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The augmented matrix of a linear system $Ax=b$ appends the right-hand side as one extra column. Elementary row operations preserve its solution set. The linear system is consistent exactly when the coefficient matrix rank equals the augmented matrix rank, because this is equivalent to $b$ lying in the span of the columns of $A$. A larger augmented matrix rank produces a zero coefficient row with a nonzero right-hand side during Gaussian elimination.

A complex nilpotent matrix of size six is determined up to matrix similarity by its matrix rank and nilpotency index. Its Nilpotent Jordan block sizes form an integer partition of six. If there are $k$ blocks and the largest size is $m$, then its matrix rank is $6-k$ and its minimal polynomial is $t^m$. The eleven possible size lists areNo two have the same pair $(m,k)$, which proves the assertion by Jordan normal form. The dimension restriction matters: in size seven, block lists $(3,3,1)$ and $(3,2,2)$ both have matrix rank four and minimal polynomial $t^3$, but give different Jordan normal forms.

The column space of a matrix $A$ is the linear span of its columns. It is the image of a linear map represented by $A$.

If an $m\times n$ matrix has columns $c_1,\dots ,c_n$, then matrix multiplication gives $Ax=x_1{c}_1+\cdots +x_n{c}_n$. Thus its column space consists of exactly the attainable output vectors in ${\mathbb{F}}^m$, and $Ax=b$ has a solution exactly when $b$ belongs to this column space. Its dimension is the matrix rank; the kernel instead describes input vectors in ${\mathbb{F}}^n$ whose contributions cancel.^[ref]

Imagine folding a sheet along its horizontal axis by the smooth map $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$, $F(x,y)=(x,y^2)$. The inputs $(a,r)$ and $(a,-r)$ have the same output. Along $y=0$, horizontal motion still changes the output, but vertical motion produces no first-order change: $F(a,\epsilon )-F(a,0)=(0,{\epsilon }^2)$. These inputs are critical points of a smooth map even though their differential is not the zero linear map. The missing direction in the target tangent space is what matters.

Precisely, let $f:X^m\to Y^n$ be a smooth map between smooth manifolds without boundary. An input $p\in X$ is a critical point of a smooth map when its differentialis not surjective, equivalently when its rank of a linear map is less than $n$. An input where it is surjective is a regular point; a map with no critical points of a smooth map is a submersion. The threshold is the target dimension, not $\min (m,n)$. In manifold charts, compute the matrix rank of the $n\times m$ Jacobian matrix; changes of charts preserve that matrix rank. ^[ref]

For the fold,Thus its set of critical points of a smooth map is the entire horizontal axis. Away from that axis, its differential can produce every infinitesimal target displacement. At the axis, it can produce only horizontal ones. This diagnoses where the two inverse branches in the picture merge.

For a scalar smooth function $f:X\to \mathbb{R}$, the target dimension is one. A linear map to $\mathbb{R}$ fails to be surjective exactly when it is zero, so here a critical point of a smooth map means $d{f}_p=0$. In Euclidean space this is precisely the familiar critical point condition $\nabla f(p)=0$: every first partial derivative vanishes. On a Riemannian manifold, it is equivalently ${\nabla }_{g}f(p)=0$ for the chosen Riemannian metric; the Riemannian gradient depends on that metric, but whether it vanishes does not. See Loring W. Tu, An Introduction to Manifolds, Proposition 8.23 for the scalar coordinate criterion. The scalar smooth function $f(t)=t^3$ has a critical point at zero without a local maximum or local minimum.

A source of lower dimension gives a different warning. For the smooth map $G:\mathbb{R}\to {\mathbb{R}}^2$, $G(t)=(t,t^2)$, the Jacobian matrix is the column $(1,2t{)}^T$, of matrix rank one at every input. Every input is a critical point of a smooth map, since the target dimension is two, even though this parametric curve has nonzero velocity everywhere and is an immersion. More generally, if $m<n$, no differential $T_{p}X\to T_{f(p)}Y$ can be surjective, so every input is critical. The terminology measures the ability to fill target directions, rather than whether the parametrized curve looks smooth.

