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Compute the two sides of the normal equations for linear least squares and divide by $n$:The first equation gives $\alpha =\overline{y}-\beta \overline{x}$. Substitute this into the second and solve:These simple linear regression coefficients are unique provided the observed distances are not all equal, sinceThe condition $n\ge 2$ alone does not guarantee this. If all $x_i=x_0$, the design matrix has matrix rank one and the valid alternative is the familyIn this rank-deficient ordinary least squares case, the fitted predictions are uniquely $\overline{y}$ but the two parameters cannot be separately determined.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

All $n$ singular values are positive, so $\Sigma$ and $V$ are invertible matrices and $A$ has matrix rank $n$. The image of a linear map is thereforewith those columns furnishing an orthonormal basis. Also $Ax=0$ implies $\Sigma V^{T}x=0$ and hence $x=0$. ThusFinally $A^{T}z=V\Sigma U^{T}z=0$ exactly when $U^{T}z=0$. The remaining columns of the full orthogonal matrix span this orthogonal complement:If $m=n$, this last null space also equals $\{0\}$ and has the empty basis.

No. Distinct singular values do not force uniqueness in the spectral norm. For an explicit example takeThe singular values of $A$ are $2>1>0$. Every $B_t$ has matrix rank one, and its error in the spectral norm isThus there are infinitely many best rank-one approximations even with a strict gap between the singular values.

The same mechanism works for every $1\le r<n$ under the question's positive singular values. Starting with $A_r$, defineWhen the singular values are distinct, ${\sigma }_1+t\ge {\sigma }_1-{\sigma }_{r+1}>0$, so every $B_t$ has matrix rank $r$, whileThe spectral norm measures only the largest remaining singular value, leaving room to change retained components without increasing that maximum. This is the nonuniqueness of spectral-norm low-rank approximation. The exceptional case $r=0$ is unique for the elementary reason given above.

Take the identity matrixBoth $B_1$ and $B_2$ have matrix rank one, andThey are distinct and both attain the minimum spectral norm error from the preceding part. Hence a best low-rank approximation of rank one need not be unique. More generally, every orthogonal projection $u{u}^T$ with $\Vert u{\Vert }_2=1$ gives another such minimizer.

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 ${\sigma }_1$.

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.

Let $r$ be the matrix rank, and partition the right singular vectors as $V=(V_r{V}_0)$, with $V_0\in {\mathbb{R}}^{n\times (n-r)}$ consisting of the columns associated with zero singular values. From $Ax=0⟺\Sigma V^{T}x=0$, these columns form an orthonormal basis of the nullspace. Therefore its orthogonal projector isIt acts on the input space ${\mathbb{R}}^n$. Symmetry and idempotence follow from $V_0^T{V}_0=I_{n-r}$. At full column rank it is zero, and at rank zero it is the identity.

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.

By the Eckart-Young-Mirsky theorem, truncate the singular value decomposition:The same $A_r$ minimizes the spectral norm and Frobenius norm errors over matrices of matrix rank at most $r$. This conventional meaning of rank-$r$ low-rank approximation includes the case $\mathrm{rank}A<r$, when $A_r=A$.

Put $P=Q{Q}^T$. Since $Q$ has $r$ columns, the rank bound for a matrix product gives $\mathrm{rank}(PA)\le r$. The Eckart-Young-Mirsky theorem and part (a) then giveOne can also see the lower bound directly. The restriction of $PA$ to the $(r+1)$-dimensional linear span of $v_1,\dots ,v_{r+1}$ has a nonzero null space, because its matrix rank is at most $r$. For a unit vector $z$ in that null space, $\Vert (A-PA)z{\Vert }_2=\Vert Az{\Vert }_2\ge {\sigma }_{r+1}$ by the singular value decomposition.

