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For the p-energythe first variation in the direction $\psi \in C_c^{\infty }(\Omega )$ isAn integration by parts therefore gives the Euler-Lagrange equationwhich is the p-Laplacian equation. In the notation of the question one takes $a^{ij}(x,z,\eta )=|\eta {|}^{p-2}{\delta }^{ij}$.

For $\eta \ne 0$, the principal coefficient matrix isIts eigenvalue in directions orthogonal to $\eta$ is $|\eta {|}^{p-2}$, while its eigenvalue parallel to $\eta$ is $(p-1)|\eta {|}^{p-2}$. The coefficients are $C^{1,\alpha }$ away from $\eta =0$, and the condition number there is at most $p-1$. On every region where $0<m\le |Du|\le M$, this gives uniform ellipticity with constants depending on $m$, $M$, and $p$. At $Du=0$ all principal eigenvalues vanish, so the operator is degenerate there and is not strictly elliptic on a domain containing a critical point.

For the HHL algorithm to have runtime polynomial in $\log N$, the Hermitian matrix $A$ must be invertible, have a condition number $\kappa$ bounded by $\mathrm{poly}(\log N)$, and be a sparse matrix with its nonzero entries efficiently accessible by an oracle. The normalized state $|b⟩$ must also be preparable in $\mathrm{poly}(\log N)$ time. With precision costs suppressed, these assumptions let HHL prepare, with high probability,

For a unit eigenvector of $A$, $(A-I)x=(\lambda -1)x$. The matrix 2-norm bound from part (a) gives $|\lambda -1|\le 2/3$. Hence every eigenvalue lies inIn particular the condition number in the Euclidean operator norm satisfies ${\kappa }_2(A)={\lambda }_{\max }/{\lambda }_{\min }\le 5$.

With $y=-c/\mu$, differentiation of the quadratic penalty method objective givesThe primal Newton system is thereforeFor bounded multipliers and regular derivatives, $H$ remains of ordinary size while ${\mu }^{-1}{A}^{T}A$ becomes large in directions normal to the constraints. If the feasible tangent space is nontrivial and retains ordinary-size curvature, this produces a large condition number as $\mu \downarrow 0$.

Introduce $w=Ap/\mu$. The same step is obtained from the symmetric augmented systemEliminating $w$ recovers the primal system exactly. This matrix contains no explicit ${\mu }^{-1}$ entries and tends to a saddle-point matrix. Under full row rank of the limiting constraint Jacobian matrix and nonsingularity of the limiting reduced Hessian, that limiting matrix is nonsingular, so this formulation avoids the artificial penalty-induced separation of scales.

These conditioning statements require regularity beyond bounded $y$ alone. A rank-deficient constraint Jacobian matrix or singular reduced Hessian can still make the augmented system ill-conditioned. If $m=n$ and $A$ is nonsingular, all directions may scale together in the primal system, so a large condition number is not automatic there either.

The least-squares normal equations areThe full column rank of $G$ gives $z^{T}Hz=\Vert Gz{\Vert }_2^2>0$ for nonzero $z$, so $H$ is a symmetric positive-definite matrix. Apply the Conjugate gradient method to this linear system. To verify that its solution is the desired one, let $x_{\ast }$ solve the displayed equations. Then $G^T(G{x}_{\ast }-b)=0$, and for any $x$ the Pythagorean identity givesThus $x_{\ast }$ is the unique minimizer of the linear least-squares problem.

One can apply $H$ without forming the Gram matrix: compute $Hp=G^T(Gp)$. More explicitly, initialize $x_0$, $r_0=c-H{x}_0$, $p_0=r_0$; until the linear-system residual vanishes, use the Conjugate gradient method recurrenceEach dense application of $H$ this way costs $O(mn)$ operations. The condition number satisfies ${\kappa }_2(H)={\kappa }_2(G{)}^2$, where ${\kappa }_2(G)={\sigma }_{\max }(G)/{\sigma }_{\min }(G)$; this is the conditioning cost of the least-squares normal equations.

Gradient descent uses only the current gradient and a suitable step size; its additional storage is order $d$. It is attractive for very large $d$ or when low per-iteration cost matters most. For a smooth strongly convex function, standard step choices give linear convergence, with a rate sensitive to the condition number. Mere convexity gives a global-minimum interpretation but does not alone imply a linear rate; for nonconvex objectives the usual guarantee concerns stationary points.

