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Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

An invertible matrix is exactly an ordered basis of its column space. The first column has $q^n-1$ choices. After $j$ independent columns have been chosen, their span has $q^j$ elements, so the next column has $q^n-q^j$ choices. Therefore

A real square QR factorization is $A=QR$ with $Q^{\mathsf{T}}Q=I$ and $R$ upper triangular. For an invertible matrix it can be made unique by requiring positive diagonal entries of $R$.

If $A$ is orthogonal, then $R=Q^{\mathsf{T}}A$ is orthogonal as well. An orthogonal upper triangular matrix is diagonal with entries $\pm 1$: its first column has just its first entry, which must have modulus one, and orthogonality then zeros the rest of the first row; repeat inductively on the lower block. Thusand under the positive-diagonal convention $\boxed{R=I}$, $Q=A$.

There are $p^n$ choices for the translation. Choose the columns of an invertible matrix successively: the first is any nonzero column, the second avoids its span, and at stage $j+1$ one avoids a $j$-dimensional span of size $p^j$. ConsequentlyThe successive linear independence conditions are exactly the criterion that the columns form a basis.

Let $v\in {\mathbb{Q}}^2$ be nonzero. If the columns $v,Av$ were linearly dependent over the rational numbers, then $Av=\lambda v$ for some $\lambda \in \mathbb{Q}$. Applying $A$ once more would giveso $({\lambda }^2+1)v=0$. Since $v\ne 0$, this forces ${\lambda }^2=-1$, impossible in $\mathbb{Q}$. Thus the two columns are linearly independent, and the square matrixis an invertible matrix.

Regard $A$ as a matrix over the rational numbers. Part (a)(ii), with $n=2$, gives its characteristic polynomial $X^2+1$. Therefore, writing $i$ for the imaginary unit,The matrix $M=iI-A$ is not zero: otherwise $A=iI$ would have noninteger diagonal entries. It consequently has rank of a matrix one over the fraction field $\mathbb{Q}(i)$ of the Gaussian integers.

By part (b)(i), $\mathbb{Z}[i]$ is a euclidean domain, hence a principal ideal domain, so apply Smith normal form. There exist $P,Q\in {\mathrm{GL}}_2(\mathbb{Z}[i])$ such thatIndeed, invertible multiplication preserves rank of a matrix, so exactly one diagonal entry is nonzero. For $e_2=(0,1{)}^{\mathsf{T}}$, multiplication on the right gives $P(iI-A)Q{e}_2=0$. Since $P$ is an invertible matrix,The Smith normal form supplies a kernel basis lemma ensures that the null vector $Q{e}_2$ is a column of an invertible matrix over $\mathbb{Z}[i]$; an arbitrary nonzero solution over $\mathbb{Q}(i)$ would not itself establish that requirement.

Over a principal ideal domain $D$, an $m$-by-$n$ matrix is in Smith normal form if all entries off the main diagonal vanish and its nonzero diagonal entries $d_1,\dots ,d_r$ occur first and satisfyAll remaining diagonal entries are zero. The zero matrix is allowed, with $r=0$. For an arbitrary matrix $M$, the Smith normal form theorem gives invertible matrices $P\in {\mathrm{GL}}_m(D)$ and $Q\in {\mathrm{GL}}_n(D)$ withwhere the right side is interpreted as a rectangular diagonal matrix when necessary. The $d_j$ are determined up to multiplication by units. Positivity can be required over $\mathbb{Z}$, but is not a condition over the Gaussian integers.

Use the cross-ratio convention compatible with the permutation identities in this paper. Represent a projective point $P_i\in {\mathbb{CP}}^1$ by a nonzero vector $p_i\in {\mathbb{C}}^2$, and write $[p_i,p_j]=\det (p_i,p_j)$. DefineThe denominator is nonzero because the four projective points are distinct. Rescaling any one representative multiplies numerator and denominator by the same nonzero scalar, so the definition is independent of the chosen homogeneous coordinates. In finite affine coordinates it readswith the point at infinity of the complex projective line handled by the homogeneous formula.

