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Since every eigenvalue is nonzero, $S$ is an invertible matrix. Multiplying $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ by the inverse matrix givesHence the eigenvectors are unchanged and the corresponding eigenvalues are $1/{\lambda }_i$.

Decompose ${\mathbb{R}}^n=\mathrm{span}\{a\}\oplus a^{\perp }$. The rank-one matrix update $a{a}^T$ acts as multiplication by $\Vert a{\Vert }^2$ along $a$ and as zero on its orthogonal complement. ThusThe perpendicular eigenvalue has multiplicity $n-1$ and is absent when $n=1$. Consequently $A^{+}$ is always an invertible matrix, while

The matrix trace is $\mathrm{Tr}A={\sum }_i{A}_{ii}$. Expanding the matrix products gives $\mathrm{Tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}={\sum }_{j,i}{B}_{ji}{A}_{ij}=\mathrm{Tr}(BA)$. Thus the cyclic property of the trace gives zero trace to every commutator. If $AB-BA=I$, taking the trace would give $0=n$, impossible for positive finite $n$ over the real or complex field.

For real $A,B$, let $p(\lambda )=\det (A+\lambda B)$, a real polynomial of degree at most $n$. Since $p(i)=\det C\ne 0$, it is not the zero polynomial. It has at most finitely many real roots of a polynomial, so $A+\lambda B$ is an invertible matrix for some real $\lambda$, in fact for all but finitely many.

Finally write the invertible intertwiner as $R=A+iB$. The relation $PR=RQ$ and the reality of $P,Q$ give $PA=AQ$ and $PB=BQ$. Choose a real $\lambda$ for which $S=A+\lambda B$ is invertible. Then $PS=SQ$, so $P=SQ{S}^{-1}$. This proves that complex similarity of real matrices implies real similarity.

An invertible matrix is a square matrix for which there is a matrix $A^{-1}$ with $A{A}^{-1}=A^{-1}A=I$. Write its assumed diagonalization as $A=P\Lambda P^{-1}$, where $P$ is invertible and $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. ThenIf the determinant is nonzero, every ${\lambda }_j$ is nonzero, so ${\Lambda }^{-1}=\mathrm{diag}({\lambda }_j^{-1})$ exists and $A^{-1}=P{\Lambda }^{-1}{P}^{-1}$. If the determinant is zero, at least one eigenvalue is zero; its nonzero eigenvector is mapped to zero, which an invertible map cannot do. HenceThe equivalence holds even without diagonalizability, but the eigenbasis proves the stated case directly.

A square matrix is a diagonalizable matrix over the chosen field when there are an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$. If $v_j$ is column $j$ of $P$ and $D_{jj}={\lambda }_j$, the identity $AP=PD$ gives $A{v}_j={\lambda }_j{v}_j$. Invertibility makes these $n$ columns linearly independent eigenvectors.

Conversely, put $n$ linearly independent eigenvectors in the columns of $P$ and their eigenvalues in $D$. Then $P$ is invertible and $AP=PD$, givingRepeated eigenvalues are allowed; the requirement is a complete eigenbasis, not distinctness of the eigenvalues.

If $I+A$ were a singular matrix, a nonzero vector $v$ would satisfy $(I+A)v=0$, so $Av=-v$. This would make $-1$ an eigenvalue of the anti-Hermitian matrix, contrary to part (a). Thus $\boxed{I+A\text{ is invertible}}$. The same argument shows $I-A$ is an invertible matrix, which will also be used below.

Put $D=A-iB$, so $D^{†}=A+iB$. Because the Hermitian matrices $A,B$ commute,For every nonzero vector $v$,since $A$ is an invertible matrix. Thus $D$ and $D^{†}$ are invertible matrices, justifying the inverse in the definition of $U$. The conjugate transpose of $U=D^{†}{D}^{-1}$ is $(D^{†}{)}^{-1}D$. Since $D$ and $D^{†}$ commute, their inverses do too, andHence $U$ is a unitary matrix.

