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A random matrix is a matrix whose entries are random variables. Its eigenvalues form a random configuration, and its empirical spectral measure records their distribution with equal mass on each eigenvalue. Entry assumptions and normalization determine the spectral regime.

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.

A divergent sequence of Möbius transformations has a subsequence converging uniformly on compact subsets of the Riemann sphere outside one exceptional point to a constant. Represent the transformations by determinant-one $2\times 2$ matrices and divide each by its matrix norm. A convergent subsequence has a nonzero rank-one limit: its determinant tends to zero because the original norms tend to infinity. The kernel line is the exceptional point, and the image line is the constant limit. This converts the limit set of a Kleinian group into a tractable boundary dynamics problem.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

For any finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ and rational positive-determinant $\gamma$, the intersection $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}\Gamma \gamma$ has finite index. Multiply $\gamma$ by a positive rational scalar to obtain an integral matrix $A$ with positive integer determinant $D$. For $h=I+DB\in \Gamma (D)$, $Ah{A}^{-1}=I+AB\mathrm{adj}(A)$ is integral, so $\Gamma (D)$ lies in $S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. This intersection has finite index. Its further intersection with ${\gamma }^{-1}\Gamma \gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. No assumption that $\Gamma$ itself is a congruence subgroup is needed.

If $ad\equiv 1(\mathrm{mod}N)$, the determinant-one matrix with rows $(a,(ad-1)/N)$ and $(N,d)$ sends $\tau -d/N$ to $a/N-1/(N^2\tau )$. The transformation of the cusp form gives the displayed identity. It turns the small-height portion of an additive-twist Mellin transform into exponentially decaying values at the inverse rational cusp.

For rational interpolation with $\mathrm{deg}q\le m$ and a monic denominator $r(x)=x^{\ell }-{\sum }_{j=0}^{\ell -1}{b}_j{x}^j$, multiplying through gives the linear systemThese equations are equivalent to the fraction conditions only when all $r(x_i)$ are nonzero. If $f_i=x_i^{m+1}$ and $m+\ell =n$, the coefficient matrix is the Vandermonde matrix with rows $(1,x_i,\dots ,x_i^n)$, so distinct interpolation nodes give a unique coefficient solution. More constructively, divide $W(x)={\prod }_{i=0}^n(x-x_i)$ by $x^{m+1}$ as $W=x^{m+1}R+S$, $\mathrm{deg}S\le m$. Then $r=R$, $q=-S$, since $x^{m+1}r-q$ is the unique monic degree-$n+1$ polynomial vanishing at all the interpolation nodes. The quotient may vanish at a node, so unique solvability of this linear system does not prove existence of a rational interpolant.

A rational matrix is a matrix whose entries are rational numbers. Multiplying by a common denominator produces an integer matrix and preserves its homogeneous solution set.

Ray transfer matrix analysis propagates paraxial ray height and slope with a two-by-two matrix. Free propagation through $L$ uses $\left(\begin{array}{cc}1& L\\ 0& 1\end{array}\right)$, a thin optical lens uses $\left(\begin{array}{cc}1& 0\\ -1/f& 1\end{array}\right)$, and an unfolded concave-mirror reflection uses $\left(\begin{array}{cc}1& 0\\ -2/R& 1\end{array}\right)$. Multiply matrices in propagation order, with the first operation at the right.

Writing complex coordinates as $z=x+iy$ converts the unitary representation $U=A+iB$ into the indicated real matrix. It preserves the Euclidean metric because $U^{†}U=I$, and its real determinant is $|{\det }_{\mathbb{C}}U{|}^2=1$. Thus even $U(n)$ embeds in $SO(2n)$. The complexified defining real representation restricts as $\mathbf{n}\oplus \bar{\mathbf{n}}$, whereas the underlying real space is the realification of a complex representation of dimension $n$.

