorem. [/step]