Minimizing $\Vert Y-X\theta {\Vert }^2$ gives the normal equations $X^{T}X\hat{\theta }=X^{T}Y$. Since $X$ has full column rank, $X^{T}X$ is positive definite and invertible, and the objective has a unique minimum. Set $A=(X^{T}X{)}^{-1}{X}^T$; then $\hat{\theta }=AY$ and $AX=I_p$. The model gives $\mathbb{E}[Y]=X\theta$ and $\mathrm{Var}(Y)={\sigma }^2{I}_n$, so $\mathbb{E}[\hat{\theta }]=\theta$ and $\mathrm{Var}(\hat{\theta })={\sigma }^{2}A{A}^T={\sigma }^2(X^{T}X{)}^{-1}$. For the alternative estimator the corresponding general formulas are $\mathbb{E}[\tilde{\theta }]=DX\theta$ and $\mathrm{Var}(\tilde{\theta })={\sigma }^{2}D{D}^T$.

Inverting the detector rotation at this time givesFor a representation with one system factor and one environment factor per displayed detector state, each parenthesized system–environment vector would have to be a product state. But the first has coefficient matrix $\mathrm{diag}(c,s)$ in the indicated orthonormal bases; the second has matrix $\mathrm{diag}(s,-c)$. Their matrix ranks are two whenever $cs\ne 0$. Equivalently, their two nonzero Schmidt coefficients are $|c|,|s|$, which no independent basis changes can remove. Thus the proposed factorization is impossible when $\sin \theta \ne 0$.

The exception matters: at $\theta =m\pi$, the detector basis is the original basis up to phases or interchange, and a factorization does exist. In particular, the literal condition $\theta \ne 0$ is insufficient if it includes $\theta =\pi$. The intended obstruction concerns genuinely mixed detector bases.

The environment has selected robust detector records: its interaction commutes with the original projectors, and orthogonal environmental records suppress their interference in the reduced density matrix. A rotated detector branch instead leaves the system and environment entangled. This is environment-induced pointer-basis selection, the issue analyzed by W. H. Zurek in his apparatus-environment model.^[ref]

This addresses the basis ambiguity but does not by itself select one realized outcome from the global superposition. Moreover, a two-state environment has recurrences: $\cos 2A$ returns to unit modulus, so the toy model does not establish irreversible macroscopic collapse. A large environment can make records effectively persistent over experimental times, but that requires further dynamical assumptions. The distinction is part of the measurement problem, not an algebraic error cured by a partial trace.

The elementary row operations preserve the determinant, and the diagonal product of the upper triangular matrix found above isConsequentlyAt $t=1$, all four rows are equal. At $t=-1$, successive rows alternate between one row and its negative. In either case the matrix rank is one, confirming the vanishing determinant.

For an $m\times n$ matrix $A$ over the real numbers, write its rows as $r_1,\dots ,r_m$ and its columns as $c_1,\dots ,c_n$. ThenThus the row space and column space consist of all linear combinations of the corresponding rows and columns; the null space is the kernel of a linear map represented by $A$. The matrix rank counts independent row or column directions, with row rank equal to column rank.

For a real $m\times n$ matrix $A$, the row space is the linear span of its rows in ${\mathbb{R}}^n$. The column space is the linear span of its columns in ${\mathbb{R}}^m$, equivalently $\{A\mathbf{x}:\mathbf{x}\in {\mathbb{R}}^n\}$. The null space isThe matrix rank is the dimension of the column space, which equals the dimension of the row space. In particular,The last equality is the rank-nullity theorem.

The reduced row echelon form gives $x_1+x_4=0$, $x_2-x_4=0$, and $x_3+x_4=0$. With free parameter $x_4=t$,The displayed nonzero vector therefore forms a basis for the null space:This agrees with the rank-nullity theorem: there are four columns and matrix rank three.

Let $c_1=(1,1,1{)}^T$ and $c_2=(2,-2,6{)}^T$. The remaining columns are $3c_1$ and $2c_2$, so these two vectors span the column space. They are linearly independent: the determinant of their first two coordinates is $-4\ne 0$. Therefore $\{(1,1,1{)}^T,(2,-2,6{)}^T\}$ is a basis of the range, and the matrix rank is two.

Equivalently the range is the plane $y_3=2y_1-y_2$: both displayed basis vectors satisfy this equation, and their two-dimensional span fills that plane.

True. A square matrix with matrix rank $n$ is invertible, and its transpose has the same rank. Applying the rank-nullity theorem to both $A$ and $A^T$ gives nullity zero in each case, so both kernels are $\{0\}$.