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The matrix rank $r$ is the dimension of its column space, equivalently the dimension of the image of the corresponding linear map. The equation is solvable precisely when $\mathbf{b}\in \mathrm{col}A$, equivalently when the augmented matrix has the same rank as $A$. If ${\mathbf{x}}_0$ is one solution, every solution is ${\mathbf{x}}_0+\mathbf{u}$ with $\mathbf{u}\in \ker A$, since subtracting any two solutions gives a kernel element. The rank-nullity theorem gives $\dim \ker A=3-r$.

Consequently, for rank three there is one solution for every $\mathbf{b}$, namely $A^{-1}\mathbf{b}$. For rank two there are no solutions outside the column plane, and an affine solution space of a linear equation of dimension one when consistent. For rank one, consistency requires $\mathbf{b}$ to lie on the column line and the solution set is an affine plane of dimension two. For rank zero, $A=0$: the solution set is all of ${\mathbb{R}}^3$ if $\mathbf{b}=0$, and empty otherwise. These descriptions include homogeneous systems, whose solution spaces pass through the origin.

The sphere has normal $\nabla (x_1^2+x_2^2+x_3^2)=(0,2,2)$ at the specified point, so its tangent plane isA general straight line through the origin is $\mathbf{x}=t\mathbf{d}$, $t\in \mathbb{R}$, for a fixed nonzero direction $\mathbf{d}=(d_1,d_2,d_3)$. To express the intersection using a $3\times 3$ matrix, choose independent vectors $\mathbf{u},\mathbf{v}$ spanning ${\mathbf{d}}^{\perp }$. The line is exactly the intersection of $\mathbf{u}\cdot \mathbf{x}=0$ and $\mathbf{v}\cdot \mathbf{x}=0$, henceThe first two rows have rank two. The third is independent of them exactly when $(0,1,1)\cdot \mathbf{d}=d_2+d_3\ne 0$. In that case the rank is three and the unique intersection isGeometrically the line crosses the tangent plane transversely. If $d_2+d_3=0$, the matrix rank is two: its third row is a linear combination of the first two, but the corresponding right-hand-side combination would be zero rather than two. The system is inconsistent and there is no intersection. The line is parallel to the tangent plane and cannot lie in it, since it contains the origin while the plane does not. Thus no choice of a genuine direction produces an infinite intersection in this particular problem.

Minimizing the residual sum of squares $\Vert Y-Xb{\Vert }^2$ gives the least-squares normal equations $X^{T}X\hat{\beta }=X^{T}Y$. Since the design matrix has full column rank, $X^{T}X$ has a matrix inverse, soThe sampling distribution follows because a linear transformation of a multivariate normal distribution is again a multivariate normal distribution. Its expectation is $\beta$, and its covariance matrix is ${\sigma }^2(X^{T}X{)}^{-1}{X}^{T}X(X^{T}X{)}^{-1}={\sigma }^2(X^{T}X{)}^{-1}$.

The hat matrix is $H=X(X^{T}X{)}^{-1}{X}^T$. It is the orthogonal projection matrix onto the column space of $X$: $H^T=H$, $H^2=H$ and $HX=X$. Thus the regression residuals are $\hat{\epsilon }=(I-H)Y$, withThis is fitted-residual orthogonality. The vectors are jointly normal, so independence of uncorrelated jointly normal variables strengthens both zero-covariance statements to independence.

For the quadratic linear regression, the columns of the design matrix are $1,x,x^2$. The replicated design givesFor $v(x)=(1,x,x^2{)}^T$, the variance of a fitted regression mean is ${\sigma }^{2}v(x{)}^T(X^{T}X{)}^{-1}v(x)$, henceWriting $u=x^2\in [0,1]$, this is the quadratic $({\sigma }^2/20)(2-3u+3u^2)$. Its derivative vanishes at $u=1/2$; it decreases before that point and increases afterwards. Therefore the maximum is ${\sigma }^2/10$ at $x=-1,0,1$, and the minimum is ${\sigma }^2/16$ at $x=\pm 1/\sqrt{2}$.

