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The Schmidt decomposition states that every finite-dimensional bipartite pure state haswhere $s_j>0$, ${\sum }_j{s}_j^2=1$, and the two displayed families are orthonormal. The integer $r$ is the Schmidt rank.

To prove it, choose product bases and writeApply the singular value decomposition $C=US{V}^{†}$. Absorbing the columns of $U$ and the complex conjugates of the columns of $V$ into new orthonormal bases gives the stated sum, with the nonzero singular values as the Schmidt coefficients. Equivalently, $s_j^2$ are the common nonzero eigenvalues of the two reduced density matrices, so $r=\mathrm{rank}{\rho }_A=\mathrm{rank}{\rho }_B$.

Let $({\sigma }_i,x_i,y_i)$ be the singular system of a compact operator. Since $P_n{x}_i=x_i$ for $i\le n$ and $P_n{x}_i=0$ for $i>n$, the truncated operator $T_n=A{P}_n$ has the finite singular value decompositionConsequentlyso $\boxed{T_n^{†}=A^{†}{Q}_n}$.

For completeness, $T_n^{†}{T}_n=P_n$ on $(\ker A{)}^{\perp }$ and $T_n{T}_n^{†}=Q_n$. These are self-adjoint orthogonal projectors, and henceThus all four Penrose equations hold, which verifies the formula independently of the singular expansion.

This statement is also always true. The same singular value decomposition gives

The statement is true without an additional rank assumption. If $A=U\Sigma V^H$ is a singular value decomposition, then

Choose an orthonormal eigenbasis $e_i$ of $G$ and set $u_i=e_i$, $v_i=A{e}_i/\sqrt{{\lambda }_i}$. Part (a) shows that the $v_i$ form an orthonormal eigenbasis of $H$. If $U$ and $V$ have these vectors as columns, then they are unitary andThis is the singular value decomposition.

Use singular value decomposition to choose unitary matrices satisfyingThe neutral currents remain flavour diagonal because the same unitary matrix occurs on both sides of each bilinear. The charged current contains the mismatchand becomes

A general unitary $3$ by $3$ matrix has nine real parameters. Independent rephasings of the six quark fields remove five phases because one common baryon-number phase changes nothing. The Cabibbo-Kobayashi-Maskawa matrix therefore has four physical parameters: three mixing angles and one CP-violating phase.

Because $A$ is a real symmetric positive-definite matrix, the finite-dimensional spectral theorem supplies an orthonormal eigenbasis $u_1,\dots ,u_n$ withIts eigendecomposition is also its singular value decomposition. If $e=y^{(\delta )}-y$, thensoThe operator norms satisfy $\Vert A{\Vert }_2={\lambda }_n$ and $\Vert A^{-1}{\Vert }_2=1/{\lambda }_1$. Therefore the worst-case relative perturbation bound isThe ratio ${\kappa }_2(A)$ is the spectral condition number of a positive-definite matrix. A large ratio means that data noise aligned with an eigenvector for the smallest eigenvalue is strongly amplified, so the inverse problem is ill conditioned.

Let $P_n:{\ell }^2(\mathbb{N})\to {\mathbb{C}}^n$ be coordinate projection and setThis finite matrix is computable from the matrix entries of $A$. Compute its singular value decompositionand define the unitary polar factor of a finite compressionThis is a unitary operator on ${\mathbb{C}}^n$, including when $T_n$ is singular.

Put $Q_n=P_n^{\ast }{P}_n$. Since $Q_n\to I$ strongly and $A$ is unitary,The continuous functional calculus for positive matrices therefore givesThe polar identity $T_n=A_n|T_n|$ now yieldsAlso $P_n^{\ast }{T}_n{P}_n=Q_{n}A{Q}_n\to A$ strongly, and henceIn particular the weak convergence required in part (c) holds. The construction uses only a finite block of the given matrix and a finite singular value decomposition, so it is an algorithm realizing all the assumptions of part (c).

Use a singular value decomposition $X=UD{V}^T$. ThenEach nonzero singular value $d$ contributes $d/(d^2+\lambda )\to d^{-1}$, while each zero one contributes zero. Thususing the Moore-Penrose inverse.

Using the signs chosen above, explicit multiplication givesThus the nonzero cases obey $A{\mathbf{v}}_m=\sqrt{{\sigma }_m}{\mathbf{u}}_m$. This relates the right singular vectors and left singular vectors in a singular value decomposition. For example, if $U=({\mathbf{u}}_1{\mathbf{u}}_2)$ and $V=({\mathbf{v}}_1{\mathbf{v}}_2{\mathbf{v}}_3)$, then $A=U\Sigma V^{\mathsf{T}}$ with $\Sigma =\left(\begin{array}{ccc}1& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right)$.

