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False. Take $A=(1,0{)}^T$, an $m=2$, $n=1$ matrix of matrix rank one, and $b=(0,1{)}^T$. Then $Ax=(x,0{)}^T$ never equals $b$. Full column rank implies injectivity, not surjectivity onto a larger codomain.

False. Triviality of $\mathcal{N}(A)$ gives $r=n$, while triviality of $\mathcal{N}(A^T)$ gives $r=m$ because transpose preserves matrix rank. Both conditions therefore force $m=n$, contradicting the stated unequal dimensions.

For the first matrix choose its single independent column:Their product has every column equal to $B_1$, and both factors have rank one.

For the second matrix, the first and third columns are independent and the second equals the first, givingMultiplication reconstructs the stated matrix. The first factor has independent columns and the second has independent rows, so both have matrix rank two, as required for a rank factorization.

The reduced row echelon form gives $u_1=2u_4/21$, $u_2=-3u_4/7$, and $u_3=-u_4/7$. Taking $u_4=21t$ gives every vector in the null space as $t(2,-9,-3,21{)}^{\top }$. Thus a basis for the kernel of the linear map isDirect multiplication gives $A(2,-9,-3,21{)}^{\top }=0$. The rank-nullity theorem also checks the count: matrix rank three plus nullity one equals the domain dimension four.

Gaussian elimination gives the row operations $R_2\leftarrow R_2-3R_1$, $R_3\leftarrow R_3-R_1$, $R_4\leftarrow R_4-2R_1$, producing the lower three rows $(0,1,4,1)$, $(0,1,4,1)$, $(0,-3,2,-1)$. Subtract the new row two from row three and add three times row two to row four; the remaining nonzero row is $(0,0,14,2)$. Move it above the zero row and finish elimination to obtainThe pivot columns are one, two and three. Row operations preserve the linear dependence relations among columns, so the corresponding columns of the original matrix form a basis for the column space, which is the image of the linear map:In particular, the matrix rank is three. Columns of the reduced matrix itself would generally give a different subspace, so the original columns must be used.

The $m$ columns of $L$ form a basis of $\ker A^{\top }$, so that null space has dimension $m$. The rank-nullity theorem applied to $A^{\top }:{\mathbb{R}}^n\to {\mathbb{R}}^n$ gives $\mathrm{rank}{A}^{\top }=n-m$. A matrix and its transpose have equal matrix rank, hence

Both $L$ and $R$ have $m$ linearly independent columns. The Gram matrix $R^{\top }R$ is invertible becauseConsequently $R^{\top }$ is a surjective linear map: for any $y\in {\mathbb{R}}^m$, the choice $x=R(R^{\top }R{)}^{-1}y$ satisfies $R^{\top }x=y$. HenceThe latter image of a linear map has dimension $m$, giving $\mathrm{rank}B=m$. If $m=0$, $B=0$ and the conclusion is immediate. This argument explains why the product retains its full possible matrix rank; a general product of two matrices need not do so.

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 $LU=A$.

Regard $A$ as a linear map ${\mathbb{R}}^4\to {\mathbb{R}}^4$. By the rank-nullity theorem and matrix rank $3$,All three are vector subspaces of ${\mathbb{R}}^4$, with the row space identified with four-component row vectors in the usual way. The null space lies in the input vector space of concentrations; the column space lies in the output vector space of concentration-change rates. Both have four coordinates, although they play different physical roles.

In this particular matrix, the row sums and column sums both vanish. Put $1=(1,1,1,1{)}^T$. Then $A1=0$ and $1^{T}A=0$, and the two three-dimensional vector subspaces coincide:This equality follows here from both sum identities and the dimensions of a vector space; it does not make $A$ a symmetric matrix and is not true for every square matrix.

A degree-two projective plane curve, allowing singular and nonreduced cases. In characteristic different from two, write its equation as $z^{\mathsf{T}}Mz$ with $M$ a symmetric matrix. It is smooth exactly when $M$ has matrix rank three. Over an algebraically closed field, rank two gives two distinct projective lines, while rank one gives a double line.

A polynomial regression of degree two is linear in its unknown regression coefficients despite being curved in the predictor. Its design matrix has columns $1,x,x^2$; three distinct predictor values give full column matrix rank.