A critical point of a smooth map is an input; a critical value is an output reached from at least one such input. For the fold $F(x,y)=(x,y^2)$, all inputs $(a,0)$ are critical points of a smooth map, and all outputs $(a,0)$ are critical values. The identical coordinate pairs live in different copies of ${\mathbb{R}}^2$. To classify an output $(a,b)$, inspect its entire fiber of a function:For $b>0$ both inputs have matrix rank two, so the output is a regular value. For $b=0$ its input has matrix rank one, so the output is a critical value. For $b<0$ the fiber of a function is the empty set, and the output is a regular value despite being outside the image of a function.

In general, for a smooth map $f:X\to Y$, write $C_f$ for its set of critical points of a smooth map. Its set of critical values is $f(C_f)\subseteq Y$. Equivalently, $q\in Y$ is a critical value if some $p\in f^{-1}(q)$ has a differential that is not surjective. A regular value requires a surjective differential at every input in that fiber of a function. The universal condition holds vacuously for an empty fiber of a function: every output outside $f(X)$ is a regular value. These quantifiers are explicit in Loring W. Tu, An Introduction to Manifolds, Definition 8.22 and its two following remarks.

The distinction is useful when solving equations. For a smooth map $f:X^m\to Y^n$ between smooth manifolds without boundary and a regular value $q$ with nonempty fiber of a function, the regular level set theorem makes $f^{-1}(q)$ an embedded submanifold of dimension $m-n$, with tangent spaceThe implicit function theorem explains the freedom count: locally, $n$ independent equations fix $n$ coordinates and leave $m-n$ free. The rank-nullity theorem gives the same count for the displayed kernel of a linear map. See Eckhard Meinrenken, Differentiable Manifolds, Theorems 1.11 and 1.15 for the local projection description and its consequence for level sets.

For example, $H:{\mathbb{R}}^2\to \mathbb{R}$, $H(x,y)=x^2+y^2$, has gradient $(2x,2y)$ and just one critical point, the origin, with critical value zero. For $c>0$, its fiber of a function is the circle of radius $\sqrt{c}$, a one-dimensional embedded submanifold; negative $c$ are regular values with empty fibers. At $c=0$, the fiber of a function is a single point instead of a circle. A critical value does not force its fiber of a function to be nonsmooth: that singleton is itself a zero-dimensional embedded submanifold, but the regular dimension conclusion $2-1=1$ does not apply.

Nor does a single regular input certify a regular value. The scalar smooth function $h(t)=t^2(t-1)$ has $h^{-1}(0)=\{0,1\}$ and derivative $h^{\prime }(t)=3t^2-2t$. The input zero is a critical point, while the input one is regular because $h^{\prime }(1)=1$. Their shared output zero is nevertheless a critical value. Conversely, different critical points of a smooth map can share one critical value: $f(t)=(t^2-1{)}^2$ sends both critical inputs $-1$ and $1$ to zero. Counting inputs and counting critical outputs are different tasks.

Finally, for $G(t)=(t,t^2):\mathbb{R}\to {\mathbb{R}}^2$, every output on the parabola is a critical value, and every output off it is a regular value with empty fiber of a function. There is no nonempty regular fiber of a function here, so the negative expression $1-2$ never becomes a claimed manifold dimension. Both diagrams show exact mathematical examples, not measured data.

For a finite matrix with singular values in nonincreasing order, truncating its singular value decomposition after $k$ terms minimizes the Frobenius norm error, whose square is the sum of the discarded squared singular values. It also minimizes the operator norm error, which equals the next singular value, or zero if none remain. See Mazeika, Section 4.3: Low rank approximation.

The problem of replacing a matrix by one of lower matrix rank is explicit in Eckart and Young's 1936 paper, ^[ref]. The publication predates the modern image compression example, showing how a general matrix question can support later applications.

An elementary matrix is obtained by applying one elementary row operation to the identity matrix. Multiplying $A$ by it on the left performs the same row operation on $A$. Such a matrix is invertible: reverse a row swap, reverse a nonzero row scaling by its reciprocal, or reverse adding a multiple of another row by adding its negative. Right multiplication gives an elementary column operation. In particular these operations preserve matrix rank.

If encoding and decoding quantum channels factor through a $k$-dimensional space, their composite Kraus operators all have matrix rank at most $k$. The rank bound for a weighted operator trace and the Kraus formula for entanglement fidelity implyThus high operation fidelity requires the leading $k$ eigenvalues of the source density operator to carry almost all its probability. The bound applies to arbitrary completely positive trace-preserving encoders and decoders.