A projective transformation is represented by an invertible matrix $A\in {\mathrm{GL}}_2(\mathbb{C})$. Since $[A{p}_i,A{p}_j]=\det (A)[p_i,p_j]$, the two factors of $\det (A)$ in the numerator cancel the two in the denominator. Therefore every projective transformation preserves this cross-ratio. This also shows that the formula descends from matrices to the projective linear group, whose representatives differ by a nonzero common scalar.

If a square matrix $A$ is an invertible matrix, multiplicativity of the determinant givesso $\det A\ne 0$.

Conversely, suppose $\det A\ne 0$. Use Gaussian elimination to reduce $A$ to row echelon form $U$. Each elementary row operation multiplies the determinant by a nonzero factor, so $\det U\ne 0$. If this square matrix had fewer than $n$ pivots, its row echelon form would have a zero row and hence zero determinant, a contradiction. Therefore it has a pivot in every row and column, and further elementary row operations reduce it to $I$.

Each elementary row operation is multiplication on the left by an invertible matrix, its inverse being the reverse operation. Thus $E_k\cdots E_{1}A=I$ implies $A=E_1^{-1}\cdots E_k^{-1}$, a product of invertible matrices. Hence

A square matrix is an invertible matrix exactly when its determinant is nonzero. The preceding recurrence relation therefore gives

The printed word “eigenvalues” must read “eigenvectors”: eigenvalues are real numbers, whereas a basis of ${\mathbb{R}}^n$ consists of vectors. The intended assertion isThis wording error is present in the original scan.^[ref]

If $X^{-1}AX=D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ for an invertible matrix $X$, write $x_i$ for its columns. The matrix identity $AX=XD$ says $A{x}_i={\lambda }_i{x}_i$. The columns of an invertible matrix are a basis, hence they give the required basis of eigenvectors.

Conversely, put a basis of real eigenvectors $x_1,\dots ,x_n$ into the columns of $X$, with corresponding eigenvalues ${\lambda }_i$. The basis property makes $X$ an invertible matrix. The equations $A{x}_i={\lambda }_i{x}_i$ give $AX=XD$, so $X^{-1}AX=D$. This is precisely diagonalizability over the real numbers.

Apply the elementary row operation $R_1\leftarrow R_1-R_2$, which preserves the determinant. Expanding along the new first row givesA square matrix is an invertible matrix exactly when its determinant is nonzero. Thereforeare precisely the three values making $A$ a singular matrix. They also have direct explanations: at $c=0$ the second row is zero, at $c=2$ the first two rows coincide, and at $c=7$ the last two columns coincide.

Suppose an invertible matrix has two LU factorizations, $A=L_1{U}_1=L_2{U}_2$, with $L_1,L_2$ unit lower triangular matrices and $U_1,U_2$ upper triangular matrices. All four factors are invertible matrices, since their square products are invertible. Multiplication by the appropriate inverse matrices givesBy the supplied closure properties, the left side is a unit lower triangular matrix and the right side is an upper triangular matrix. A matrix of both types is diagonal; its diagonal entries are all one. Thus both sides are the identity matrix, whenceQualification of the printed wording: this establishes uniqueness if the specified LU factorization exists. Invertibility alone does not guarantee an unpivoted LU factorization. For example $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ is invertible, but in a product $\left(\begin{array}{cc}1& 0\\ \ell & 1\end{array}\right)\left(\begin{array}{cc}{u}_{11}& u_{12}\\ 0& u_{22}\end{array}\right)$, the zero top-left entry forces $u_{11}=0$, making the bottom-left entry $\ell u_{11}=0$ as well, contrary to the required value one. A permutation matrix is needed in general, as in PLU factorization.

Use multiplicativity of the determinant and the determinant criterion for an invertible matrix:Neither factor can therefore be zero, so both matrices are nonsingular. The determinant of a triangular matrix is the product of its diagonal entries:Each product is nonzero, forcing every diagonal factor to be nonzero. These explicitly stated determinant properties justify both conclusions.

