The product $AB$ is $m\times p$, so multiplying it by the $p\times q$ matrix $C$ gives an $m\times q$ matrix. The product $BC$ is $n\times q$, so multiplying it on the left by the $m\times n$ matrix $A$ also gives an $m\times q$ matrix. Both multiplications have matching inner dimensions.

Suppose a vector $v$ lies outside the span of $S$. Then $S\cup \{v\}$ is linearly independent: in an equation $\alpha v+{\sum }_i{\beta }_i{s}_i=0$, a nonzero $\alpha$ would express $v$ in that span, and with $\alpha =0$ the independence of $S$ forces every ${\beta }_i=0$.

This would give $n+1$ linearly independent vectors in a space with an $n$-vector basis, contrary to the supplied claim. Therefore $S$ spans $V$. Independence plus spanning makes $S$ a basis.

For the equal-cardinality basis criterion, because $S$ spans $V$, repeatedly remove a vector whenever it is a linear combination of the others. Such a removal does not change the span. The finite process ends with a linearly independent spanning subset $T$, hence a basis of $V$.

If the original $S$ were linearly dependent, at least one removal would occur and $|T|<n$. But a basis of an $n$-dimensional vector space has $n$ vectors. Equivalently, apply the allowed claim to the basis $T$ and any fixed $n$-vector basis of $V$: the latter would wrongly be linearly dependent. Thus no removal was possible. The original spanning set is linearly independent and therefore a basis.

The supplied factors give $Q^{T}b=(0,2{)}^T$. Back substitution in $Rx=Q^{T}b$ yields $2y=2$ and $10x+4y=0$, soMultiplying by the original matrix gives $(1,1,1,1{)}^T$, so this least-squares minimizer has residual zero: this particular overdetermined system is exactly consistent. Full column rank still guarantees uniqueness of the minimizer.

Substitution gives $R^{T}Rx=R^T{Q}^{T}b$. Since $R^T$ is invertible, cancel it on the left to obtain the nonsingular triangular systemThis is the reduced QR least-squares solution. Its geometric justification is the orthogonal decompositionThe second term is independent of $x$, while invertibility of $R$ lets the first term be zero at exactly one point. Although $Q^{T}Q=I_M$, one must not replace $Q{Q}^T$ by $I_N$: $Q$ is tall and its product $Q{Q}^T$ is the orthogonal projection onto its column space.

The normal equations for linear least squares areThe supplied QR decomposition gives full column rank because $R$ is invertible and $Q^{T}Q=I$. Hence, for every nonzero $u$, $u^T{A}^{T}Au=\Vert Au{\Vert }_2^2>0$. Thus $A^{T}A$ is a symmetric positive-definite matrix and the displayed square system is nonsingular.

From the zero initial vector, the linear-system residual is $r_0=(1,1{)}^T$. Then $r_0^T{r}_0=2$, $A{r}_0=(5,13{)}^T$, and $r_0^{T}A{r}_0=18$. The step is ${\alpha }_0=1/9$, givingAs a check, the new residual is $(4/9,-4/9{)}^T$, orthogonal to the preceding residual as expected after this exact line search. It is nonzero, so one iteration has not already solved the system.

Real symmetry and positive definiteness are sufficient, and the preceding parts establish both properties for this matrix. If a residual becomes zero, stop: the solution has been reached and the displayed step quotient should not be evaluated as $0/0$.

For a convergence justification, let $x_{\ast }$ solve $A{x}_{\ast }=b$, let $e=x_{\ast }-x$, and let $r=Ae$. The quadratic energy error is $E=e^{T}Ae/2$. Moving by $\alpha r$, with $\alpha =(r^{T}r)/(r^{T}Ar)$, is exact line search for a positive-definite quadratic and givesWriting ${\lambda }_{\min },{\lambda }_{\max }>0$ for the extreme eigenvalues, one has $r^{T}Ar\le {\lambda }_{\max }{r}^{T}r$ and $r^{T}r=e^T{A}^{2}e\ge {\lambda }_{\min }{e}^{T}Ae$. ThereforeThe factor lies below one, so the energy tends to zero and positive definiteness forces $x\to x_{\ast }$. Here the bound is $E_{\mathrm{new}}\le (10/11)E$, since the eigenvalues are 1 and 11. This proves convergence for every initial vector, rather than merely checking one numerical iteration.

