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During radiation domination in a flat expanding universe, $a(t)\propto t^{1/2}$ and $H=1/(2t)$.

If a smooth component dominates the expansion but does not cluster, pressureless matter still satisfiesAs ${\bar{\rho }}_m$ becomes negligible, Hubble friction leaves a constant mode and a decaying mode, so structure growth freezes.

For a barotropic fluid, the squared sound speed is $c_s^2=\partial P/\partial \rho$. A Fourier density mode obeysin linear Newtonian perturbation theory.

The Jeans wavenumber separates pressure-supported modes from gravitationally unstable density modes.

The density contrast is the fractional perturbation $\delta =(\rho -\bar{\rho })/\bar{\rho }$.

An upper or lower triangular matrix has all entries on one side of the main diagonal equal to zero. Its eigenvalues, with algebraic multiplicity, are its diagonal entries.

A diagonal matrix has zero entries away from its main diagonal. Its eigenvalues are its diagonal entries.

Let $A$ be a Hermitian matrix, $M$ a positive semidefinite matrix, and $s\ge 0$ with $\mathrm{tr}M\le s$. ThenThe Hermitian assumption gives real eigenvalues and makes this an ordered real inequality.

By the finite-dimensional spectral theorem, choose an orthonormal basis of eigenvectors for $A$. In that basis, write the diagonal entries of $M$ as $m_i\ge 0$. Then $\mathrm{tr}(MA)={\sum }_i{m}_i{\lambda }_i$, while ${\sum }_i{m}_i=\mathrm{tr}M\le s$. Replacing each ${\lambda }_i$ by $\max \{{\lambda }_{\max }(A),0\}$ proves the first bound; the operator norm proves the second.

The first bound is optimal. If $A$ has a positive largest eigenvalue with unit eigenvector $u$, choose $M=su{u}^{\ast }$. If all eigenvalues are nonpositive, choose $M=0$.

The tensor product of linear maps is determined by $(A\otimes B)(x\otimes y)=Ax\otimes By$ and linear extension.

The real orthogonal group $O(n)$ consists of the real $n$-by-$n$ matrices $R$ satisfying $R^{T}R=I$, with matrix multiplication as its group operation. These are precisely the linear maps preserving the standard dot product, sinceIts identity element is $I$, and the inverse of $R$ is its matrix transpose $R^T$. Multiplication preserves the defining relation, so these matrices form a group.

For example, in two dimensions,is a rotation matrix, while $\mathrm{diag}(1,-1)$ is a reflection. Both preserve lengths and angles. By contrast, the shear mapping with matrix $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ sends $(0,1)$ to $(1,1)$ and changes its Euclidean norm, so it is not orthogonal.

Taking the determinant in $R^{T}R=I$ gives $(\det R{)}^2=1$. The determinant-one elements form the special orthogonal group $SO(n)$; determinant-minus-one elements reverse orientation. The distinction separates rotations from transformations involving a reflection.

The Lie algebra describes infinitesimal motion near the identity. Differentiating $R(t{)}^{T}R(t)=I$ at $R(0)=I$ gives $A^T+A=0$ for $A=R^{\prime }(0)$. Thus the tangent space at the identity is the vector space of skew-symmetric matrices, of dimension $n(n-1)/2$.

The Hadamard product of two complex positive semidefinite matrices of the same size is a positive semidefinite matrix. Write their spectral decompositions as sums $A={\sum }_r{u}_r{u}_r^{†}$ and $B={\sum }_s{v}_s{v}_s^{†}$, absorbing each nonnegative eigenvalue into its vector. Thenwhere $u_r\circ v_s$ is the componentwise product. Every term is a positive semidefinite matrix of matrix rank at most one, so their sum is a positive semidefinite matrix. This proof includes the real case by replacing the conjugate transpose with the transpose. Iterating gives positivity of every nonnegative integer Hadamard power; the zeroth power is the all-ones matrix.

A quadratic form on a vector space over a field $F$ is a function $Q:V\to F$ satisfying $Q(av)=a^{2}Q(v)$ whose polar bilinear form of a quadratic form $B_Q(v,w)=Q(v+w)-Q(v)-Q(w)$ is a bilinear form. In finite dimension, it is represented in coordinates by a homogeneous degree-two polynomial, or by $Q(v)=v^{T}Av$ with a suitable matrix $A$.

