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A nonzero pivot element can still be a poor numerical choice. Partial pivoting bounds the absolute values of the elimination multipliers by one in exact arithmetic, because each numerator is no larger than the selected denominator. This usually improves numerical stability, but does not prevent every possible growth of intermediate entries or guarantee accurate answers for every linear system.^[ref]

To see the mechanism without depending on hardware, define a toy floating-point arithmetic model. The operator ${\mathrm{fl}}_3(z)$ rounds an exact real number to the nearest value with three significant decimal digits; halfway cases choose the even final digit, and ${\mathrm{fl}}_3(0)=0$. Allow an unbounded exponent range. Round the inputs and the result of every addition, subtraction, multiplication and division separately; a product followed by a subtraction rounds twice. This is a prescribed arithmetic model, not a measurement of a physical computer. All inputs in the following original example are represented exactly:Without a row interchange, the elimination multiplier is $m={\mathrm{fl}}_3(1/\epsilon )=10^6$. The last coefficient and right-hand side becomeTwo distinct exact values now have the same rounded representation. Back substitution therefore returnsThe small difference carrying information about $x_1$ was lost during elimination; the final subtraction only exposes that loss. Dividing by the tiny pivot element amplifies sensitivity to the lost information.

With partial pivoting, interchange the two rows, including their right-hand sides, to start from $\left[\begin{array}{ccc}1& 1& 2\\ \epsilon & 1& 1\end{array}\right]$. Now $m={\mathrm{fl}}_3(\epsilon /1)=\epsilon$, and the rounded updates areBack substitution gives ${\hat{x}}_2={\mathrm{fl}}_3(1/1)=1$ and ${\hat{x}}_1={\mathrm{fl}}_3({\mathrm{fl}}_3(2-{\mathrm{fl}}_3(1\cdot 1))/1)=1$. Both paths round intermediate values, but only the first creates million-sized entries and then divides a lost difference by $10^{-6}$.

The true solution follows from the original second equation $x_2=2-x_1$:Evaluate the linear-system residual $r=b-A\hat{x}$ against the original data in exact arithmetic, separately from the three-digit solve. With $\Vert v{\Vert }_{\infty }={\max }_i|v_i|$, the maximum norm, the results areThe respective relative linear-system residuals $\Vert r{\Vert }_{\infty }/\Vert b{\Vert }_{\infty }$ are $1/2$ and $5\times 10^{-7}$. Computing the diagnostic in higher precision matters: a check made only against the rounded reduced equations would miss the original error.

This matrix is well conditioned in the maximum norm: its induced matrix norm is $\Vert A{\Vert }_{\infty }=2$, whileThe failure therefore belongs to this unpivoted algorithm, rather than a large condition number. In general, however, partial pivoting cannot remove sensitivity inherent in the data. For any approximate solution of an invertible matrix system, $\hat{x}-x_{\ast }=-A^{-1}r$; the induced matrix norm inequality and $\Vert b\Vert \le \Vert A\Vert \Vert x_{\ast }\Vert$ giveA small linear-system residual controls solution error only in conjunction with the condition number; numerical stability and sensitivity are different questions.^[ref]

A linear map $T:V\to W$ has a trivial kernel when $\ker T=\{0_V\}$: the zero vector is the only input producing the zero output. The kernel is a singleton set, not the empty set, because every linear map sends zero to zero. Equivalently, its nullity is zero and $T$ is injective. No finite-dimensional hypothesis is required.^[ref]

The reason this zero-output test controls all outputs is linearity:With a trivial kernel, equal outputs force equal inputs. Therefore every equation $T(x)=b$ has at most one solution. Existence is a separate question: $b$ must belong to the image of a linear map. In the fiber description under nullity, $x_0+\{0\}$ is just the single point $x_0$.

For example, the linear map $J:{\mathbb{R}}^2\to {\mathbb{R}}^3$, $J(x,y)=(x,y,0)$, has trivial kernel and full column rank:The output $(3,4,0)$ comes from exactly $(3,4)$, while $(3,4,1)$ is unreachable. Conversely, $P:{\mathbb{R}}^3\to {\mathbb{R}}^2$, $P(x,y,z)=(x,y)$, is surjective and has full row rank, but its kernel is $\mathrm{span}\{(0,0,1)\}$: every output $(a,b)$ comes from the solution straight line $(a,b,s)$.

For an $m\times n$ matrix, a trivial kernel is equivalent to matrix rank $n$ and linear independence of its columns. It can therefore occur only if $n\le m$. If $m=n$, the rank-nullity theorem also forces surjectivity, giving an invertible matrix; if $m>n$, the example $J$ shows why unique recovery of reachable inputs does not imply that every output is reachable. Finally, a trivial kernel guarantees exact uniqueness without guaranteeing numerical stability: the singular matrix example $A(t)$ has trivial kernel for every $t\ne 0$, even as its condition number grows without bound when $t$ approaches zero.