Complete the square in the Hermitian matrix polynomial:For every nonzero vector $v$,Thus $H$ is a positive-definite matrix, so $\boxed{0\text{ is not an eigenvalue of }H}$. In finite-dimensional spectral theorem terms, each eigenvalue isIf ${\alpha }^2\ge 4\beta$, the scalar polynomial has real roots of a polynomialAn allowed nonzero root can be an eigenvalue of $A$, a Hermitian matrix which is also an invertible matrix, in which case $H$ has a zero eigenvalue. For example, $\alpha =2$, $\beta =1$, $A=-I$ gives $H=0$. The weaker inequality does not force failure for every $A$; it removes the general positive lower bound. There is also the harmless corner $\alpha =\beta =0$, for which $H=A^2$ stays positive definite because $A$ is invertible, even though the strict discriminant hypothesis is absent.

The anticommutator condition and invertibility give the matrix similaritiesSince matrix trace is invariant under matrix similarity,Hence $\boxed{\mathrm{Tr}A=\mathrm{Tr}B=0}$. Taking determinants in the first matrix similarity also givesSince $A$ is an invertible matrix, $\det A\ne 0$, and $\boxed{n\text{ is even}}$. In terms of eigenvalues, the same matrix similarity pairs each nonzero eigenvalue $\lambda$ with $-\lambda$, with the same algebraic multiplicity, so an odd number of them is impossible.

Expanding the characteristic polynomial givesTherefore $a=-2z$ and $b=z^2-x^2-y^2$. Both are polynomial functions of the three real coordinates, so they are continuous functions, including on the open subset $X$ of invertible matrices.

On $X$, $b=\det A\ne 0$. Its image contains positive values, for example $b(I)=1$, and negative ones, for example $b(\mathrm{diag}(1,-1))=-1$. If $X$ were connected, its continuous real image would be an interval containing both signs and hence zero. That is impossible. Thus $X$ is not connected. More precisely its determinant image is $\mathbb{R}\setminus \{0\}$: take $x=y=0$ for a positive value and $z=y=0$ for a negative one.

The first row has off-diagonal sum $a_{12}<1/4$. Rows $2,\dots ,n-1$ have sums less than $1/2$, and the last row has sum $c+a_{n,n-1}<5/4$, since $0<c<1$. Every row except the second therefore has its Gershgorin disc strictly away from zero: its centre is at least one in the first row and at least three in the remaining rows.

For the second row, an exact sufficient condition is $|b|>a_{21}+a_{23}$. Since the sum is strictly below $1/2$, the coefficient-independent boundsguarantee that zero is outside every Gershgorin disc. Consequently zero is not an eigenvalue and $A$ is an invertible matrix. Equality at $\pm 1/2$ is allowed because the off-diagonal entry bounds are strict. These are sufficient strict diagonal dominance bounds; values of $b$ between them can also give invertible matrices.

Gaussian elimination uses elementary row operations to remove entries below successive pivot elements, reducing a linear system to one with an upper triangular matrix. Back substitution then solves that system. It also computes a reusable matrix factorization, so additional right-hand sides can be handled by forward substitution and back substitution without repeating the elimination.

For elimination without row exchanges, let $A^{(1)}=A$. At step $k$, provided the pivot element $a_{kk}^{(k)}$ is nonzero, form the elimination multipliersand perform ${\text{row}}_i\leftarrow {\text{row}}_i-m_{ik}{\text{row}}_k$, applying the same operation to the right-hand side. More explicitly the resulting entries satisfy $a_{ij}^{(k+1)}=a_{ij}^{(k)}-m_{ik}{a}_{kj}^{(k)}$ for $i>k$, $j\ge k$.

Let $E_k=I-{\sum }_{i>k}{m}_{ik}{e}_i{e}_k^T$, where $e_i$ are the standard basis vectors. ThenThe matrix $L$ is a unit lower triangular matrix storing the elimination multipliers, while $U$ is an upper triangular matrix:Solving $Ax=b$ becomes $Ly=b$ followed by $Ux=y$. If a required pivot element vanishes, an unpermuted LU decomposition can fail even for an invertible matrix. Row exchanges, typically selected by partial pivoting, instead produce $PA=LU$ with a permutation matrix $P$; then solve $Ly=Pb$, $Ux=y$.