The requested order is $A=P^{-1}DP$. To obtain this order directly, put the transposes of the two eigenvectors in the rows of $P$, using the fact that $A$ is symmetric:Then $\det P=\phi -\psi =\sqrt{5}\ne 0$, so $P$ is an invertible matrix, withMultiplication gives $PA=\left(\begin{array}{cc}\phi +1& \phi \\ \psi +1& \psi \end{array}\right)=DP$, since ${\phi }^2=\phi +1$ and ${\psi }^2=\psi +1$. Multiplying by $P^{-1}$ proves the required diagonalization. Putting the eigenvectors in columns would instead give the equally standard order $A=QD{Q}^{-1}$, so the convention must be checked rather than copied mechanically.

Rearrange the identity intoTaking determinants gives $\det A\det (B-A-I)=1$. Hence $\det A\ne 0$, so $A$ is an invertible matrix, with $A^{-1}=B-A-I$. This uses that the matrices are finite and square: a right inverse then implies a two-sided inverse. Applying the identity to a vector in the kernel without first establishing commutation would not justify cancelling an $A$ on the wrong side. The argument is a case of invertibility forced by a polynomial product identity.

The matrix $C=A-b{b}^T$ is an invertible matrix. Indeed, on the line spanned by $b$ it has eigenvalue $-1/2$, and on $b^{\perp }$ it agrees with the positive-definite restriction of $A$. Therefore the full Newton step is well defined and givesThis is the unique stationary point because $C$ is an invertible matrix. The algebra applies to every starting point, but it does not make the answer a minimum: for $d\ge 2$ it is the saddle point just classified, and for $d=1$ it is a maximum. A safeguarded minimization algorithm that replaces an indefinite matrix used as its Hessian matrix or damps the step need not take this exact step; the assertion concerns the ordinary full Newton step formula.

If $(I+A)x=0$, take its dot product with $x$ and apply (e):Hence $x=0$. A square matrix with zero null space is an invertible matrix by the rank-nullity theorem. Therefore $I+A$ is nonsingular. This proves invertibility of identity plus a real skew-symmetric matrix.

Apply Gaussian elimination to the supplied nonsymmetric matrix,The first pivot element is $-2$, with elimination multiplier $l_{21}=2/(-2)=-1$. Perform $R_2\leftarrow R_2+R_1$, givingThe second pivot element is $-1$, with elimination multiplier $l_{32}=-1$. Perform $R_3\leftarrow R_3+R_2$, then use the third pivot element $-2$ and elimination multiplier $l_{43}=-1$ to perform $R_4\leftarrow R_4+R_3$. The LU factorization isHere $L$ is a unit lower triangular matrix, hence an invertible matrix, and $U$ is an upper triangular matrix. No row interchange is needed: all three pivot elements used in elimination are nonzero. The final diagonal entry is zero, with an entirely zero last row; there is no further division by it. The three nonzero rows of $U$ are linearly independent, so the matrix rank is $3$ and $A$ is a singular matrix. Direct matrix multiplication gives $LU=A$.

Since $L$ is an invertible matrix, $Av=0$ is equivalent to $Uv=0$. The last nontrivial row gives $v_3=v_4$; the second gives $v_2=v_4$; and the first gives $2v_1=v_2+v_4$, so $v_1=v_4$. A basis of the null space is thereforeIts single basis vector has Euclidean norm $2$, so the unit null space basis vector is $(1,1,1,1{)}^T/2$.

The pivot elements of $U$ lie in columns $1,2,3$. Take these columns from the original matrix $A$ to obtain a basis of its column space:The fourth column is minus the sum of these three columns, and the leading three-by-three matrix minor has nonzero determinant $-4$, confirming linear independence.

Elementary row operations preserve the row space. Thus the nonzero rows of $U$ give a basis of the row space of $A$:These basis vectors need not have unit Euclidean norm or be orthogonal. Their Euclidean norms are $\sqrt{6},\sqrt{2},2\sqrt{2}$, respectively; the three column basis vectors have Euclidean norms $2\sqrt{2},\sqrt{6},2\sqrt{2}$.