An $m\times n$ matrix has full column rank when its matrix rank is $n$, equivalently when its columns are linearly independent vectors. This requires $n\le m$. Its kernel is $\{0\}$, so the represented linear map is injective and every solvable linear system $Ax=b$ has a unique solution. It need not reach every output.^[ref]

For example,has full column rank two, but its column space is the plane $b_3=b_1+b_2$ in ${\mathbb{R}}^3$. An attainable output fixes the input uniquely as $(b_1,b_2{)}^T$; an output outside this plane has no preimage.

An $m\times n$ matrix has full row rank when its matrix rank is $m$, equivalently when its rows are linearly independent vectors. This requires $m\le n$. Its column space is all of ${\mathbb{F}}^m$, so the represented linear map is surjective and every linear system $Ax=b$ is solvable. If $n>m$, its nullity is $n-m>0$, so solutions are not unique.^[ref]

For example, the matrix transpose of the matrix in full column rank isIt has full row rank two. Every output $(p,q{)}^T$ has solutions $(p-t,q-t,t{)}^T$, with $t\in \mathbb{R}$. A square matrix of size $n$ has full row rank and full column rank together exactly when its matrix rank is $n$; then it is an invertible matrix. A nonsquare matrix cannot have both properties, and has no two-sided inverse: such an inverse would make its linear map a linear isomorphism between vector spaces of different dimensions.

The real $m$ by $n$ matrices of matrix rank $m\le n$ form an open smooth manifold $X_{m,n}\subseteq {\mathbb{R}}^{mn}$. Indeed, it is the union of the open sets on which a chosen $m$ by $m$ minor has nonzero determinant, so its dimension is $mn$.

A linear system is inconsistent when it has no solution. In Gaussian elimination, a row $0=c$ with $c\ne 0$ proves inconsistency. Equivalently the matrix rank of the augmented matrix is greater than the matrix rank of the coefficient matrix.

An $n$-by-$n$ square matrix $A$ over a field is invertible when it has a two-sided matrix inverse. Equivalently, its linear map $x\mapsto Ax$ is a bijection: every output comes from exactly one input. For a square matrix, the following conditions are equivalent:Here matrix rank $n$ means both full column rank and full row rank. The square hypothesis matters: a rectangular matrix can have full column rank without reaching every output, or full row rank without determining every input.^[ref]

The equivalences explain why invertibility is a question about retaining information. A trivial kernel makes the linear map injective; the rank-nullity theorem then gives matrix rank $n$, so its columns form a basis and the linear map is also surjective. Assigning to each output its unique input gives a linear map: the preimage of $\alpha b+\beta c$ is $\alpha A^{-1}b+\beta A^{-1}c$. The determinant is nonzero precisely when the columns are linearly independent. Thus algebraic solvability and independent directions describe the same phenomenon; over the real numbers, nonzero oriented volume supplies its geometric interpretation.

For a matrix whose entries depend on a parameter, a nonzero maximal minor certifies a fixed matrix rank away from its zeros. At a zero, other minors must still be checked: one vanishing minor does not itself imply a rank drop. Once the matrix rank is determined, the rank-nullity theorem gives the dimension of the kernel of a linear map. Exceptional parameters can then be solved separately to exhibit the additional kernel of a linear map directions.

For a real matrix $T$, a Gram product has the displayed null space. Here Gram product means a product $T{T}^{\top }$ or $T^{\top }T$.

Proof. The inclusion from right to left is immediate. Conversely, if $T{T}^{\top }x=0$, thenso $T^{\top }x=0$. Similarly $\ker (T^{\top }T)=\ker T$. Thus these positive semidefinite matrices retain the relevant matrix rank of $T$. The argument uses the positive Euclidean norm; a corresponding statement for an arbitrary indefinite bilinear form need not hold.

Let $0\le s\le k\le n$ and form the integer matrix with rows indexed by $S\subseteq [n]$, $|S|\le s$, columns indexed by $k$-subsets $T$, and entry $M_{S,T}=1_{S\subseteq T}$. Its matrix rank over any finite field ${\mathbb{F}}_p$ is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Over the rational numbers, the row of a $j$-subset $S$ is ${\left(\genfrac{}{}{0ex}{}{k-j}{s-j}\right)}^{-1}$ times the sum of the rows of all $s$-subsets containing $S$. Thus the rows of degree exactly $s$ span, giving the bound over the rational numbers. Every larger square minor therefore has integer determinant zero, and remains zero modulo $p$. The matrix rank cannot increase under this reduction. This argument avoids dividing by an integer that might vanish in the finite field, and gives the sharp uniform-layer step in the Frankl-Wilson theorem.

