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Use the distance from a point to a plane formula with the unit normal $\hat{\mathbf{n}}$. Since the plane passes through the origin, the distance isFor a direct check, the closest point on a plane is $\mathbf{p}-(\mathbf{p}\cdot \hat{\mathbf{n}})\hat{\mathbf{n}}=(1,\sqrt{2},-1{)}^T$, whose displacement from $\mathbf{p}$ has Euclidean norm $\sqrt{2}$.

The defining condition for an antisymmetric matrix forcesIts dot product with $(x,y{)}^T$ is zero, and its Euclidean norm is $|a|\sqrt{x^2+y^2}$. Thus it combines a right-angle rotation with a uniform dilation by $|a|$. More precisely, the rotation is clockwise for $a>0$ and anticlockwise for $a<0$. A direction-independent description is a clockwise right-angle rotation followed by multiplication by the signed scalar $a$.

A second application givesGeometrically this is a half-turn combined with a dilation by $a^2$. Because $A$ is nonzero, $a\ne 0$.

Contract with $v_j$ and use its squared Euclidean norm:Thus $v$ lies in the kernel of this orthogonal projection.

In Cartesian coordinates, $\nabla \Phi =({\Phi }_x,{\Phi }_y,{\Phi }_z)$. For orthogonal coordinates, let $\mathbf{r}(q_1,q_2,q_3)$ be the position, $h_i=|\partial \mathbf{r}/\partial q_i|$ the scale factors of orthogonal coordinates, and ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$ the corresponding unit vectors. Since $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$ and $d\Phi =\nabla \Phi \cdot d\mathbf{r}={\sum }_i{\Phi }_{q_i}d{q}_i$, comparison of independent increments gives the gradient in orthogonal coordinates:For the hyperbolic parametrization of oblate spheroidal coordinates, the three coordinate tangent vectors areTheir pairwise dot products are zero. Their squared Euclidean norms are ${\sinh }^{2}R+{\sin }^2\theta$, ${\sinh }^{2}R+{\sin }^2\theta$, and ${\cosh }^{2}R{\cos }^2\theta$, respectively. Thus the coordinate surfaces meet orthogonally wherever this is a regular coordinate chart, andThe last expression uses the usual latitude range $|\theta |<\pi /2$; without that range a scale factor is its absolute value. Axial or ring degeneracies are excluded from a regular chart.

A square complex matrix $U$ is a unitary matrix when $U^{†}U=I$, so it preserves the Euclidean norm. If $Uv=\mu v$ with $v\ne 0$, then $\Vert v\Vert =\Vert Uv\Vert =|\mu |\Vert v\Vert$, giving $|\mu |=1$. For a Hermitian matrix $A=A^{†}$ and $A{e}_j={\lambda }_j{e}_j$,because the numerator equals its complex conjugate. Hence no eigenvalue of $A+iI$ is zero, proving it is invertible.

For $V=(A-iI)(A+iI{)}^{-1}$, the Hermitian conjugate is $V^{†}=(A-iI{)}^{-1}(A+iI)$. The factors and their inverses commute, since they are rational functions of $A$. Thus $V^{†}V=I$. Its action on the same eigenvectors isThese eigenvalues have modulus one. This is the Cayley transform of a Hermitian matrix in the convention that avoids the unitary eigenvalue one.

For the inverse direction, $U^{†}=U^{-1}$ givesTaking the conjugate transpose of $B=i(I+U)(I-U{)}^{-1}$ therefore yieldsThe last equality uses commutativity of the two factors. This proves that the inverse Cayley transform of a unitary matrix without eigenvalue one is Hermitian. Solving for $U$ gives $U=(B-iI)(B+iI{)}^{-1}$. An orthonormal eigenbasis of $B$ therefore diagonalizes $U$, with diagonal entries $(b_j-i)/(b_j+i)$.

The result extends even when a unitary matrix has eigenvalue one: choose $|\alpha |=1$ so that $\alpha U$ does not. There are only finitely many forbidden phases, since the matrix has finitely many eigenvalues. Diagonalizing $\alpha U$ also diagonalizes $U$ in the same basis. Thus every unitary matrix is unitarily diagonalizable.