BFGS uses successive gradients to update an approximation to inverse curvature, incorporating information that can correct poor scaling. Full BFGS uses order $d^2$ storage and update work. With suitable line searches and local smoothness and positive-curvature assumptions near a nondegenerate minimum, it has superlinear convergence. This often favours it in moderate dimensions when gradients are not extremely cheap relative to the extra work.

Neither gradient descent nor BFGS requires an explicitly evaluated Hessian matrix. Thus an expensive Hessian matrix is compatible with both, rather than a reason to reject BFGS. If an accurate Hessian matrix and its factorization are exceptionally cheap, Newton's method in optimization becomes an additional reasonable option. Convexity helps identify a stationary point as a global minimum; without it, both methods need suitable descent and curvature safeguards. The BFGS update learns curvature from the Quasi-Newton secant equation.^[ref]

We fix the convention $PA=LU$: the permutation matrix $P$ acts on the rows of $A$, and the same row ordering must be applied to $b$ before the Triangular solves for LU decomposition. Equivalently, $A=P^{T}LU$, since $P^{-1}=P^T$. LAPACK documents its factors as $A=PLU$ instead, so check the defining equation rather than just the letter $P$.^[ref] Partial pivoting selects the largest absolute candidate in the current column, among rows $k,\dots ,n$, before forming elimination multipliers.^[ref]

A zero first pivot element does not imply a singular matrix. For example,Unpivoted Gaussian elimination would divide $2$ by zero. Indeed, no $A_0=LU$ with a unit lower triangular matrix $L$ exists: $u_{11}=0$ would force $(LU{)}_{21}=l_{21}{u}_{11}=0$, contradicting $(A_0{)}_{21}=2$. One row interchange resolves the obstruction. For $b_0=(1,5{)}^T$, use $P{b}_0=(5,1{)}^T$; back substitution then gives $x=(1,1{)}^T$. Using the original $b_0$ with these factors would solve a different linear system.

Partial pivoting bounds each elimination multiplier by one in absolute value in exact arithmetic. This helps control roundoff error, but intermediate entries can still grow; it is not an unconditional guarantee of numerical stability for every matrix.^[ref] It also does not remove the original problem's condition number. For the diagonal matrix $D=\mathrm{diag}(1,\epsilon )$ with $0<\epsilon <1$, the condition number in the Euclidean norm is ${\kappa }_2(D)=1/\epsilon$. Row permutations preserve this condition number, because they preserve Euclidean norms. Even an exact solve changes from $(1,1{)}^T$ to $(1,1+\eta /\epsilon {)}^T$ when the right-hand side changes from $(1,\epsilon {)}^T$ to $(1,\epsilon +\eta {)}^T$. This sensitivity is a property of the linear system, not a zero-pivot element failure.^[ref]

Suppose $A\in {\mathbb{C}}^{n\times n}$ is a diagonalizable matrix, $n\ge 2$, with an eigenbasis $v_1,\dots ,v_n$ and spectral decompositionAssume the dominant eigenvalue ${\lambda }_1$ is a simple eigenvalue andThen the normalized power iteration approaches the one-dimensional eigenspace of ${\lambda }_1$. Its asymptotic direction-error bound has factor $\gamma =|{\lambda }_2/{\lambda }_1|<1$. This spectral gap is a gap in modulus: if $g=|{\lambda }_1|-|{\lambda }_2|$, then $\gamma =1-g/|{\lambda }_1|$. A small relative gap gives slow convergence.^[ref]

The short proof is the exact eigenbasis identityThe remaining sum has norm bounded by a constant times ${\gamma }^k$. This proves convergence of direction, allowing a changing sign or complex factor. A tiny $c_1$ or a poorly conditioned, nonorthogonal eigenbasis can delay visible convergence: the condition number of $V$ affects constants, while $\gamma$ controls the eventual rate. The bound need not be an equality; if the initial coefficient of the second-largest eigenvalue vanishes, the active spectrum of the power method gives a faster rate. When $\gamma =0$, all unwanted components disappear after one multiplication.^[ref]

The hypothesis is $c_1=(V^{-1}{x}_0{)}_1\ne 0$, not an ordinary nonzero inner product with the right eigenvector. Let $w_1^{\ast }$ be the first row of $V^{-1}$, a left eigenvector normalized so $w_1^{\ast }{v}_j$ is $1$ for $j=1$ and $0$ otherwise; $\ast$ denotes conjugate transpose. Then $c_1=w_1^{\ast }{x}_0$. For a Hermitian matrix, choose an orthonormal eigenbasis and the criterion becomes $v_1^{\ast }{x}_0\ne 0$: the starting vector must be nonorthogonal to the dominant eigenvector. For a general diagonalizable matrix, the nonorthogonal eigenbasis makes these two tests different, as illustrated under power iteration failure cases.