Choose a random sampling matrix $X$, form $Y=AX$, and find an orthonormal basis $Q$ for the sampled column space by a thin QR decomposition. The low-rank approximation $\hat{A}=Q{Q}^{T}A$ keeps the component of $A$ in that space. Since $Q{Q}^{T}AX=AX$, it reproduces the sampled action exactly. If $Y$ has deficient matrix rank, one can use only a basis of its column space; alternatively an orthonormal completion must be specified.

Take independent uniform samples from a $d$-dimensional vector space over the two-element field. At current matrix rank $r$, the existing span contains $2^r$ of the $2^d$ possible vectors, so a new sample increases the matrix rank with probability $1-2^{r-d}$. The mean waiting time for that increase is $1/(1-2^{r-d})$ by the geometric distribution. Adding the waiting times gives $\mathbb{E}M=d+{\sum }_{j=1}^d(2^j-1{)}^{-1}<d+2$. For $d\ge 1$, the bound follows by separating $j=1$ and using $(2^j-1{)}^{-1}<2^{1-j}$ for $j\ge 2$; for $d=0$, no samples are needed.

After $m=d+t$ samples, failure to span implies that some nonzero linear functional vanishes on all samples. Each fixed nonzero functional does so with probability $2^{-m}$; the union bound over the $2^d-1$ possibilities gives failure less than $2^{-t}$. This provides both an expected sample count and a fixed-budget high-success version of Simon's algorithm.

If $\rho$ is a density operator with decreasing eigenvalues ${\lambda }_j$ and $T$ has matrix rank at most $k$, let $\Pi$ project onto the image of $T$. The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product givesThe Hermitian effect variational principle bounds $\mathrm{Tr}(\rho \Pi )$ by ${\sum }_{j<k}{\lambda }_j$, proving the claimed estimate without assuming $\rho$ is invertible.

For a matrix of matrix rank $r$, choose an ordered basis of its column space as the columns of $B$. Express every column of $A$ in that basis and put its coordinates in $C$. Then $A=BC$, $B$ has full column rank and $C$ has full row rank. Choosing independent original columns makes the corresponding columns of $C$ the standard basis vectors, directly proving its rank.

The basis choice is not unique: for any invertible $r\times r$ matrix $S$, the factors $BS$ and $S^{-1}C$ have the same product. Once $B$ is fixed, its coordinate matrix is unique. For rank zero, the empty factors of the displayed shapes represent the zero matrix.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

Two constant deformation gradients are rank-one connected when their difference is an outer product $a\otimes N$. For a nonzero jump its matrix rank is one; equality $A=B$ is the trivial compatible case. The vector $N$ specifies a reference interface normal, and $a$ is the jump amplitude. The Hadamard jump condition identifies this algebraic condition with continuity of an affine deformation across a planar interface. The normal can be normalized by rescaling $a$ without changing the product.

For coefficient column $q=(q_0,\dots ,q_d{)}^T$, $M_{ij}=q_i\overline{q_j}$ is a Hermitian matrix and a positive semidefinite matrix because $x^{\ast }Mx=|q^{\ast }x{|}^2$. Expanding $|q(z){|}^2$ on the unit circle gives coefficient $p_k={\sum }_{i-j=k}{M}_{ij}$. Thus $M$ has matrix rank one when $q\ne 0$ and zero otherwise. General feasible Gram matrices need not have matrix rank one.

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.

Let $A_i$ be real symmetric matrices with sum $I_n$. Their matrix ranks sum to at least $n$, because their ranges span ${\mathbb{R}}^n$. If the ranks sum to $n$, those ranges form a direct sum. For $y$ in the range of $A_i$, the identity $y={\sum }_j{A}_{j}y$ and uniqueness of its range decomposition imply $A_{j}y={\delta }_{ij}y$. Thus $A_i^2=A_i$ and $A_i{A}_j=0$ for $i\ne j$: the matrices are mutually orthogonal orthogonal projections. In a combined orthonormal basis, the forms $Z^T{A}_{i}Z$ for $Z\sim N_n(0,I_n)$ use disjoint sets of independent standard normal coordinates. They are independent chi-squared distributions with degrees of freedom equal to the ranks. Conversely, if all these forms have those chi-squared laws, their expectations and their sum give ${\sum }_i\mathrm{rank}{A}_i=n$.