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For an $n$-by-$n$ square matrix $A=(a_{ij})$ over a field, its matrix trace is the sum of its diagonal entries:For example, $A=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ has trace $2+1=3$. Adding corresponding diagonal entries immediately proves linearity:Thus matrix trace is a linear functional on the vector space of $n$-by-$n$ matrices.^[ref]

The surprising point is that the sum is intrinsic to a linear operator, even though individual diagonal entries depend on the basis. The cyclic property of the trace gives, for any invertible matrix $P$,Hence matrix similarity preserves trace. For the example above, choose the change-of-basis matrix $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. ThenThe diagonal entries have changed but their sum has not. This makes trace useful for extracting a feature of the underlying linear operator without finding a preferred basis.

Over the complex numbers, or over the real numbers after allowing complex eigenvalues, the trace equals their sum with algebraic multiplicity:No diagonalizable matrix hypothesis is needed.^[ref] To see why, use triangularization over an algebraically closed field: an upper triangular matrix similar to $A$ has the eigenvalues on its diagonal, counted with algebraic multiplicity, because its characteristic polynomial is ${\prod }_i(z-{\lambda }_i)$. Similarity preserves trace, so summing that diagonal proves the formula.

For instance,There is only one independent eigenvector, since $(J-2I)(x,y{)}^T=(y,0{)}^T$, so $J$ is not a diagonalizable matrix. Nevertheless its repeated eigenvalue contributes twice. For our earlier $A$, the characteristic polynomial is $z^2-3z+1$, its eigenvalues are $(3+\sqrt{5})/2$ and $(3-\sqrt{5})/2$, and their sum is $3$. An inexpensive diagonal sum therefore gives spectral information before solving the characteristic polynomial.

The trace of a real matrix does not give that linear map's volume factor: that factor is the absolute value of its determinant. For example, $2I_2$ has trace $4$ and area factor $4$, but $2I_3$ has trace $6$ and volume factor $8$. Their occasional numerical agreement in two dimensions is a coincidence. The connection between trace and local volume change gives a geometric interpretation near the identity matrix, and describes the volume rate of a flow map generated by a linear system of differential equations.

A square matrix $A$ is invertible exactly when its minimal polynomial has nonzero constant term. If $m_A(t)=tq(t)$, then $Aq(A)=0$, and invertibility would contradict the minimality of $m_A$. Conversely, if $m_A(t)=tq(t)+c$ with $c\ne 0$, then $A[-q(A)/c]=I$.

The Moore-Penrose inverse extends the useful part of a matrix inverse to rectangular and singular matrices. A linear system $Ax=b$ may have no exact solution, or may have many. The pseudoinverse resolves both ambiguities: $x_{\ast }=A^{+}b$ is the minimum-norm least-squares solution. First minimize the Euclidean norm $\Vert Ax-b{\Vert }_2$; among all vectors attaining that minimum, choose the one with smallest Euclidean norm $\Vert x{\Vert }_2$. It does not pretend that inconsistent equations can all be satisfied.

For $A\in {\mathbb{C}}^{m\times n}$, its Moore-Penrose inverse is the unique $n\times m$ matrix satisfying the four Penrose equations. The singular value decomposition gives a direct construction:Here $U,V$ are unitary matrices, the star is the conjugate transpose, and ${\Sigma }^{+}$ is the $n\times m$ rectangular diagonal matrix obtained by replacing each nonzero singular value ${\sigma }_j$ by $1/{\sigma }_j$ and leaving zeros at zero. Over the real numbers, use orthogonal matrices and transposes. The pseudoinverse reverses the stretches that can be reversed; it discards output components the original linear map cannot produce and does not invent an input component in its kernel. ^[ref]

The two products have especially clear meanings:They are orthogonal projection matrices: $A{A}^{+}$ selects the producible part of an output, while $A^{+}A$ selects the detectable part of an input. They are generally not identity matrices. For an invertible matrix, both are identities and $A^{+}=A^{-1}$.