Translate the generalized coordinates so that the equilibrium is at zero. Its potential gradient vanishes, and its quadratic potential energy is $q^T\mathsf{V}q/2$, where $\mathsf{V}$ is the symmetric Hessian matrix. The Euler-Lagrange equations for small motion areA normal mode has fixed component ratios, $q(t)=v\cos (\omega t+\varphi )$, withThe angular frequency $\omega$ is its eigenfrequency. A zero mode instead admits a constant or linear time factor. For modes $v_{\alpha },v_{\beta }$, symmetry givesHence mass-matrix orthogonality of normal modes isWithin a degenerate eigenspace, a basis can be chosen orthogonal in this positive weighted inner product.

For three masses connected pairwise, the exact energies areA fixed common translation $x_i\mapsto x_i+a$ preserves all differences and velocities. A fixed rigid rotation $x_i\mapsto Q{x}_i$ preserves their Euclidean norms. Thus both energies have these symmetries. Here the transformation is fixed in time; a moving reference frame would change the kinetic energy. Every equilateral configuration of side $\sqrt{3}$ has zero spring extensions, hence zero force and minimum potential, proving that it is an equilibrium.

For the specified displacement coordinates, writeThe exact potential is $V=(k/2){\sum }_{ij=12,13,23}(d_{ij}-\sqrt{3}{)}^2$. For an equilibrium edge vector $d$ and small perturbation $\delta d$,Thus only each first-order extension is needed for the quadratic potential. These extensions, with an immaterial sign choice for the horizontal edge, areConsequently $V_2=(k/2)(s_1^2+s_2^2+s_3^2)$, with remainder $O(|q{|}^3)$. If $s=Bq$, then $\mathsf{V}=k{B}^{T}B$. Expanding the squares givesThe small-oscillation mass and stiffness matrices are $\mathsf{T}=m{I}_6$ and the displayed $\mathsf{V}$. Hence

The two common translations give zero-mode vectorsAn infinitesimal counterclockwise rotation sends each equilibrium vertex $r_i$ to $J{r}_i$, where $J(x,y)=(-y,x)$. Therefore a third zero-mode vector isEach vector makes all three $s_j$ vanish, so lies in the kernel of a linear map $B$ and therefore in the kernel of $\mathsf{V}$. The rows of $B$ are independent: the $q_4$ and $q_6$ entries first force the coefficients of its last two rows to vanish, then the first coefficient vanishes. Thus $\mathrm{rank}B=3$, and the rank-nullity theorem proves that these are exactly the three zero modes.

For radial motion takeso that $x_i(t)=(1+a(t))r_i$ to first order. The force on mass $i$ is $-ka{\sum }_{j\ne i}(r_i-r_j)=-3ka{r}_i$, since ${\sum }_i{r}_i=0$. Newton's equation therefore gives $m\ddot{a}=-3ka$. Equivalently $\mathsf{V}{v}_b=3k{v}_b$. Hence the breathing normal mode hasAll three masses move radially outward together, then inward together, preserving the triangular shape. The remaining two normal modes of an equilateral spring triangle are shear deformations with squared frequency $3k/(2m)$; together with the breathing mode and three rigid zero modes they account for all six degrees of freedom.

The columns of $R_z$ are an orthonormal basis, since ${\cos }^2\varphi +{\sin }^2\varphi =1$ and the first two columns have zero dot product. Thus $R_z^T{R}_z=I$: it is an orthogonal matrix. Consequentlyand hence $\boxed{|{\mathbf{v}}^{\prime }|=|\mathbf{v}|}$. This is the preservation of the Euclidean norm by a rotation matrix.

Expand the quadratic form of this Gram matrix:For $\mathbf{c}\ne 0$, linear independence of the real vectors implies ${\sum }_i{c}_i{\mathbf{x}}_i\ne 0$, so its squared Euclidean norm is strictly positive. Thusproving the positive definiteness of a Gram matrix in the independent-vector case.