When the characteristic is not two, one may choose $A$ to be a symmetric matrix and recover it from $B_Q/2$. In characteristic two, the symmetric-matrix restriction would lose examples: $Q(x,y)=xy$ has an upper-triangular representing matrix $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, while $v^{T}Av$ for a symmetric matrix has no $xy$ cross term. Its polar form is $B_Q((x,y),(u,v))=xv+yu$, an alternating bilinear form. The distinction matters when studying quadratic forms over finite fields.^[ref]

An inner product assigns a scalar $⟨u,v⟩$ to two vectors and supplies their lengths and orthogonality. Using the convention linear in the first argument, it satisfiesThe resulting norm is $\Vert v\Vert =\sqrt{⟨v,v⟩}$. Over the real numbers, conjugation has no effect and the form is symmetric and bilinear. Over the complex numbers, it is sesquilinear; the convention linear in the second argument is also common, so formulas for coefficients must match the stated convention.

The Euclidean example is $⟨u,v⟩={\sum }_i{u}_i\overline{v_i}$. For real $u=(2,1)$ and $v=(1,3)$ this gives $⟨u,v⟩=5$, while $\Vert u{\Vert }^2=5$ and $\Vert v{\Vert }^2=10$. If $r=(-1,2)$, then $⟨r,u⟩=0$: perpendicularity becomes an algebraic equation. The Cauchy-Schwarz inequality gives $|⟨u,v⟩|\le \Vert u\Vert \Vert v\Vert$, and in a real inner product space it makes the angle formula $\cos \theta =⟨u,v⟩/(\Vert u\Vert \Vert v\Vert )$ meaningful for nonzero vectors.

Different inner products express different measurements. The weighted example $⟨u,v⟩=4u_1{v}_1+u_2{v}_2$ on ${\mathbb{R}}^2$ gives $\Vert (x,y){\Vert }^2=4x^2+y^2$, so its unit-length vectors lie on an ellipse. For functions, an integral such as $⟨f,g⟩=\int f\overline{g}$ supplies the geometry of the L2 space. This lets orthogonal projection and least-squares solutions apply to curves as well as coordinate lists.

Every complex square matrix is a product of two symmetric matrices, one invertible.

An orthogonal matrix similarity changes coordinates by an orthogonal matrix. It preserves eigenvalues, singular values and Euclidean geometry; it also preserves symmetry when applied to a symmetric matrix. A sequence of Householder transformations can use this freedom to produce a Hessenberg reduction, while a symmetric matrix can be reduced further to tridiagonal form.

For a block matrix $M=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ with $A$ an invertible matrix, the Schur complement of $A$ in $M$ is $S=D-C{A}^{-1}B$. It is the effective coefficient matrix for the remaining variables after block elimination. The definition requires invertibility of $A$, but not of $S$ or $M$. Its size is that of $D$, even when $B$ and $C$ are rectangular matrices.

In the original example under block matrix inverse, $D=5$ becomes $S=19/6$ because changing the remaining variable also changes the two variables solved through $A$. This is what makes a Schur complement more informative than simply deleting rows and columns. It retains their mediated effect, and its matrix inverse is the corresponding block of $M^{-1}$ when the block matrix inverse exists.

The same elimination appears when minimizing a quadratic form. The Schur complement as a minimized quadratic form gives a geometric example; the Schur complement determinant identity gives an exact test for an invertible matrix.

For linear maps $B:U\to V$ and $A:V\to W$ with finite-dimensional $V$, the rank inequality for a composition isRestrict $A$ to $\mathrm{im}B$. The rank-nullity theorem gives $\mathrm{rank}(AB)=\dim \mathrm{im}B-\dim (\mathrm{im}B\cap \ker A)$, and the intersection has dimension at most $\dim \ker A$.

For example, the product of two rank-four endomorphisms of a five-dimensional vector space has rank at least three. The bound controls how much rank can be lost when the image of one map meets the kernel of the next.

Equivalent matrices represent the same linear map after independent changes of domain and codomain bases.

The adjugate is the transpose of the cofactor matrix and satisfies $A\mathrm{adj}(A)=(\det A)I$.

The finite matrix geometric series telescopes as stated in the title. If its partial sums converge, then their successive differences show $D^{m+1}\to 0$. Passing to the limit gives $(I-D){\sum }_{j\ge 0}{D}^j=I$, so the sum is the matrix inverse $(I-D{)}^{-1}$. Conversely, existence of the inverse alone does not imply series convergence: the scalar matrix $D=2I$ is a counterexample. The Neumann series gives a sufficient condition from a matrix norm smaller than one.

For a nilpotent matrix with $D^r=0$, convergence is automatic and the inverse is the finite polynomial $I+D+\cdots +D^{r-1}$. Multiplying by $I-D$ directly verifies this inverse without computing a determinant or diagonalizing.