This gives a useful order of operations for least-squares solutions: project the requested output onto the column space, then invert the restriction of $A$ from $(\ker A{)}^{\perp }$ to its column space. Adding a kernel vector leaves the residual unchanged but increases the squared Euclidean norm by the Pythagorean theorem in an inner-product space, because $A^{+}b$ is orthogonal to that kernel. The pseudoinverse therefore provides one canonical answer without hiding either lost information or irreconcilable data.

The name records a history with several contributors. E. H. Moore gave an earlier generalization in 1920; M. H. Pearl's 1966 paper records that work, Arne Bjerhammar's 1951 rediscovery for maximal-rank rectangular matrices, and Roger Penrose's independent 1955 treatment. ^[ref] Penrose's original paper proves existence and uniqueness from the four Penrose equations for arbitrary complex matrices, including rectangular ones. ^[ref]

A bilinear form $B$ is nondegenerate when its left radical and right radical are both zero: every nonzero vector can be detected by some pairing, whichever input slot it occupies. On a finite-dimensional vector space, either condition implies the other, and the following tests are equivalent:Here $A$ is the matrix of a bilinear form, and $V^{\ast }$ is the dual space. The linear map $w\mapsto B(-,w)$ has matrix $A$ in the chosen basis and its dual basis, so its kernel is the right radical. The map $v\mapsto B(v,-)$ has matrix $A^T$. An invertible matrix has zero kernel on both sides, proving the finite-dimensional equivalence. In infinite dimensions, zero left radical and right radical imply these maps are injective, but do not generally make them surjective.^[ref]

For the real symmetric bilinear form $B((x,y),(s,t))=xs-yt$, the determinant is $-1$. Its isotropic vector $(1,1)$ has zero self-pairing but pairs to $1$ with $(1,0)$, so it is detected. The alternating bilinear form with matrix $J$ has determinant $1$ and detects every nonzero vector even though all self-pairings vanish. Neither zero self-pairings nor the absence of positive definiteness implies degeneracy.

Nondegeneracy is useful because it turns a linear functional into a unique vector. For $A=\mathrm{diag}(1,-1)$, the linear functional $\ell (s,t)=2s+3t$ is $B((2,-3),(s,t))$; solving for that representing vector amounts to solving an invertible system of linear equations. The analogous identification underlies bilinear orthogonal complements and the dual isomorphisms induced by a nondegenerate pairing.

A critical point of a real function with continuous second partial derivatives is nondegenerate when its Hessian matrix is an invertible matrix there, equivalently when none of its eigenvalues is zero. A positive-definite matrix gives a strict local minimum, a negative definite Hessian gives a strict local maximum, and opposite-sign eigenvalues give a saddle point of a scalar function. A nondegenerate critical point need not be an extremum: $x^2-y^2$ has Hessian matrix $\mathrm{diag}(2,-2)$, which is an invertible matrix.

Such critical points are the local building blocks of Morse theory.

A stationary point $x_0$ of a smooth phase $\varphi$ is nondegenerate when its Hessian matrix at $x_0$ is an invertible matrix. In one dimension, this condition is ${\varphi }^{\prime \prime }(x_0)\ne 0$.

For a real symmetric invertible matrix $H$, the full undamped Newton step for the displayed quadratic function sends every starting point to the unique stationary point:This follows because the gradient is affine and its first-order expansion is exact. If $H$ is a positive-definite matrix the result is the unique global minimum; a negative-definite matrix gives a global maximum, and an indefinite matrix $H$ gives a saddle point. Thus exact arrival does not by itself show that a minimization method has minimized anything. A line search, damping, or a positive-definite matrix replacing the Hessian matrix changes this exact one-step assertion.

Let $A$ be a real $n\times n$ matrix of matrix rank $n-m$, and let the columns of $L,R$ be bases of $\ker A^{\top }$ and $\ker A$, respectively. Then $B=L{R}^{\top }$ has matrix rank $m$, $A^{\top }B=A{B}^{\top }=0$, and $M=A+B$ is an invertible matrix.