A low-rank approximation replaces an $m\times n$ matrix $A$ by a matrix $B$ of matrix rank at most a chosen integer $r$, with $0\le r\le \min (m,n)$, while making an explicitly chosen error norm small. The point is to describe many entries using a few shared patterns: an image compression may retain broad brightness patterns while discarding finer ones. The approximation is useful when the saved storage or computation justifies the lost detail. A small matrix rank alone says nothing about the error.

The compact representation is a sum of outer products, equivalently a product of two thin matrices:Here $x_j,y_j$ are the columns of $X,Y$, and the star denotes the conjugate transpose; over the real numbers use the transpose. The rank bound for a matrix product gives $\mathrm{rank}B\le r$. Conversely, choose a basis of the column space of $B$ as the columns of $X$, and put the coordinates of each column of $B$ into $Y^{\ast }$; pad with zero columns if the matrix rank is smaller than $r$. Each nonzero outer product is a rank-one matrix: it repeats a single column pattern at different strengths across the columns.

Keeping the two factors stores $r(m+n)$ scalars instead of $mn$. This saves space only when $r(m+n)<mn$. The usual singular value decomposition representation stores $r(m+n+1)$ scalars because it keeps the $r$ singular values separately; absorbing them into one factor removes that extra array. These are representation sizes at equal scalar precision, before file overhead. See Uschmajew and Vandereycken, §9.2: thin factors and their storage, and Illinois CS 357, Low-rank approximation: storing singular vectors and values.

The factors also give $Bz=X(Y^{\ast }z)$ without forming the full matrix $B$. Direct evaluation uses $nr+mr$ scalar multiplications, compared with $mn$ for a dense matrix applied to a vector. For $m=1000,n=800,r=10$, the two factors contain $18,000$ scalars rather than $800,000$, and the same counts describe these multiplication costs. Constructing the factors is a separate cost. Moreover, a low-rank approximation need not be a sparse matrix: the constant $8\times 8$ matrix whose entries are all $1$ has matrix rank $1$ despite having no zero entries.

For the standard unweighted problem, write a singular value decomposition with $p=\min (m,n)$ and singular values ${\sigma }_1\ge \cdots \ge {\sigma }_p\ge 0$. Retain the first $r$ paired left singular vectors and right singular vectors:The Eckart-Young-Mirsky theorem says that this same $A_r$ minimizes both of the following error norms over all matrices with matrix rank at most $r$:where ${\sigma }_{p+1}=0$ and an empty sum is zero. Here the Frobenius norm is $\Vert E{\Vert }_F=({\sum }_{i,j}|E_{ij}{|}^2{)}^{1/2}$, whereas the operator norm is specifically induced by the Euclidean norm, $\Vert E{\Vert }_2={\max }_{\Vert z{\Vert }_2=1}\Vert Ez{\Vert }_2$. The first aggregates errors in all entries; the second bounds the worst output error for a unit vector input. These formulas apply to real and complex matrices. ^[ref]

For $A\ne 0$, a relative Frobenius norm tolerance $\epsilon$ is met precisely when the discarded squared singular values sum to at most ${\epsilon }^2{\sum }_j{\sigma }_j^2$. Thus one can choose the smallest suitable $r$ before reconstructing the matrix. Many individually small discarded singular values can still have a large combined error. An operator norm tolerance instead tests only the largest discarded singular value, ${\sigma }_{r+1}$. The two questions can therefore select different ranks.

The following stipulated $8\times 8$ matrix makes the tradeoff visible. Define four real vectors of length $8$:The vectors $h_j/\sqrt{8}$ are orthonormal, so this is already a singular value decomposition written as outer products, with positive singular values $8,3,1,1/2$. All entries lie between $7/16$ and $25/16$ and can be displayed as brightnesses. The first component is uniform brightness, the second distinguishes broad blocks, and the last two supply finer alternating patterns. This is a designed arithmetic example, not measured image data.