Because $M$ commutes with $M^2$,Its squared Euclidean norm isA scalar equals its transpose, while $M^T=-M$, so $u_1^{T}M{u}_1=-u_1^{T}M{u}_1=0$. ConsequentlyThus each negative eigenvalue has at least this pair of orthonormal eigenvectors.

The quadratic form is ${\mathbf{x}}^{T}Q\mathbf{x}$ for the real symmetric matrixThe $y$ axis has eigenvalue $-3$. On the $xz$ plane, the characteristic polynomial isThus the orthogonal diagonalization from part (a) yieldsFor later use, one possible orthonormal basis isThen $u=(3x+z)/\sqrt{10}$, $v=y$, $w=(x-3z)/\sqrt{10}$. This explicit choice also confirms that the Euclidean norm is preserved: $x^2+y^2+z^2=u^2+v^2+w^2$.

Here uniform ellipticity means that there is a constant $\lambda >0$, independent of $x$, such thatThe assumption that the entries are essentially bounded functions also gives an upper bound $|a(x)\xi |\le \Lambda |\xi |$ with a finite constant $\Lambda$. For this symmetric divergence-form operator, the lower bound makes its energy control the squared Euclidean norm of the gradient.

The unit circle is the inverse image of the closed set $\{1\}$ under the continuous map $(x,y)\mapsto x^2+y^2$, so it is closed. It is also bounded, since every point has Euclidean norm one. The Heine-Borel theorem gives compactness.

A Givens rotation acts as the identity matrix outside two selected coordinates $i,j$, and has the following block on those coordinates:The convention with the opposite signs of the sines is equivalent after replacing $\theta$ by $-\theta$. A Householder reflection with nonzero normal vector $w\in {\mathbb{R}}^n$ isIt reverses the component parallel to $w$ and fixes the perpendicular hyperplane. Both are orthogonal matrices, so their transposes are their inverses and they preserve the Euclidean norm.

A real square matrix $Q$ is an orthogonal matrix when $Q^{T}Q=I$, equivalently its columns form an orthonormal basis. Here $Q^T$ is the transpose and $I$ the identity matrix. For every vector $x$,Taking nonnegative square roots provesThus an orthogonal matrix preserves the Euclidean norm.

Let $r=\sqrt{x_i^2+x_j^2}$. If $r>0$, choose $\theta$ so thatSuch an angle exists because the two ratios have squared sum one. Then the changed coordinates areAll other coordinates satisfy $y_k=x_k$. If $r=0$, both input coordinates are zero; take $\theta =0$, giving the identity matrix on this pair. In all cases,This fixes a nonnegative retained coordinate and handles division by zero explicitly. For floating-point computation, the stable construction of a Givens rotation uses scaled Euclidean norm evaluation to avoid unnecessarily overflowing or underflowing the squares.

Put $C=Q_{1}A$ and write its second column as $v$. For $n\ge 3$, takewhere each angle is computed from the current second column, in the right-to-left application order. These Givens rotations eliminate its entries in rows $3,\dots ,n$ by accumulating their Euclidean norm in row two.

Each rotation acts only on rows with indices at least two. The first column of $C$ is $\Vert a{\Vert }_2{e}_1$, whose entries in all those rows are zero. Mixing two zeros leaves them zero, and row one is untouched. Thus the first column is unchanged. The resulting matrix has the formThis verifies both required properties, including the case of zero second-column tail. When $n=2$ there are no rows to eliminate: take $Q_2=I_2$, the empty product. No sign change of $v_2$ is required in that case. Each Givens row-pair update preserves previously eliminated columns, which is the central invariant of QR decomposition by Givens rotations.

Since every $u_{jj}>0$, the diagonal matrix $D$ is invertible. SetThe division is by the diagonal entry of row $i$. Thus $\hat{U}$ is an upper triangular matrix with diagonal entries one and $A=LD\hat{U}$.

To use symmetry without assuming the conclusion, multiply by the matrix inverse of $L$ and its transpose:The matrix $M$ is symmetric because $A$ is a symmetric matrix. It is also an upper triangular matrix, since every factor on the right is an upper triangular matrix. A symmetric upper triangular matrix is diagonal. Its diagonal is exactly $D$, because $\hat{U}$ and $L^{-T}$ have diagonal entries one. Hence $M=D$, givingThis is the LDL decomposition.