At $100^{\circ }\mathrm{C}$, $x=0$ and the expected yield is $\alpha$. The unbiased residual variance estimate is $s^2=2.43/(30-3)=0.09$. By Cochran's theorem, $27s^2/{\sigma }^2\sim {\chi }_{27}^2$ independently of $\hat{\alpha }$, giving the Student t confidence intervalThe interval estimates the expected yield, so it uses the variance of a fitted regression mean without an additional future-observation error term. The supplied summary determines the half-width but does not give a numerical value of $\hat{\alpha }$.

The system of linear equations is consistent precisely when $\mathbf{b}$ belongs to the column space of $A$. If ${\mathbf{x}}_0$ is one solution, all solutions form the affine subspacebecause subtracting two solutions produces a vector in the kernel, and adding any kernel vector preserves the equation. Therefore the necessary and sufficient cases are:

For the matrix equation, each column of $X$ solves the same system of linear equations with the corresponding column of $B$. Thus a unique $X$ exists if and only if $A$ is invertible, and then $X=A^{-1}B$. Necessity does not depend on $B$: if $A$ is singular and one $X$ exists, put any nonzero kernel vector in one column of a matrix $Y$ and zeros in its other columns. Then $AY=0$ and every $X+tY$ is another solution.

For the specified data, $\det A=3$. If ${\mathbf{x}}_j$ denotes row $j$ of $X$ and ${\mathbf{b}}_j$ row $j$ of $B$, row elimination yields ${\mathbf{x}}_3=(2{\mathbf{b}}_1-{\mathbf{b}}_2-{\mathbf{b}}_3)/3$, ${\mathbf{x}}_1={\mathbf{b}}_2-{\mathbf{x}}_3$, and ${\mathbf{x}}_2={\mathbf{b}}_1-{\mathbf{b}}_2-{\mathbf{x}}_3$. HenceMultiplying $AX$ gives the prescribed $B$, which both checks the arithmetic and confirms the unique answer.

Full column matrix rank makes $X^{T}X$ a positive-definite matrix, and its matrix inverse is symmetric. Thus the hat matrix satisfiesAlso $PX=X$ and the image of $P$ is contained in the column space of $X$. These identities show that $P$ is the orthogonal projection matrix onto the column space of $X$.

The ordinary least squares objective has gradient $2X^{T}X\beta -2X^{T}Y$. The normal equation therefore givesThe Gram matrix $X^{T}X$ is a positive-definite matrix because $X$ has full column rank. More explicitly, for any $b$,since $X^T(Y-X\hat{\beta })=0$. This proves that the displayed solution is the unique global minimum.

The fitted values are $\hat{Y}=X\hat{\beta }=HY$, where the hat matrix isThe inverse of the symmetric matrix $X^{T}X$ is symmetric, so $H^T=H$. Multiplication gives $H^2=X(X^{T}X{)}^{-1}(X^{T}X)(X^{T}X{)}^{-1}{X}^T=H$. Thus $H$ is the orthogonal projection onto the column space of the design matrix.

The regression residual vector is $e=Y-\hat{Y}=GY$ with $G=I_n-H$. Consequently $G^T=G$, $G^2=I_n-2H+H^2=G$, and the residual sum of squares isBy fitted-residual orthogonality, $GH=0$. Using $\mathrm{Cov}(Y)={\sigma }^2{I}_n$, the cross-covariance matrix isBoth vectors are jointly normal, so they are also independent by independence of uncorrelated jointly normal variables. The zero covariance calculation itself only needs the common error variance and lack of error correlations.

For the first wind fit, let $v_i$ be wind velocity and $Y_i$ electrical output. The simple linear regression is $Y_i=\alpha +\beta v_i+{\epsilon }_i$, with independent ${\epsilon }_i\sim N(0,{\sigma }^2)$. It has two fitted mean parameters and $25-2=23$ residual degrees of freedom. The missing analysis of variance entries are therefore velocity degrees of freedom $1$, velocity mean square $8.9296$, and residual degrees of freedom $23$. The residual mean square is $1.2816/23\simeq 0.05572$, and $8.9296/(1.2816/23)\simeq 160.25$, consistent with the printed value after rounding.

Because the model includes a regression intercept, the coefficient of determination is the explained sum of squares divided by the total centered sum of squares:A large coefficient of determination does not rule out a wrong mean function. In the PDF's first residual-versus-fitted plot, residuals are negative at both ends and positive in the middle. This curved pattern agrees with the visibly flattening output-versus-velocity relationship and motivates a polynomial regression with a quadratic term.