Write ${\mathcal{P}}_n$ for the vector space of polynomials of polynomial degree at most $n$, and put $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$. As the question calls the weighted integral an inner product, its finiteness is implicit; for example, a positive integrable weight $w$ suffices for all the continuous functions involved here.

For every $q\in {\mathcal{P}}_n$, differentiating the squared weighted norm $\Vert f-p_n-tq{\Vert }_w^2$ at its minimum $t=0$ gives the orthogonality equationsIn particular, the requested linear system isIts Gram matrix $G_{kj}={\int }_{-1}^{1}w(x)x^{k+j}dx$ is a symmetric positive-definite matrix: for any nonzero vector $a$,Indeed a nonzero polynomial is nonzero on some open interval, and the weight is positive. Therefore $G$ is an invertible matrix, proving uniqueness. Conversely, if these equations hold, the Pythagorean theorem in an inner-product space gives, for every $q\in {\mathcal{P}}_n$,so the solution really is the unique best approximation.

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 $n$.

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.

Write $A=D+L+R$, where $D$ is the diagonal matrix of diagonal entries of $A$, and $L,R$ are its strictly lower- and upper-triangular parts. Since $A$ is symmetric, $R=L^T$. Subtract the two factors in the SOR iteration:Therefore, for $\omega \ne 0$,Positive definiteness makes each $D_{ii}>0$, so the triangular matrix $B(\omega )$ is an invertible matrix for every $\omega \ne 0$. The original formula gives $C(0)=I$; the expression for $B(\omega )$ itself is undefined there. The affine term ensuring that the desired solution is a fixed point is $c=B(\omega {)}^{-1}b$.

The claim is true, withIt is a symmetric positive-definite matrix: for every nonzero vector $z$, the fact that $A$ is an invertible matrix gives $z^{T}Mz=\Vert Az{\Vert }_2^2>0$. Since $A{x}_{\ast }=b$, every candidate $x$ satisfiesThe generalized minimal residual method therefore minimizes precisely this weighted norm of the error over the linear span of $Q_k$. The same matrix $M$ works for every $k$ and does not depend on $b$.

Write $x=Q_{k}y$ and $\beta =\Vert b{\Vert }_2$. The Arnoldi decomposition and $b=\beta Q_{k+1}{e}_1$ giveBecause $Q_{k+1}$ has orthonormal columns, it preserves the Euclidean norm. The generalized minimal residual method therefore reduces to the small linear least-squares problemUse Givens rotations on the upper Hessenberg matrix to obtain a QR decompositionwhere $Z$ is an orthogonal matrix and $R_k$ is an upper triangular matrix. Then solve $R_k{y}_k=g_{1:k}$ by backward substitution in a triangular system and form $Q_k{y}_k$. The resulting minimum linear-system residual has Euclidean norm $|g_{k+1}|$.

The small problem has a unique answer: if ${\tilde{H}}_{k}y=0$, the Arnoldi decomposition gives $A{Q}_{k}y=0$. The invertible matrix $A$ implies $Q_{k}y=0$, and the orthonormality of $Q_k$ then gives $y=0$. Thus ${\tilde{H}}_k$ has full column rank and $R_k$ is nonsingular.

Take the invertible matrix $A=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, whose determinant is $-1$. Suppose it had an LU factorization with any lower triangular $L$ and upper triangular $U$:The $(1,2)$ entry gives ${\ell }_{11}{u}_{12}=1$, so ${\ell }_{11}\ne 0$. The $(1,1)$ entry gives ${\ell }_{11}{u}_{11}=0$, so $u_{11}=0$. But then the $(2,1)$ entry would be ${\ell }_{21}{u}_{11}=0$, contradicting $a_{21}=1$. Thus this nonsingular matrix has no unpermuted LU factorization, even without imposing a unit diagonal on $L$. A row permutation resolves the zero pivot, explaining the role of partial pivoting.

There are $p^n$ possible translation vectors. An invertible matrix is specified by an ordered basis of ${\mathbb{F}}_p^n$. Its first column can be any nonzero vector, giving $p^n-1$ choices. After choosing $i$ independent columns, their span has $p^i$ elements, so the next column has $p^n-p^i$ choices. Therefore