For a real symmetric positive-definite matrix $A$, choose an invertible change of variables $P$ and solveThe transformed matrix is symmetric positive definite because $y^{T}PA{P}^{T}y=(P^{T}y{)}^{T}A(P^{T}y)>0$ for nonzero $y$. The aim is to improve its condition number or eigenvalue clustering while keeping the transformation inexpensive. The change-of-variables matrix $P$ itself need not be symmetric.

A common convention starts instead from a symmetric positive-definite approximate matrix $M\approx A$ and uses $P=M^{-1/2}$ conceptually. If $M=L{L}^T$ is a Cholesky decomposition, choosing $P=L^{-1}$ gives an equivalent symmetric transformation. Practical iterations can apply solves with $M$ or its factors without explicitly constructing a matrix square root. Applying ordinary Euclidean conjugate gradients directly to the generally nonsymmetric matrix $M^{-1}A$ would require accounting for the appropriate transformed inner product.

Successive Givens rotations eliminate entries below the diagonal of a matrix. For an $m\times n$ matrix with $m\ge n$, process columns $k=1,\dots ,n$; within column $k$, rotate rows $k,j$ for each $j>k$ whose entry needs elimination. Apply each Givens row-pair update to columns $k,\dots ,n$. Both rows are zero in earlier columns, so these operations preserve earlier eliminations. No assumption of full column rank is needed to perform the eliminations.

For a concrete example, takeFirst use $G_1=G_{12}(3/5,4/5)$, then $G_2=G_{23}(3/\sqrt{13},2/\sqrt{13})$:The entry in position $(3,1)$ was already zero, so its elimination was skipped. The second Givens rotation combines the remaining pair $(3,2{)}^T$ in column two. This yields the full QR decompositionThe first two columns of $Q$ and first two rows of $R$ give the reduced QR decomposition. Direct multiplication verifies both $QR=A$ and $Q^{T}Q=I$. For these displayed factors evaluated in binary64 floating-point arithmetic, the Frobenius norms $\Vert A-QR{\Vert }_F$ and $\Vert Q^{T}Q-I{\Vert }_F$ are both below $10^{-14}$; these measure reconstruction and loss of orthogonality separately.

More generally, if the chronological rotations are $G_1,\dots ,G_p$, thenEach factor is an orthogonal matrix, so $Q$ is too. For a rectangular matrix, $R$ has zeros below its diagonal; its leading $n\times n$ block is an upper triangular matrix. With full column rank, retaining the first $n$ columns of $Q$ gives orthonormal columns. To impose a positive diagonal when necessary, flip matching columns of $Q$ and rows of $R$, preserving their product.

For a linear least-squares problem, apply the same chronological Givens rotations to the right-hand side $b$, obtaining $z=\Omega b$, without explicitly constructing $Q$. Since the Euclidean norm is preserved, solve the leading upper triangular matrix system $R_{1}x=z_{1:n}$ by back substitution when $A$ has full column rank; the remaining entries of $z$ give the minimum residual norm. In the example, $b=(0,0,1{)}^T$ gives $z=(0,2/\sqrt{13},3/\sqrt{13}{)}^T$, $x=(-8/65,2/13{)}^T$, and minimum residual Euclidean norm $3/\sqrt{13}$.

The locality is useful for a sparse matrix or an existing QR decomposition needing an extra row: only two rows change at each step, and the new row can be eliminated against the stored upper triangular matrix. For an $n\times n$ factor, this extra-row procedure takes $n$ rotations and $O(n^2)$ arithmetic operations. Locality does not guarantee preservation of sparsity: mixing rows can create nonzeros wherever either row had one. For a fully populated $m\times n$ input, the straightforward elimination uses ${\sum }_{k=1}^n(m-k)$ rotations, each touching $n-k+1$ columns, giving $O(m{n}^2)$ work. Storage layout, zero patterns and implementation affect speed; this count alone establishes no universal advantage over other QR decomposition algorithms.

The Euclidean norm preservation explains the appeal for numerical stability: exact rotations do not amplify a perturbation's length. Rounded coefficients and Givens row-pair updates still introduce roundoff error. Small reconstruction and orthogonality residuals check the computed factors, while the condition number of the original problem still controls how accurately a linear least-squares problem determines its solution.