Here $A_1=h_0{h}_0^T$ has every entry equal to $1$, while $A_2=h_0{h}_0^T+3h_1{h}_1^T/8$ has entries $11/8$ within the two diagonal blocks and $5/8$ in the other two blocks. For instance, $A_{11}=25/16$ becomes $(A_2{)}_{11}=11/8$: its error is $3/16$. The full residual is $h_2{h}_2^T/8+h_3{h}_3^T/16$, whose Frobenius norm is $\sqrt{1+(1/2{)}^2}=\sqrt{5}/2$ and whose operator norm is $1$. Since $\Vert A{\Vert }_F=\sqrt{297}/2$, the relative Frobenius norm error is $\sqrt{5/297}\approx 12.97\%$. Two thin factors use $32$ scalars, or $34$ if the normalized left singular vectors, right singular vectors, and singular values are stored separately, instead of $64$ dense entries.

Retaining three components reduces the Frobenius norm error to $1/2$ while using $48$ scalars in two factors. Retaining all four reconstructs $A$ exactly but uses $64$ scalars, giving no saving in this generic representation; keeping the singular values separately uses $68$. Our particular sign patterns could be encoded more economically, but that would exploit an additional structure beyond the general low-rank approximation. The striking feature is that numerical error and storage can both be read from the decomposition, rather than judged only by how smooth the picture looks.

The constraint is deliberately matrix rank at most $r$. The truncation has $\mathrm{rank}{A}_r=\min (r,\mathrm{rank}A)$. If $\mathrm{rank}A<r$, the best approximation is $A$ itself, with zero error. Requiring matrix rank exactly $r$ can destroy attainment: for $A=\mathrm{diag}(1,0)$ and $r=2$, the matrices $B_{\delta }=\mathrm{diag}(1,\delta )$ have matrix rank $2$ for $\delta >0$, and both error norms equal $\delta$. Their error has infimum zero, but no rank-two matrix attains it. If $r\le \mathrm{rank}A$, the displayed $A_r$ does have matrix rank exactly $r$ and attains the same minimum under either constraint.

Optimality also does not always mean uniqueness. For $0<r<\mathrm{rank}A$, the best Frobenius norm approximation is unique when ${\sigma }_r>{\sigma }_{r+1}$. ^[ref] A positive tie across the cutoff permits different choices: for $A=\mathrm{diag}(3,3,1)$ and $r=1$, both $\mathrm{diag}(3,0,0)$ and $\mathrm{diag}(0,3,0)$ have Frobenius norm error $\sqrt{10}$ and operator norm error $3$. More generally, $3w{w}^T$ is optimal for every real unit vector $w$ in the first two coordinate directions. Rotating which tied direction is kept changes the approximation. Repeated singular values entirely inside the retained group do not change its sum of outer products. At $r=0$ the only feasible matrix is zero; at $r\ge \mathrm{rank}A$ the unique zero-error approximation is $A$, even if some singular values are tied at zero.

A strict gap does not guarantee uniqueness for the operator norm. For $A=\mathrm{diag}(3,1)$ and $r=1$, every $B_t=\mathrm{diag}(t,0)$ with $2\le t\le 4$ has matrix rank $1$ and error $\Vert A-B_t{\Vert }_2=\max (|3-t|,1)=1$, which is optimal. Only $t=3$ minimizes the Frobenius norm, since $\Vert A-B_t{\Vert }_F=\sqrt{(3-t{)}^2+1}$. An unchanged worst direction can hide extra error in another direction under the operator norm.

Finally, these guarantees concern the stated norms and the sole matrix rank constraint. They do not establish that a discarded pattern is noise or irrelevant to a particular task: the finer patterns in our constructed example were put there intentionally. Adding entrywise weights or restrictions on the factors changes the problem, so the Eckart-Young-Mirsky theorem above does not automatically solve it. For historical context, the original 1936 matrix problem is documented in the existing Eckart-Young-Mirsky theorem section.

For a resolved magnetic field, the unaveraged tensor $B_i{B}_j$ describes magnetic tension; the isotropic magnetic pressure is treated separately in the magnetohydrodynamic total pressure. Its spatial or ensemble average need not have matrix rank one. Statistical isotropy in three dimensions gives $⟨B_i{B}_j⟩=⟨B^2⟩{\delta }_{ij}/3$, by taking the trace. In Alfvén velocity units this tensor has dimensions of squared velocity.

