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For a real symmetric positive-definite matrix $A$, the energy norm is $\Vert x{\Vert }_A=(x^{T}Ax{)}^{1/2}=\Vert A^{1/2}x{\Vert }_2$. In the defining supremum of the induced matrix norm, substitute $y=A^{1/2}x$. The matrix square root is an invertible matrix, so this ranges over every nonzero vector and proves the displayed identity. This matrix similarity is useful because contraction in the energy norm becomes an ordinary Euclidean operator norm estimate after a change of coordinates.

A real square orthogonal matrix satisfies $Q^{T}Q=Q{Q}^T=I$, so it preserves the Euclidean norm. The columns of $Z_h$ are orthonormal, hence $\Vert Z_{h}v{\Vert }_2=\Vert v{\Vert }_2$. The induced matrix norm therefore givesFor the left truncation, $\Vert Q_{\ell }^{T}A{\Vert }_2\le \Vert A{\Vert }_2$, and equality is guaranteed for arbitrary $A$ only when $\boxed{\ell =m}$. Indeed, if $\ell <m$, take $A=q_{\ell +1}{e}_1^T$, where $q_{\ell +1}$ is an omitted column of $Q$. Then $\Vert A{\Vert }_2=1$ but $Q_{\ell }^{T}A=0$. Merely taking $\ell \ge n$ does not guarantee preservation for an arbitrary $A$.

A full QR decomposition of a real $m\times n$ matrix is $A=QR$, with $Q$ orthogonal and $R$ upper trapezoidal; for a square matrix, $R$ is upper triangular. Fix the following sign convention for a Givens rotation: it is the identity except for its $(i,k)$ blockThe block transpose times the block is $I_2$ because $c^2+s^2=1$, proving orthogonality. Writing $r=\sqrt{x_i^2+x_k^2}$, the choice $\boxed{c=x_i/r,s=x_k/r}$ gives $(Jx{)}_k=0$ and $(Jx{)}_i=r$. To zero the other coordinate take $\boxed{c=x_k/r,s=-x_i/r}$, giving $(Jx{)}_i=0$. If $r=0$, both entries are already zero and the identity suffices.

For a tridiagonal matrix, successively rotate adjacent row pairs $(1,2),(2,3),\dots ,(n-1,n)$, each time zeroing the current entry $(k+1,k)$. Earlier columns remain triangular because both affected rows are already zero there. The result $R=J_{n-1}\cdots J_{1}A$ is upper triangular, so $Q=J_1^T\cdots J_{n-1}^T$ gives its QR decomposition. Only $n-1$ rotations are required; the resulting $R$ has at most two superdiagonals.

For a full square matrix, process columns $j=1,\dots ,n-1$. Within column $j$, for $i=n,n-1,\dots ,j+1$, rotate rows $(i-1,i)$ to eliminate the current $(i,j)$ entry. Previously cleared columns stay clear. With $J_1,\dots ,J_s$ in application order, $s=n(n-1)/2$, one has $\boxed{R=J_s\cdots J_{1}A,Q=J_1^T\cdots J_s^T}$. This is QR decomposition by Givens rotations.

The vector Euclidean norm is $\boxed{\Vert x{\Vert }_2=(x^{T}x{)}^{1/2}=(\sum _i{x}_i^2{)}^{1/2}}$, the ordinary length from the origin. The induced matrix norm isGeometrically this is the largest stretch of a unit vector. A linear map sends the unit sphere to an ellipsoid, possibly degenerate; its largest semiaxis has length $\Vert A{\Vert }_2$. Equivalently the matrix 2-norm is the largest singular value.

The induced matrix norm associated with the given energy norm is a supremum of length ratios. The invertible change of variables $y=A^{1/2}x$ givesThe last expression is the Euclidean operator norm. This is the matrix norm induced by an energy norm: transforming to coordinates with Euclidean lengths transforms the operator by the corresponding matrix similarity.

For each row, the weighted Cauchy-Schwarz inequality gives $|(Bx{)}_i{|}^2\le ({\sum }_j|b_{ij}|)({\sum }_j|b_{ij}||x_j{|}^2)$. Summing over rows bounds the result by the maximum absolute row sum times the maximum absolute column sum times $\Vert x{\Vert }_2^2$. Taking square roots and the supremum proves the displayed induced matrix norm inequality. It gives a Euclidean estimate directly from entries without computing singular values.