For constraints $h=0$, $g\ge 0$, minimize the displayed merit function for progressively smaller $\mu >0$, where $g_i^{-}=\min (g_i,0)$. Increasing the quadratic penalty discourages constraint violation. In a violated inequality, differentiation gives the multiplier estimate ${\lambda }_i=-g_i/\mu$; an inactive inequality has estimate zero. Thus obtaining small violations by this method alone typically requires a very small $\mu$, which may cause a large condition number in the inner minimization.

For an invertible matrix $A$, the rank-one matrix update $B=A+u{v}^T$ is a singular matrix exactly when $1+v^T{A}^{-1}u=0$. This follows directly from the matrix determinant lemma. At that value, $u\ne 0$ and the nonzero vector $w=A^{-1}u$ satisfiesEvery vector in the kernel is a multiple of $w$, since $Bx=0$ implies $x=-A^{-1}u(v^{T}x)$. The matrix rank therefore drops by exactly one.

For $B(t)=I+tu{v}^T$ with $u=(1,2{)}^T$ and $v=(3,-1{)}^T$, the update becomes a singular matrix at $t=-1$, and $B(t)u=(1+t)u$. At $t=-0.99$, this direction is compressed by a factor $0.01$ and the inverse amplifies it by a factor 100. The actual spectral norm of the matrix inverse, which maximizes amplification over every direction, is about $700.17853409$.

The exact formula can expose a large condition number in the updated linear system even when the original one had a small condition number. It does not guarantee numerical stability merely because its evaluation uses fewer arithmetic operations. The denominator condition certifies existence; the condition number and floating-point behavior address different questions about using the resulting answer.

The linear-system residual $r^{(k)}=b-A{x}^{(k)}$ is computable without knowing $x_{\ast }$. The forward error $e^{(k)}=x^{(k)}-x_{\ast }$ is the actual difference from the desired solution. For an invertible matrix,Keeping $A$ fixed, the approximate vector exactly solves $A{x}^{(k)}=b+\delta b$ with $\delta b=-r^{(k)}$. Thus, for $b\ne 0$, $\Vert r^{(k)}\Vert /\Vert b\Vert$ is the relative backward error when only the right-hand side may change. For a vector norm and its induced operator norm,A small relative linear-system residual ensures a small relative forward error when the condition number is moderate; an ill-conditioned matrix can amplify it. The bound follows directly from $e=-A^{-1}r$ and $\Vert b\Vert =\Vert A{x}_{\ast }\Vert \le \Vert A\Vert \Vert x_{\ast }\Vert$.^[ref]

For the Jacobi iteration for a three-variable tridiagonal system, the first linear-system residual is $(1,3,1{)}^T$, while the first forward error is $(-1/2,-1,-1/2{)}^T$. In the Euclidean norm, the relative forward error is $\sqrt{3/28}\approx 0.3273$ and the relative linear-system residual is $\sqrt{11/120}\approx 0.3028$. Its spectral condition number of a positive-definite matrix is $(4+\sqrt{2})/(4-\sqrt{2})\approx 2.094$, so the bound is about $0.6340$. The bound is a guarantee, not an equality. In floating-point arithmetic, recompute the linear-system residual from the current vector when checking termination; a small update alone need not certify a small forward error.

Right preconditioning substitutes $x=M^{-1}u$ into the linear system $Ax=b$. Applying its transformed linear map first solves $Mz=v$, then computes $Az$. Recover the solution by solving $Mx=u$. Its linear-system residual $b-A{M}^{-1}u$ is exactly the original linear-system residual $b-Ax$, which is useful when monitoring the imbalance in the original equations.

The left preconditioning and right preconditioning transformed matrices are similar matrices:They therefore have the same eigenvalues, although their Euclidean norm condition numbers and residual histories need not agree. A similarity transformation preserves eigenvalues; a nonorthogonal change of coordinates need not preserve lengths.

Round-off error arises when numbers or arithmetic results are rounded to a finite set of representable values. Its scale is associated with machine epsilon, and its effects depend on subsequent operations. Even a formula with zero truncation error can lose information. In decimal arithmetic rounded to three significant digits, $1.234$ becomes $1.23$ and $1.233$ becomes $1.23$: subtracting the stored numbers gives zero although the exact difference is $0.001$. The subtraction exposes errors already present in its nearly equal operands.