The second wind model is $Y_i=\alpha +{\beta }_1{v}_i+{\beta }_2{v}_i^2+{\epsilon }_i$, again with independent common-variance normal errors. Its fitted mean is $-1.155898+0.722936v_i-0.038121v_i^2$. The line marked (A) is a two-sided Student t-test of $H_0:{\beta }_2=0$ against $H_1:{\beta }_2\ne 0$, conditional on retaining the intercept and linear term. Under the null hypothesis,Its two-sided p-value is $6.59\times 10^{-8}$. This is far below $0.05$, so reject a purely linear mean in favour of the quadratic fit. Equivalently, the extra-term nested-model F-test has $F=t^2\simeq 63.15$ and null law $F_{1,22}$.

The third fit is a reciprocal-predictor regression, $Y_i=a+b/v_i+{\epsilon }_i$, with estimated mean $2.9789-6.9345/v_i$. It has two mean parameters rather than the quadratic model's three. Its residual standard error is smaller, $0.09417$ versus $0.1227$, and its coefficient of determination is larger, $0.9800$ versus $0.9676$. The corresponding residual sums of squares are approximately $23(0.09417{)}^2=0.20396$ and $22(0.1227{)}^2=0.33122$. The reciprocal fit is preferable on both these fit measures and parsimony. The two models are not nested, so an ordinary extra-term F-test between them is inappropriate. On the common normal-error likelihood, the difference ${\mathrm{AIC}}_3-{\mathrm{AIC}}_2=25\log (0.20396/0.33122)-2\simeq -14.12$ also favours the third model.

Neither lower residual sum of squares nor a higher coefficient of determination establishes adequate assumptions. The quadratic residual-versus-fitted plot removes the original pronounced curvature; the reciprocal plot also has no comparably obvious mean trend. The low-output residuals appear somewhat more spread out, so check scale-location plots and residuals against velocity for heteroscedasticity. Q-Q plots assess normality; residuals against observation order assess serial dependence; regression leverage and Cook's distance identify influential observations. Further cross-validation, replicate observations at comparable velocities, and prediction errors would help choose between their extrapolation behaviours. Both fitted shapes should be judged principally over the observed positive-velocity range.

There are two defects. The mixed fit supplies a restricted maximum likelihood, whereas the ordinary fit supplies an ordinary log-likelihood, so their difference is not a likelihood-ratio test statistic. Also, $H_0:{\tau }^2=0$ lies on the boundary of ${\tau }^2\ge 0$; the usual Wilks theorem does not give a ${\chi }_1^2$ null law.

A valid approach first refits the mixed model by ordinary maximum likelihood estimation, setting REML to false, and fits the same fixed effects under $H_0$. For $V={\tau }^{2}Z{Z}^T+{\sigma }^{2}I$, maximizeover $\beta ,{\sigma }^2>0,{\tau }^2\ge 0$, and separately over $\beta ,{\sigma }^2>0$ with ${\tau }^2=0$. Set $T=2({\hat{\ell }}_1-{\hat{\ell }}_0)\ge 0$.

For finite-sample calibration, use the location-scale invariant simulation test for a Gaussian variance component. With the actual $X,Z$ fixed, simulate $B$ independent vectors $Y^{(b)}\sim N(0,I_{176})$, refit both models by ordinary maximum likelihood estimation to each, and calculate $T_b$ in exactly the same way. Under $H_0$, $Y=X\beta +\sigma z$. Translating by a vector in the column space of $X$ and multiplying by a positive scalar preserves the statistic: both maximized log-likelihoods acquire the same scale constant. Thus this simulation has the correct null law without knowing $\beta$ or $\sigma$. A conservative Monte Carlo p-value isIn the usual regular limit with increasing independent groups, a variance-component likelihood-ratio test at a boundary instead uses $\frac{1}{2}{\delta }_0+\frac{1}{2}{\chi }_1^2$: for $T>0$, its approximate tail probability is $\frac{1}{2}\mathrm{Pr}({\chi }_1^2\ge T)$, and at $T=0$ the nonrandomized p-value is one. This approximation is not an exact guarantee for sixteen rats. Alternatively one can compare consistently defined restricted likelihoods with the same $X$ and simulate their restricted-ratio null distribution, as in the RLRsim documentation.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.