Proof. Since $R$ has independent columns, its Gram matrix $R^{\top }R$ is invertible, and $R^{\top }$ maps onto ${\mathbb{R}}^m$. Thus $\mathrm{im}B=\mathrm{im}L=\ker A^{\top }$ and $\mathrm{rank}B=m$. The basis conditions give the two zero products. In particular,If $M{M}^{\top }x=0$, taking its dot product with $x$ gives $\Vert A^{\top }x{\Vert }^2+\Vert B^{\top }x{\Vert }^2=0$. Hence $x\in \ker A^{\top }=\mathrm{im}B$ and simultaneously $x\in \ker B^{\top }=(\mathrm{im}B{)}^{\perp }$. Therefore $x=0$, and $M$ is invertible. The case $m=0$ is immediate.

For a common eigenvector of $A{A}^{\top }$ and $B{B}^{\top }$, the two eigenvalues have zero product and positive sum: exactly one is zero. Thus the added term fills the missing orthogonal output directions without interfering with the original ones.

Suppose an $s$-stage Runge-Kutta method has order $2s-1$, an invertible matrix $A$, and $b^T{A}^{-1}1=1$. Its stability function is $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. The adjugate matrix formula gives a rational function with denominator degree $s$ and numerator degree at most $s$; its zero limit at infinity lowers the latter degree to at most $s-1$. The order of a numerical method gives $R(z)-e^z=O(z^{2s})$, identifying the $[s-1/s]$ Padé approximant. Uniqueness follows because the cross-multiplied difference of two candidates has degree at most $2s-1$ but a zero of order at least $2s$, so is identically zero. The A-stability of near-diagonal exponential Padé approximants then proves A-stability, and the zero limit proves L-stability. These conditions identify the stability function; they do not alone claim uniqueness of all entries of the Runge-Kutta method tableau.

Parameter continuation changes $\lambda$ by a small amount and solves again for $u$, using the previous solution as the initial guess for Newton method. If $F_u$ is an invertible matrix, the implicit function theorem gives a local solution branch $u(\lambda )$, withLocal convergence of Newton method also needs regularity and a sufficiently small step: the cited result assumes a Lipschitz continuous derivative and an invertible $F_u$ at the seed.^[ref]

For the cubic polynomial,The derivative becomes unbounded at either continuation fold, and the scalar Newton method divides by $F_x=0$ there. Just beyond the upper continuation fold, no nearby real solution exists at the increased $\lambda$, although a distant negative-$x$ solution still exists. Converging to that solution would jump away from the local solution branch. Decreasing the parameter step cannot make $x(\lambda )$ a valid smooth local description at the bend.

Choose a basis and let $A$ be the matrix of $T$. Use the monic conventionUnder a change of basis, $A$ becomes $B=P^{-1}AP$ for an invertible matrix $P$. Thenso multiplicativity of the determinant gives $\det (xI-B)=\det (xI-A)$. Therefore the characteristic polynomial is independent of the chosen basis and is well-defined for the linear transformation itself.

A diagonalizable matrix over the field $F$ admits an invertible matrix $P$ with entries in $F$ such thatEquivalently, the vector space $F^n$ has a basis of eigenvectors of $A$. Indeed, $AP=PD$ says that each column of $P$ is an eigenvector with the corresponding diagonal entry as its eigenvalue. Requiring the basis and eigenvalues to lie in $F$ makes the choice of field essential.

If $g=\nabla f(x)\ne 0$, the positive-definite symmetric matrix $B$ is an invertible matrix and $s=-B^{-1}g\ne 0$. Multiplying $Bs=-g$ by $s^T$ givesHence $s$ is a descent direction. If $g=0$, then $s=0$ and the algorithm has already reached a stationary point; the strict descent assertion applies before termination.