Take the positive square roots of the diagonal entries:The matrix $\hat{L}$ is a lower triangular matrix with positive diagonal, andThis is a Cholesky decomposition. For a nonzero vector $x$, the vector ${\hat{L}}^{T}x$ is nonzero because ${\hat{L}}^T$ is invertible. ThereforeHere $\Vert \cdot {\Vert }_2$ is the Euclidean norm. Thus $A$ is a positive-definite matrix.

For a real square matrix, a QR factorization writes $B=QR$ with $Q$ an orthogonal matrix, $Q^{T}Q=I$, and $R$ an upper triangular matrix. If $B$ is invertible, so is $R$, andCompute $Q^{T}b$ and use back substitution; no matrix inverse needs to be formed.

Apply the Gram-Schmidt process to the columns $b_1,b_2,b_3$. The first column has Euclidean norm two, giving $q_1=(1,0,0{)}^T$ and $r_{11}=2$. For the second column,For the third column,This residual already has Euclidean norm one. Set $q_3=(0,-3/5,4/5{)}^T$ and $r_{33}=1$. ThusThe columns of $Q$ are orthonormal, and matrix multiplication verifies $QR=B$.

Using precisely this QR factorization, transform the right-hand side:Back substitution in $Rx=Q^{T}b$ givesTherefore

The Euclidean norm is invariant under an orthogonal matrix, so the full singular value decomposition givesThe second term is independent of $x$; the first attains its unique minimum zero because $\Sigma$ is an invertible matrix. The linear least squares solution is thereforeHere $A^{+}=V{\Sigma }^{-1}{U}^T$ is the Moore-Penrose inverse, and the fitted vector is the orthogonal projection $A{x}_{\ast }=U{U}^{T}b$ onto the image of a linear map defined by $A$.

With the singular value decomposition supplied, forming $U^{T}b$ costs $O(mn)$ arithmetic operations, diagonal scaling costs $O(n)$, and multiplication by $V$ costs $O(n^2)$. Since $m\ge n$,One need not form $U_{\perp }$, $A^{+}$ or $A^{T}A$ to calculate $x_{\ast }$. Computing the needed thin singular value decomposition first has standard dense cost $O(m{n}^2)$; explicitly forming an entire $m\times m$ orthogonal completion can require additional work when $m$ is much larger than $n$.

We use the spectral theorem to construct a sufficiently accurate approximate eigenpair. Complete $x$ to an orthonormal eigenbasis $x,x_2,\dots ,x_n$ of the symmetric matrix $B$, with corresponding eigenvalues $\lambda ,{\lambda }_2,\dots ,{\lambda }_n$. Simplicity of $\lambda$ means that every ${\lambda }_j-\lambda$ below is nonzero. PutThe orthogonal decomposition of $Ex$ shows that $x^{T}v=0$ andWith $u_{\epsilon }=x+\epsilon v$ and ${\mu }_{\epsilon }=\lambda +\epsilon a$, direct expansion cancels every first-order term:Also $\Vert u_{\epsilon }{\Vert }_2^2=1+{\epsilon }^2\Vert v{\Vert }_2^2\ge 1$. Thus its eigenpair residual has Euclidean norm $O({\epsilon }^2)$ in Big O notation.

For completeness, a real symmetric matrix $M$ always has an eigenvalue within $\Vert (M-\mu I)u{\Vert }_2/\Vert u{\Vert }_2$ of any real $\mu$ and nonzero $u$. Indeed, writing $u={\sum }_j{c}_j{q}_j$ in an orthonormal eigenbasis with eigenvalues ${\theta }_j$ givesApply this symmetric eigenvalue residual bound to $M=B+\epsilon E$. There is an eigenvalue ${\lambda }_{\epsilon }$ such thatwhich provesThis establishes the first-order perturbation of a simple symmetric eigenvalue without presupposing differentiability of the perturbed eigenvector. If $n=1$, the sum defining $v$ is empty and the displayed eigenvalue formula is exact.