The minimal polynomial of a real matrix, initially computed over the complex numbers, has real coefficients. Its coefficientwise complex conjugate is also a monic polynomial of the same degree annihilating the matrix. Uniqueness of the minimal polynomial forces the two polynomials to agree.

A real matrix is a matrix whose entries are real numbers. The vector space of $m\times n$ real matrices is denoted $M_{m,n}(\mathbb{R})$.

A character is real-valued when all its values are real numbers, equivalently $\chi =\overline{\chi }$. By the character inversion identity, this is equivalent to $\chi (g)=\chi (g^{-1})$ for every $g$. This condition alone does not assert that the underlying complex representation can be realized by real matrices.

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 reduced column echelon form of a matrix $A$ is defined by taking its transpose, forming the reduced row echelon form, and transposing back:It is obtained from $A$ by elementary column operations. Its nonzero columns precede its zero columns, the topmost nonzero entry of each nonzero column is $1$, these leading entries move strictly downward as the column number increases, and each leading entry is the only nonzero entry in its row.

Every elementary column operation preserves the column space, because it and its inverse express each new column as a linear combination of the old columns. Thus the nonzero columns of the reduced form give a basis of the original column space. In contrast, elementary row operations preserve the row space. There is no general identity $\mathrm{rcef}(A)=\mathrm{rref}(A{)}^T$: the definition requires reducing $A^T$, not $A$.

For an invertible matrix of size $n$, both reduced forms are $I_n$. A singular example showing the distinction is

A row echelon form is in reduced row echelon form when every pivot element is one and is the only nonzero entry in its column.^[ref] This extra reduction displays each pivot unknown directly in terms of the free variables in a linear system. The zero matrix also satisfies the definition: it has no pivot elements.

Every matrix has exactly one reduced row echelon form under elementary row operations, with its size and column order held fixed.^[ref] Ordinary row echelon form is not unique: multiplying a nonzero row by two over the rational numbers keeps the staircase but changes the matrix. The reduced conditions remove that freedom.

For a fixed number of columns, each row space has a unique reduced matrix after zero rows are removed; at a fixed full size, the number of appended zero rows is also determined. Thus reduced row echelon form represents a vector subspace independently of the starting basis. This is the uniqueness used by Gaussian binomial coefficients to count vector subspaces over a finite field.

The fixed column order matters. Elementary column operations can change the row space, the kernel and the resulting reduced row echelon form. For instance, swapping the columns of $(12)$ over the rational numbers gives $(21)$, whose reduced row echelon form is $(11/2)$ instead of $(12)$. When solving a linear system, a column interchange also requires tracking the corresponding interchange of unknowns.

A gauge condition $F(x)=0$ is regular at a representative $x_0$ if its derivative along the gauge orbit is an invertible matrix. The inverse function theorem then makes the gauge-condition values valid local coordinates along that orbit. A unique zero without this regularity is insufficient: the translation action on the line with $F(x)=x^3$ has a unique zero but a zero derivative there. Multiple regular intersections give a Gribov ambiguity rather than a globally unique slice.

A perturbation of a renormalization-group fixed point is relevant if it grows as the coarse-graining length increases. For a continuous flow $d\delta g/d\ell =M\delta g$, an eigenvector of the linearized matrix with eigenvalue $\sigma$ grows as $e^{\sigma \ell }$; a positive real part is relevant, a negative real part is irrelevant and a zero real part requires nonlinear analysis. For a discrete transformation with length factor $b>1$, its Jacobian matrix multiplier $\lambda$ instead gives $\delta g_n={\lambda }^n\delta g_0$: $|\lambda |>1$ is relevant, $|\lambda |<1$ is irrelevant and $|\lambda |=1$ requires nonlinear analysis. For a positive multiplier the scaling exponent is $y=\log \lambda /\log b$. Thus a positive discrete eigenvalue smaller than one is an irrelevant operator direction, while the sign criterion belongs to the continuous generator.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.