Let $x_1,\dots ,x_N$ be the interior nodes, and let ${\varphi }_1,\dots ,{\varphi }_N$ be the nodal basis of the common conforming finite element space, with ${\varphi }_j(x_i)={\delta }_{ij}$ and zero boundary values. These piecewise-linear hat functions give the expansionsTesting the first equation with ${\varphi }_i$ and then the second equation with ${\varphi }_i$ gives the linear systemHere $S$ is the diffusion stiffness matrix and $M$ the mass matrix. Their entries are symmetric in $i,j$, so both are symmetric matrices.

For $c\in {\mathbb{R}}^N$, put $v_c={\sum }_j{c}_j{\varphi }_j$. The two quadratic forms areIf $c\ne 0$, the linear independence of the basis ensures $v_c\ne 0$, so $c^{\mathsf{T}}Mc>0$. If $c^{\mathsf{T}}Sc=0$, the Poincaré inequality and zero Dirichlet boundary condition force $v_c=0$, contradicting $c\ne 0$. Thus both $S$ and $M$ are positive-definite matrices.

For $X=(p,q)\in {\mathbb{R}}^N\times {\mathbb{R}}^N$, symmetry of $M$ cancels the cross terms:ThereforeThis is the positivity of a quadratic form intended by the last sentence of the question. For $N>0$, $A$ is a nonsymmetric matrix, since $M\ne 0$. Under the convention that “positive-definite matrix” includes symmetry, that last sentence needs this qualification: the symmetric part of a matrix is positive definite. The quadratic lower bound for a nonsymmetric matrix still implies that $A$ is an invertible matrix: $AX=0$ would give $X^{\mathsf{T}}AX=0$, hence $X=0$.

Write $f=m_T$. Part (ii) identifies the first superdiagonal block exactly:where $f^{\prime }$ is the formal derivative. Thus the candidate is $g=f^{\prime }$. If the irreducible polynomial $f$ has $f^{\prime }\ne 0$, its polynomial degree makes $f$ and $f^{\prime }$ coprime. Equivalently, $f$ is a separable polynomial. This includes every irreducible polynomial over a field of characteristic zero, and every irreducible polynomial over a perfect field.

Under this separability hypothesis, Bézout's identity for polynomials gives $a(x)f(x)+b(x)f^{\prime }(x)=1$. Evaluating at $A$ proves that $C=f^{\prime }(A)$ is an invertible matrix. Moreover, $f(B)$ has zero diagonal blocks, because $f(A)=0$. It is therefore a nilpotent matrix with $f(B{)}^N=0$. The only product of $N-1$ strictly upward block steps that reaches block $(1,N)$ uses the first superdiagonal every time, soFor $N=1$, the corresponding assertion is simply $f(B{)}^0=I\ne 0$. The block upper triangular matrix $B$ hasThe minimal polynomial divides this characteristic polynomial. Since $f$ is an irreducible polynomial, it must be a power of $f$; the exact nilpotency index just found forces

The arbitrary-field qualification is necessary. In the rational function field $F={\mathbb{F}}_2(t)$, take $f(x)=x^2-t$. This is an irreducible polynomial: a square of a rational function has even order of vanishing at $t=0$, while $t$ has odd order. Its formal derivative is zero. With $N=2$,The off-diagonal block in $f(B)$ is zero, and $m_U=f$, rather than $f^2$. No alternative polynomial $g$ coprime to $f$ could represent that block: $g(A)=0$ would imply $m_T=f\mid g$. Thus the printed coprimality claim fails as stated over a general field.

There is also a precise general replacement. In characteristic $p>0$, let $q=p^r$ be the largest power for which $f(x)=h(x^q)$, so that $h^{\prime }\ne 0$; set $q=1$ in characteristic zero. Irreducibility of $f$ implies that $h$ is an irreducible polynomial, so $h$ and $h^{\prime }$ are coprime. The binomial theorem in characteristic $p$ gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$, with the coefficients equivalently supplied by the Hasse derivatives, giveswhere $H$ commutes with $K$ and is a block upper triangular matrix whose diagonal blocks are $h^{\prime }(A^q)$. These blocks are invertible by Bézout's identity for polynomials, since $h(A^q)=f(A)=0$. Thus $H$ is invertible, andThe same characteristic polynomial and divisibility argument now prove the corrected arbitrary-field answer:This is the minimal polynomial of a companion block chain; $q=1$ recovers the intended separable case.