A matrix decomposition expresses a matrix as a matrix product of factors with simpler structure. An LU decomposition uses triangular factors, a QR decomposition uses an orthogonal matrix and a triangular factor, and a singular value decomposition separates input directions, stretching factors and output directions. The factorization is an identity; the chosen structure determines what it helps compute.

For an original two-by-two example,The outer factors are a lower triangular matrix and an upper triangular matrix, each with unit diagonal as in a unit triangular matrix; the middle factor is a diagonal matrix. Their matrix product is $M$, and the final diagonal entry $5/2$ is the Schur complement of the first entry $2$. This is the scalar version of the factorization under block elimination.

To solve $Mz=(3,4{)}^T$, solve with the three factors in their displayed order from left to right: first the lower triangular factor, then the diagonal factor, then the upper triangular factor. The intermediate vectors are $(3,5/2{)}^T$ and $(3/2,1{)}^T$, giving the concise answer $z=(1,1{)}^T$. A new right-hand side uses the same factors. This is why Reusing LU factors for multiple right-hand sides is useful: factorization records work that otherwise would be repeated, while forward substitution and back substitution apply it cheaply.^[ref]

Different decompositions reveal different features. A Cholesky decomposition uses positive definiteness of a symmetric matrix, while a singular value decomposition also applies to rectangular matrices and exposes matrix rank. The existence and numerical behaviour of the factors depend on their own hypotheses; for example, an unpivoted LU decomposition can fail when a chosen pivot element is zero even though the matrix is an invertible matrix.

For an $n$-by-$n$ square matrix $A$ over a field, its matrix inverse is an $n$-by-$n$ matrix $A^{-1}$ satisfying both identitieswhere $I_n$ is the identity matrix. It exists exactly when $A$ is an invertible matrix, equivalently when its matrix rank is $n$ or its determinant is nonzero. It reverses the linear map: applying $A$ and then $A^{-1}$ recovers the input, while applying $A^{-1}$ and then $A$ recovers the output.^[ref]

The matrix inverse is unique. If $B$ and $C$ both satisfy the identities, then $B=B(AC)=(BA)C=C$. Although the definition is two-sided, for finite square matrices either identity alone suffices: $BA=I$ forces a trivial kernel, whereas $AB=I$ forces surjectivity; either makes $A$ an invertible matrix by the rank-nullity theorem. Rectangular matrices may have only one of these properties.

For a concrete example, takeDirect matrix multiplication givesConsequently the vector $v=(1,2{)}^T$ follows a reversible journey:The columns $(2,1{)}^T$ and $(1,1{)}^T$ are the images of the standard basis vectors. They turn the unit square into a parallelogram; the matrix inverse restores both the square and its surrounding grid. Since $\det A=1$, this example changes shape while preserving area, illustrating that invertibility does not require preserving lengths or angles.

For comparison, the singular matrixcannot have a matrix inverse: the output $(1,0{)}^T$ cannot specify which of the two inputs occurred. Their difference $(1,-1{)}^T$ belongs to its kernel. Indeed $S(x,y{)}^T=(x+y,0{)}^T$ collapses the plane to a line, its matrix rank is one, and its determinant is zero. No subsequent function of the output alone can reconstruct every input.

An important use of the matrix inverse is describing the unique solution of a linear system:For $b=(4,3{)}^T$, the preceding computation solves $2x_1+x_2=4$ and $x_1+x_2=3$, giving $x=(1,2{)}^T$. In computation, usually solve the linear system using Gaussian elimination and a factorization such as an LU decomposition, rather than forming every entry of the matrix inverse. The inverse formula specifies the answer; a factorization provides a reusable way to compute it for each right-hand side.^[ref]

Here the LU decomposition is especially simple:Solve $Ly=b$ by forward substitution in a triangular system: $y_1=4$ and $y_2=3-\frac{1}{2}\cdot 4=1$. Then solve $Ux=y$ by backward substitution in a triangular system: $x_2=2$ and $x_1=(4-2)/2=1$. The same factors work when $b$ changes. If the whole matrix inverse is actually needed, its $j$th column is the solution of $Ax=e_j$, where $e_j$ is the $j$th standard basis vector.

Reversing several operations means undoing the last one first. For invertible matrices of the same size,because $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=I$, and the product in the other order is also $I$. This is why a change-of-basis matrix uses its matrix inverse to reverse the direction of coordinate conversion.