The rank of a matrix is the dimension of its column space, equivalently the dimension of the image of a linear map represented by the matrix. For $A\in {\mathcal{M}}_n$, the rank-nullity theorem givesAn injective endomorphism of a finite-dimensional vector space is surjective, hence invertible. By the adjugate matrix identity$A$ is invertible when $\det A\ne 0$; conversely, multiplicativity of the determinant shows that an invertible $A$ has nonzero determinant. Therefore

Let $E_{ij}$ be the matrix unit with its only nonzero entry at $(i,j)$. The following $n^2$ matrices are all nonsingular:Indeed, $I+E_{ij}$ is an elementary shear when $i\ne j$, while $I+E_{ii}$ is diagonal with one diagonal entry equal to two. Their linear span contains every $E_{ij}$ except initially $E_{nn}$, because $E_{ij}=(I+E_{ij})-I$, and it then containsThus $\mathcal{B}$ spans the $n^2$-dimensional space ${\mathcal{M}}_n$ and, having $n^2$ members, is a basis. This also covers $n=1$, when $\mathcal{B}=\{I\}$.

Now let $A$ be a nonsingular zero-one matrix. If it had fewer than $n-1$ zero entries, at least two rows would contain no zero at all. Those two rows would both be the all-one row, contradicting linear independence. Hence every such matrix has at mostones. The bound is attained. Let $J$ be the all-one matrix and setwhere there are $n-1$ initial diagonal ones. If $A\mathbf{x}=0$ and $S={\sum }_j{x}_j$, the first $n-1$ row equations give $x_i=S$, while the last gives $S=0$. Hence every $x_i=0$, so $A$ is nonsingular and has exactly $n^2-n+1$ ones. Therefore

Because $X$ has full column rank, $X^{T}X$ is invertible. The ordinary least squares estimator isIt satisfies the normal equationsso the residual is orthogonal to the column space of $X$. For any $\beta$,The two terms are orthogonal, and the Pythagorean theorem in an inner-product space givesThus $\hat{\beta }$ minimizes the least-squares objective.

A pivot column of a matrix is a column whose position contains a leading nonzero matrix entry in its row echelon form. The corresponding columns of the original matrix form a basis of its column space. The column positions survive row operations, although the column vectors themselves generally change.^[ref]

A reduced QR decomposition of an $m\times k$ matrix over the real numbers with full column rank, $m\ge k$, writes $A=QR$, where $Q$ is $m\times k$ with orthonormal columns and $R$ is a $k\times k$ upper triangular matrix with positive diagonal. Applying the Gram-Schmidt process to the columns $v_j$ of $A$ constructs $Q=(q_1\cdots q_k)$ and records the coefficientsThis last column identity is exactly the matrix multiplication $A=QR$. The upper triangular matrix records that the $j$th input uses only the first $j$ new directions, while its positive diagonal records the norms of the nonzero orthogonal projection residuals. Since $Q^{T}Q=I$, one also has $R=Q^{T}A$; $Q^T$ denotes the transpose and $I$ the identity matrix. Over the complex numbers, use the conjugate transpose $Q^{\ast }$ in these identities.

For the three-dimensional Gram-Schmidt process example, the exact matrices areFor example, the last column reconstructs asThe diagonal entries are approximately $1.414214,1.224745,1.154701$. Each is a norm of the new direction before it was scaled to length one.

For a linear least-squares problem, the orthogonal projection of $b$ onto the column space of $A$ is $Q{Q}^{T}b$. Hence the minimizing coefficients solve the linear system $Rx=Q^{T}b$ by back substitution. Let $A_2=(v_1{v}_2)$ consist of the first two columns of the example; its reduced factors are $Q_2=(q_1{q}_2)$ and the leading $2\times 2$ block $R_2$ of $R$. For $b=(0,0,1{)}^T$, the linear system $R_{2}x=Q_2^{T}b$ isThus $A_{2}x=(1/3,-1/3,2/3{)}^T$, agreeing with the orthogonal projection computed above, and the minimum Euclidean norm of $A_{2}x-b$ is $1/\sqrt{3}$. The QR decomposition separates building an orthonormal basis from recovering coefficients in the original basis.^[ref]