Set $y=P^{T}x$. Because $P$ is an orthogonal matrix, $x=Py$ and $x=0$ exactly when $y=0$. The preceding diagonalization givesConsequently $x^{T}Ax\ge 0$, with equality only when $x=0$. The matrix is a symmetric positive-definite matrix, not merely positive semidefinite. In the original coordinates the same identity iswhich also makes positivity transparent.

The characteristic polynomial isThus the eigenvalues are 1 and 11. Corresponding eigenvectors are $(3,-1{)}^T$ and $(1,3{)}^T$. Their dot product is zero and each has Euclidean norm $\sqrt{10}$. Normalizing givesThe columns are orthonormal, so $P^{T}P=I$. The eigenvector equations give $AP=PD$ and hence $P^{T}AP=D$, proving the claimed orthogonal diagonalization of a real symmetric matrix.

Not unique in general. For any invertible $r\times r$ matrix $S$,is another rank factorization, with the same full ranks. This is a change of basis in the intermediate $r$-dimensional space. For example, even $[1]=[1][1]=[2][1/2]$ has different full-rank factors. Once a particular ordered basis $B$ is fixed, its coordinate matrix $C$ is unique; it is the basis choice that creates the freedom.

Choose $r$ independent columns of $A$ that form a basis of its column space, and put them in $B\in {\mathbb{R}}^{m\times r}$. For each column $a_j$ of $A$, let $c_j$ be its coordinate vector in that basis; put these coordinate vectors in $C\in {\mathbb{R}}^{r\times n}$. The column equations $a_j=B{c}_j$ give $A=BC$.

By construction $B$ has full column rank $r$. The columns of $C$ corresponding to the selected basis columns are the standard vectors of ${\mathbb{R}}^r$, so $C$ has full row rank $r$. This proves the requested rank factorization. When $r=0$, $A=0$ and the corresponding empty-column and empty-row factors have the required shapes and rank zero.

False. Triviality of $\mathcal{N}(A)$ gives $r=n$, while triviality of $\mathcal{N}(A^T)$ gives $r=m$ because transpose preserves matrix rank. Both conditions therefore force $m=n$, contradicting the stated unequal dimensions.

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 $\{0\}$.

False. Take $A=(1,0{)}^T$, an $m=2$, $n=1$ matrix of matrix rank one, and $b=(0,1{)}^T$. Then $Ax=(x,0{)}^T$ never equals $b$. Full column rank implies injectivity, not surjectivity onto a larger codomain.

True. The rank-nullity theorem gives $\dim \mathcal{N}(A)=n-r=0$. The only vector in this zero-dimensional kernel of a linear map is the zero vector.

True. Using the same transpose identity and $X^T=-X$ givesThe final equality follows by multiplying $-X$ and $-X^{-1}$ in either order and obtaining the identity. Thus the inverse is a skew-symmetric matrix as well. The invertibility hypothesis is retained throughout.

True. Transposing the inverse identities shows $(X^{-1}{)}^T=(X^T{)}^{-1}$. If $X^T=X$, this becomes $(X^{-1}{)}^T=X^{-1}$, exactly the symmetric matrix condition. The reasoning uses the matrix inverse and transpose identities, without assuming that different matrix factors commute.

The system is consistent for every $b\in {\mathbb{R}}^m$: take $x=Yb$, since $A(Yb)=(AY)b=b$. Thus $Y$ is a right inverse giving an explicit solution for every right-hand side.

All solutions are $Yb+u$ with $u$ in the kernel of a linear map represented by $A$. The rank-nullity theorem gives kernel dimension $n-m$; the solution is unique when $n=m$, and is nonunique when $n>m$.

If $A$ is $m\times n$, then $Y$ is $n\times m$. The equation $AY=I_m$ makes every standard basis vector of ${\mathbb{R}}^m$ a linear combination of the columns of $A$: take the corresponding column of $Y$ as coefficients. Hence the column space is all of ${\mathbb{R}}^m$, giving $\mathrm{rank}(A)=m$. In particular, $n\ge m$ and $A$ has full row rank.