By Jordan normal form, each complex nilpotent matrix of size six can be put by matrix similarity into a block diagonal matrix whose Nilpotent Jordan blocks have sizesA Nilpotent Jordan block of size $s$ has matrix rank $s-1$ and minimal polynomial $t^s$. Therefore, for the whole matrix,Thus the given matrix rank determines the number $k$ of Jordan blocks, and the given minimal polynomial determines the largest size $s_1$. For size six these two data determine the entire integer partition. Here is the complete list, arranged by the largest size:To see that the list is exhaustive, choose the largest part $s_1$ between one and six and partition the remainder $6-s_1$ into parts at most $s_1$. The resulting rows all have distinct pairs $(s_1,k)$. Hence $A$ and $B$ have the same Jordan normal form $J$.

Explicitly, choose invertible matrices $X,Y$ such that $A=XJ{X}^{-1}$ and $B=YJ{Y}^{-1}$. ThenThis proves matrix similarity and illustrates the classification of six-dimensional nilpotent matrices by rank and nilpotency index.

We construct a cube root of a matrix on each Jordan block. Since $A$ is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block $J_m(\lambda )=\lambda I+N$, where $N=J_m(0)$ is a Nilpotent Jordan block and $N^m=0$. Choose a complex cube root $\beta$ of $\lambda$, so ${\beta }^3=\lambda$.

We need a polynomial $q(z)$ with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$. Once $c_1,\dots ,c_{j-1}$ have been chosen, the coefficient of $z^j$ in $q(z{)}^3$ is $3c_j$ plus an expression depending only on those earlier coefficients. Choose $c_j$ to make this coefficient one at $j=1$ and zero at $2\le j<m$. Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently $c_j=\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)$, where the generalized binomial coefficient is $\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$ and the empty product is one.

Substitute $z=N/\lambda$. The polynomial remainder is a multiple of $N^m$, hence zero. It follows thatFor $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the block diagonal matrix $C={⨁}_r{C}_{m_r}({\lambda }_r)$. Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.

For a diagonalizable matrix, write $M=P\mathrm{diag}({\lambda }_j)P^{-1}$. Relative to an existing fixed point,Convergence to the same fixed point for every initial vector therefore holds exactly whenHere $\rho (M)$ is the spectral radius. Necessity follows by choosing an initial displacement in each eigenspace; sufficiency follows because every displayed scalar power tends to zero. The condition also makes $I-M$ an invertible matrix, so that fixed point exists uniquely for every $c$.

If instead different initial vectors may converge to different constant solutions, an eigenvalue $1$ is allowed: all other eigenvalues must have modulus less than one, and the component of $c$ in the eigenspace for $1$ must vanish. That component of the initial vector then remains unchanged. No other unit-modulus eigenvalue is compatible with convergence for arbitrary initial vectors.

Repeated substitution in the affine iteration givesTherefore the required matrix polynomials areThese formulas do not require an inverse. If $I-M$ is an invertible matrix, multiplying the finite geometric sum by $I-M$ gives $g(M)=(I-M^{k-1})(I-M{)}^{-1}$.

A fixed point obeys $(I-M)X=c$. Thus the necessary and sufficient existence condition isIf $1$ is not an eigenvalue of $M$, then $I-M$ is an invertible matrix and the unique constant solution is $\boxed{X=(I-M{)}^{-1}c}$. If $I-M$ is singular but the equation is consistent, all constant solutions are $X_0+\ker (I-M)$ for any particular solution $X_0$. For an existing fixed point, $x_{k+1}-X=M(x_k-X)$; the successive difference in the question likewise obeys $\delta x_{k+1}=M\delta x_k$.