For row reduction with a parameterized right-hand side at $\tau =5$, the nonzero reduced equations areColumn 3 has no pivot element, so $x_3$ is a free variable in a linear system. Taking $x_3=t$ gives every solution:There is no further restriction on $t$. Independent substitution into the original linear system givesConversely, the two reduced equations force exactly these coordinate formulas, so no solutions have been omitted. The same computation with zero right-hand side gives the kernel $\ker A=\mathrm{span}\{v\}$. Its dimension is one. The rank-nullity theorem appears here as $2+1=3$: two pivot columns and one freely chosen coordinate account for the three unknowns.^[ref]

In general a consistent linear system has solution set $x_0+\ker A$. Subtracting $A{x}_0=b$ from $Ax=b$ proves this description in both directions. The particular solution places the affine subspace, and the kernel supplies its directions.^[ref] Over the real numbers this example is an entire line; over the rational numbers the same formula uses rational $t$. A free variable in a linear system over a finite field instead has only finitely many choices.

The pivot columns identify a different object: the original column space. Here they selectwhich form its basis, while $a_3=-7a_1+3a_2$. The corresponding reduced columns are $(1,0,0,0{)}^T$ and $(0,1,0,0{)}^T$; their linear span is a different plane. For example, every original output satisfies $y_3=y_1+y_2$, but $(1,0,0,0{)}^T$ does not. Since $E$ is an invertible matrix, column linear relations survive $EA=R$ even though the output vectors move. This is why choosing original columns at the reduced pivot positions works.

Consider the linear map $A:{\mathbb{R}}^3\to {\mathbb{R}}^2$ represented by the rectangular matrixTo compute its kernel, solve the homogeneous linear system $x+2y=0$, $y+z=0$. Taking $y=t$ gives $x=-2t$, $z=-t$, soThus $v$ represents a null direction of a linear map: decreasing the first input by two, increasing the second by one, and decreasing the third by one changes neither output. Solving the homogeneous linear system is the concrete computation behind a kernel; the column space instead decides which outputs are attainable.^[ref]

Now prescribe the nonzero output $b=(3,2{)}^T$. The linear system $Au=b$ becomes $x+2y=3$, $y+z=2$. Again take $y=t$:Direct substitution gives $A{u}_0=(3,2{)}^T$, and $A(u_0+tv)=A{u}_0+tAv=b$. Conversely, the two linear equations force exactly this form, so no solutions have been omitted. This fiber is an affine subspace of dimension one in the three-dimensional input vector space. It is parallel to the kernel, but does not pass through the origin. For instance, $u_0=(3,0,2{)}^T$ and $u_1=(1,1,1{)}^T$ both give $b$, while their difference $u_1-u_0=v$ gives zero. Adding $u_0$ and $u_1$ gives output $2b$, which is another way to see that this nonzero-output fiber is not a vector subspace.

The first and third columns of $A$ are $(1,0{)}^T$ and $(0,1{)}^T$, a basis of the output vector space. Hence $A$ has matrix rank two and reaches every output. The rank-nullity theorem predicts nullity $3-2=1$ before we solve any linear system. It provides a useful completeness check: one nonzero kernel vector already forms a basis here, so a second independent null direction of a linear map would signal a calculation error. In general, matrix rank $r$ for an $m\times n$ matrix leaves $n-r$ independent kernel directions, rather than $m-r$.^[ref]

The row space of a matrix is the span of its rows, and its column space is the span of its columns. Both have dimension equal to the matrix rank, though they lie in different ambient coordinate spaces for a rectangular matrix.

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.

Two matrices of the same size over the same field are row-equivalent matrices if a sequence of elementary row operations transforms one into the other.^[ref] Every operation is reversible, so they have the same row space and the same column linear relations. Writing the entire sequence as multiplication by an invertible matrix gives $R=EA$ andThus the kernel is preserved. For a nonzero right-hand side, perform every operation on the whole augmented matrix $[A\mid b]$: changing the coefficients while keeping $b$ fixed generally changes the solutions. The column space instead transforms to $E\mathrm{col}A$, which need not equal $\mathrm{col}A$.

The row rank of a matrix is the dimension of the vector space spanned by its rows. Elementary row operations preserve this row space; multiplication on the right by an invertible matrix preserves all linear relations among the rows.