This example also has a large condition number: independently perturbing the two inputs of $a-b$ can produce a relative amplification $(|a|+|b|)/|a-b|=2467$. High sensitivity belongs to the input-output problem; numerical stability concerns the arithmetic used to solve it. Conversely, on a common binary system with 53 significant bits and rounding to nearest, $(10^{16}+1)-10^{16}$ evaluates to zero while $(10^{16}-10^{16})+1$ evaluates to one. Exact arithmetic makes both one: changing summation order changes the accumulated roundoff error. MIT's summation lecture explains why numerical stability and the condition number must be assessed separately.^[ref]

A matrix over a field $\mathbb{F}$ is in row echelon form when all zero rows follow the nonzero rows, each nonzero row's first nonzero entry lies strictly to the right of the preceding row's first nonzero entry, and all entries below each such leading entry vanish. These leading entries are its pivot elements. They need not be one, and entries above them need not vanish.^[ref]

The staircase turns a linear system into a sequence of simpler equations. For example,is in row echelon form, with pivot columns 1 and 2. Its second equation determines $x_2$ once $x_3$ is chosen; its first then determines $x_1$ by back substitution. It is not in reduced row echelon form, because the second pivot column contains a nonzero entry above its pivot element. Neither being square nor having a nonzero entry in every row is required.

Elementary row operations bring every matrix to row echelon form: choose a nonzero entry in the leftmost available nonzero column, interchange rows, clear entries below it, and repeat on the remaining rows and later columns. Skip columns with no remaining nonzero entry. Clearing above each pivot element and scaling it to one completes reduced row echelon form.^[ref]

The nonzero rows of an row echelon form are linearly independent: in a vanishing linear combination, inspecting the first pivot column forces the first coefficient to be zero, then the next forces the second, and so on. They give a basis of the original row space. The number of pivot elements is both the row rank and the column rank. To recover a basis of the original column space, select the corresponding original pivot columns; the reduced column vectors generally belong to a different column space.

Here the arithmetic is exact in the specified field. For floating-point arithmetic, deciding whether a computed entry counts as zero requires a numerical criterion; see Tiny-pivot roundoff in Gaussian elimination and condition number. An exact row echelon form classification alone does not settle numerical stability.

The displayed Thomas algorithm requires every modified pivot element to be nonzero in exact arithmetic. Nonzero original diagonal entries alone are insufficient: later pivot elements can vanish after elimination. Two useful sufficient contexts are a real symmetric positive-definite matrix, and strict diagonal dominance by rows:For the symmetric positive-definite matrix case, elimination leaves positive-definite remaining blocks by the Schur complement, so each $p_i>0$. Equivalently, the initial principal minors ${\Delta }_i=\det A_{1:i,1:i}$ are positive, and $p_i={\Delta }_i/{\Delta }_{i-1}$ with ${\Delta }_0=1$.

For strict diagonal dominance, a short induction shows $|p_i|>|c_i|$. The first row gives $|p_1|=|b_1|>|c_1|$. Assuming the preceding bound, the triangle inequality givesIn particular $p_n\ne 0$. These are structural guarantees against a zero pivot element, not universal bounds on forward error in floating-point arithmetic. A poorly conditioned linear system can still amplify roundoff error; numerical stability and the condition number must be distinguished.

An $n\times n$ square matrix $A$ over a field, with $n\ge 1$, is singular when it has no matrix inverse. Geometrically, its linear map loses at least one input direction: some nonzero vector $v$ satisfies $Av=0$, so $A(x+v)=Ax$. Over the real numbers, the following family makes that loss visible:

For this diagonal matrix, $\det A(t)=t$. Whenever $t\ne 0$, the input can be recovered as $(x,y)=(b_1,b_2/t)$ and the kernel is $\{0\}$. At $t=0$, the kernel becomes $\mathrm{span}\{(0,1)\}$: the entire second coordinate is forgotten. Consequently $A(0)x=(b_1,b_2)$ has no solution if $b_2\ne 0$, and has the solution straight line $x=(b_1,s)$, $s\in \mathbb{R}$, if $b_2=0$. A singular matrix therefore need not make a particular linear system inconsistent; it prevents unique recovery when that linear system is consistent.

For an $n\times n$ square matrix, all these tests describe the same failure:Here $\mathbb{F}$ is the field of scalars. Equivalently, the columns are linearly dependent, the linear map is neither injective nor surjective, and some outputs cannot be reached. See the invertible matrix equivalences and Strang's MIT 18.06SC lecture summary, Multiplication and inverse matrices, p. 2. Over an infinite field, every consistent linear system for a singular matrix has infinitely many solutions; over a finite field with $q$ elements and nullity $d$, it has exactly $q^d$ solutions. This follows by adding each kernel vector to one solution.