Throughout Question 2(a) and 2(b), write $U=[u_1,\dots ,u_n]$, $V=[v_1,\dots ,v_n]$, and $\Sigma =\mathrm{diag}({\sigma }_1,\dots ,{\sigma }_n)$. Then the supplied full singular value decomposition can also be written $A=U\Sigma V^T$, with $U^{T}U=V^{T}V=I_n$ and $[U,U_{\perp }]$ an $m\times m$ orthogonal matrix.

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 $n$.

For $m=n$, $U$ and $V$ are square orthogonal matrices. Multiply the linear system successively by $U^T$, ${\Sigma }^{-1}$ and $V$ to obtainWith the singular value decomposition already supplied, compute $c=U^{T}b$, divide $c_j$ by ${\sigma }_j$, then form $x=Vc$. The two dense matrix-vector products each cost $O(n^2)$ arithmetic operations in Big O notation, and the diagonal scaling costs $O(n)$. ThusIf the singular value decomposition first has to be computed, its usual dense cost is $O(n^3)$; that is a separate setup cost and can be reused for further right-hand sides.

Retain the first $r$ terms in the singular value decomposition:The Eckart-Young-Mirsky theorem says that this truncated singular value decomposition is a best low-rank approximation in the spectral norm. The remaining singular values of the difference are ${\sigma }_{r+1},\dots ,{\sigma }_n$, together with zeros, soFor $r\ge 1$ the displayed matrix has matrix rank exactly $r$, since every retained singular value is positive. For $r=0$, the only matrix of matrix rank zero is the zero matrix, and the same formula gives error ${\sigma }_1$.

The singular value decomposition $A=U\Sigma V^T$ givesTherefore $\boxed{{\sigma }_i(A)=\sqrt{{\lambda }_i(A^{T}A)}}$, counting multiplicities. These are the nonnegative square roots: zero singular values must be included if $A$ is singular. Strict positivity requires full rank, which the question does not assume.

Write the lower bidiagonal matrix entries as $B_{ii}=d_i$ and $B_{i+1,i}=e_i$. Its columns $i$ and $j$ have disjoint supports if $|i-j|>1$, so $T=B^{T}B$ has no entries outside the three central diagonals. More explicitly,This proves both symmetry and the tridiagonal matrix structure.

A full real singular value decomposition iswhere $\Sigma \in {\mathbb{R}}^{m\times n}$ has only diagonal entries, with ${\sigma }_1\ge \cdots \ge {\sigma }_{\min (m,n)}\ge 0$. These are the singular values. The number of positive singular values is the matrix rank. The columns of $V$ are right singular vectors and those of $U$ are left singular vectors, giving orthonormal input and output coordinate systems.

The nullspace is the vector subspace $\boxed{\ker A=\{x\in {\mathbb{R}}^n:Ax=0\}}$. A full real singular value decomposition iswhere $U,V$ are orthogonal matrices and ${\Sigma }_{ij}=0$ for $i\ne j$, with ${\Sigma }_{jj}={\sigma }_j\ge 0$ for $1\le j\le n$. We may order ${\sigma }_1\ge \cdots \ge {\sigma }_n\ge 0$. The columns of $V$ form an orthonormal basis of the input space ${\mathbb{R}}^n$, and the columns of $U$ form one of the output space ${\mathbb{R}}^m$. The positive singular values number exactly the matrix rank.

A real singular value decomposition has the form $A=U\Sigma V^T$, where $U\in {\mathbb{R}}^{m\times m}$ and $V\in {\mathbb{R}}^{n\times n}$ are orthogonal matrices, and $\Sigma$ is a rectangular diagonal matrix with entries ${\sigma }_1\ge \cdots \ge {\sigma }_{\min (m,n)}\ge 0$. These entries are the singular values. Their squares are the eigenvalues of $A^{T}A$, omitting the additional zero eigenvalues when $n>m$. The positive singular values number exactly the matrix rank.

Use the singular value decomposition and the orthogonal invariance of the Euclidean norm. Put $z=V^{T}x$ and $c=U^{T}b$. Since precisely ${\sigma }_1,\dots ,{\sigma }_r$ are positive,The second sum is independent of $x$ and cannot be removed. The first sum vanishes exactly when $z_j=c_j/{\sigma }_j$ for $1\le j\le r$, while $z_{r+1},\dots ,z_n$ are arbitrary. Thus every least-squares solution isThe unavoidable linear-system residual is the component of $b$ orthogonal to the column space of $A$:The minimum-norm least-squares solution sets all ${\alpha }_j=0$ and equals $A^{+}b$, where $A^{+}$ is the Moore-Penrose inverse. The squared minimum, rather than the minimum itself, is the second sum in the first display.

Let $u_j$ and $v_j$ be the columns of the orthogonal matrices $U$ and $V$ in the singular value decomposition. Multiplying by the rectangular diagonal matrix retains its $r$ nonzero diagonal entries:The families $(u_j)$ and $(v_j)$ are orthonormal, hence in particular orthogonal. Each summand is a rank-one outer product, and the remaining terms vanish because their singular values are zero.