To see why row rank equals column rank, write a reduced row echelon form as $R=EA$, where $E$ is a product of elementary matrices and is an invertible matrix. Reversibility of the row operations gives equal row spaces for $R$ and $A$. Also $Rz=0$ if and only if $Az=0$, so the columns have exactly the same linear relations. If $R$ has $r$ nonzero rows, its pivot columns are the first $r$ standard basis vectors of ${\mathbb{F}}^m$ and span its column space. Its nonzero rows are linearly independent, since their matrix entries in the pivot positions form an identity matrix. Both dimensions therefore equal $r$, and both counts transfer back to $A$.

The row space of a matrix is the linear span of its rows. It is the column space of the matrix transpose.

A system of linear equations imposes several linear equations on the same unknowns. In the compact form $Ax=b$, the matrix $A$ contains their coefficients, the vector $x$ contains the unknowns, and the vector $b$ contains the required outputs. Over the real numbers, the following examples exhibit the three possible outcomes: one solution, infinitely many solutions, or no solution.

For $x+y=3$ and $x-y=1$, adding the two linear equations gives $2x=4$, hence $x=2$ and $y=1$. Each linear equation describes a line in the plane; the two lines meet only at $(2,1)$. Substitution checks both requirements: $2+1=3$ and $2-1=1$. The two measurements together determine both unknowns.

Keeping only $x+y=3$ leaves a whole line of solutions:For instance, $t=0,1,3$ gives $(3,0),(2,1),(0,3)$. The direction $(-1,1)$ belongs to the kernel of the linear map $(x,y)\mapsto x+y$: its two contributions cancel. Moving in that direction changes the unknowns without changing their measured sum. The solution line is an affine space, translated away from the origin.

For $x+y=3$ and $2x+2y=7$, the second linear equation instead demands $x+y=3.5$. The parallel lines never meet. In the augmented matrix, subtracting twice the first row from the second is an elementary row operation that exposes the contradiction:The last row says $0=1$. A zero row of coefficients is harmless only when its appended entry is also zero.

The column viewpoint unifies these pictures. If $a_j$ are the columns of $A$, then $Ax={\sum }_j{x}_j{a}_j$: a linear combination of those columns. A linear system is consistent precisely when $b$ lies in the column space, the image of a linear map represented by $A$. Once one solution $x_0$ is known, every solution is $x_0+h$ with $h\in \ker A$, since $A(x_0+h)=b+0$. Conversely, subtracting any two solutions gives an element of the kernel. Thus the kernel measures freedom, while the image of a linear map decides existence. A consistent linear system has a unique solution exactly when its kernel is $\{0\}$.^[ref]

Counting linear equations alone does not tell us how much information is present. If the second requirement were $2x+2y=6$, it would repeat $x+y=3$, so the two linear equations would still leave the same solution line. The matrix rank counts independent requirements. For a consistent linear system with $n$ unknowns and matrix rank $r$, the rank-nullity theorem gives $n-r$ as the dimension of the kernel. This is the number of independent directions in which a solution can move. In the single-measurement example, one measured combination leaves one direction of freedom; in the sum-and-difference example, neither unknown can change without changing an output.

The two-factor tensor rank is the smallest number of pure tensors whose sum equals $t$, with zero allowed as the empty sum. For two finite-dimensional vector spaces over the same field, it equals the matrix rank of the tensor-product coefficient matrix. A sum of $r$ outer products has column space contained in the span of their $r$ first factors, so its matrix rank is at most $r$. Conversely, choose a basis $u_1,\dots ,u_r$ for the column space, express each column in that basis, and collect the coefficients into row vectors $w_k^T$. This yields $C={\sum }_{k=1}^r{u}_k{w}_k^T$, hence a sum of $r$ pure tensors.

The pure tensor $(2,1)\otimes (-1,3)$ therefore has two-factor tensor rank one, whereas $e_1\otimes f_1+e_2\otimes f_2$ has two-factor tensor rank two. Both have two factors: the number of factors and the minimum number of product summands are different counts. Over the real numbers, the singular value decomposition supplies a useful decomposition $C={\sum }_k{\sigma }_k{u}_k{v}_k^T$, with one nonzero term per matrix rank. The two-factor rank equality does not extend to three or more factors by merely choosing one matrix layout.