The square hypothesis is essential. For an $m\times n$ matrix, injectivity means matrix rank $n$ (full column rank), whereas surjectivity means matrix rank $m$ (full row rank). A rectangular matrix can satisfy one condition without the other; its kernel and image of a linear map give the useful tests without calling it a singular matrix.

Exact singularity also differs from sensitivity. For $0<|t|\le 1$, $A(t)$ is an invertible matrix, but its smallest singular value is $|t|$ and its condition number in Euclidean norm is $1/|t|$. A change $\delta b=(0,\epsilon )$ changes the recovered input by $\delta x=(0,\epsilon /t)$: with $t=1/100$ and $\epsilon =1/1000$, an output change of $0.001$ causes an input change of $0.1$. These are exact toy values, not measurements. The condition number quantifies this sensitivity before a direction is lost; see Steve Marschner, Condition numbers, §2. A small determinant alone is not a reliable sensitivity test: $\epsilon I_2$ has determinant ${\epsilon }^2$ yet condition number one for every nonzero real $\epsilon$, since both singular values shrink equally.

The singular values of a matrix measure stretching of Euclidean norm. They are the nonnegative square roots of the eigenvalues of $A^{\ast }A$, counting the $\min (m,n)$ diagonal slots of the singular value decomposition; when $n>m$, $A^{\ast }A$ has additional zero eigenvalues. Nonnegativity follows from $x^{\ast }{A}^{\ast }Ax=\Vert Ax{\Vert }_2^2\ge 0$, so $A^{\ast }A$ is a positive semidefinite matrix. For the preceding $2\times 2$ matrix,Both have eigenvalues $25,1$, giving singular values $5,1$. In particular,Here the operator norm uses Euclidean norm: no unit vector stretches more than fivefold. Zero singular values identify lost directions; the number of positive singular values equals the matrix rank. This summarizes both amplification and information loss with nonnegative numbers.^[ref]

Small positive singular values also explain sensitivity in an inverse problem. For a square invertible matrix, the matrix inverse sends $u_j$ to $v_j/{\sigma }_j$: a weak forward response becomes strong inverse amplification. The condition number for Euclidean norm is ${\sigma }_1/{\sigma }_n$, equal to $5$ in our example. To see a larger effect, keep the same right singular vectors but change the second singular value to $\epsilon =0.01$:The output perturbation has Euclidean norm $0.001$, but the recovered input changes by $0.1$. Its hundredfold amplification follows from the direction of the perturbation, not just its size. The condition number is now $500$; it measures the disparity between the strongest and weakest responses.^[ref]

Let $A$ and $M$ be real symmetric positive-definite matrices. The principal square root of a positive semidefinite matrix construction gives an invertible matrix for each. SetThen $Ax=b$ is equivalent to $By=c$. This is a matrix congruence, and $B$ is a symmetric positive-definite matrix: $B^T=B$, and for every nonzero vector $v$,since $S^{-1}v\ne 0$. This short proof establishes both requirements for applying the conjugate gradient method to the transformed linear system.

The left transformed matrix has the same eigenvalues becauseIts correct geometry uses the weighted inner product $⟨u,v{⟩}_M=u^{T}Mv$: $⟨u,M^{-1}Av{⟩}_M=u^{T}Av=⟨M^{-1}Au,v{⟩}_M$. Thus it is symmetric in this inner product, even when it is not a symmetric matrix in ordinary coordinates. The spectral condition number governing the transformed symmetric problem is ${\kappa }_2(B)$, not generally ${\kappa }_2(M^{-1}A)$.

The Preconditioned conjugate gradient method realizes this transformation using auxiliary solves $M{z}_k=r_k$, without constructing $S$ or its matrix inverse. With fixed $M$, initialize $r_0=b-A{x}_0$ and $p_0=z_0$; while $r_k\ne 0$, usethen solve $M{z}_{k+1}=r_{k+1}$ and setStop at a zero linear-system residual before forming another ratio. Positive definiteness makes the denominators positive before termination.^[ref]

An invertible matrix alone is insufficient: with $A=I$ and $M=\mathrm{diag}(-1,1)$, the left transformed matrix has eigenvalues $-1,1$, and $M$ has no real symmetric positive-definite matrix square root. Even a valid symmetric positive-definite matrix can worsen conditioning if its scaling is unsuitable, as the next worked example shows.