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A basis is a linearly independent spanning set of vectors in a vector space. Every vector is a unique finite linear combination of basis vectors. Independence supplies uniqueness, and spanning supplies existence. An ordered basis turns the coefficients into a coordinate vector.

For $\mathcal{B}=(b_1,b_2)=((2,1),(1,2))$, the matrix with these vectors as columns isHence the coordinate vector of $v=(5,4)$ is $[v{]}_{\mathcal{B}}=B^{-1}v=(2,1)$, and $B[v{]}_{\mathcal{B}}=v$ recovers the usual coordinates. The determinant $3$ is nonzero, so this recovery works for every vector in ${\mathbb{R}}^2$. A basis need not have perpendicular or unit-length vectors; an orthonormal basis adds those geometric conveniences when an inner product is available.

The point of changing basis is to simplify the operation being studied. Diagonalization uses an eigenbasis to turn a coupled linear map into independent scalings. In a space of polynomials, the basis $1,x,x^2$ makes differentiation easy to encode, while orthogonal polynomials can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.^[ref]

A basis vector is one of the vectors in a specified basis of a vector space. Its coordinate column in that basis has one entry equal to one and all other entries zero. Columns of a matrix of a linear map are the images of domain basis vectors, expressed in the chosen codomain basis.

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

A change of basis rewrites a vector using a different ordered basis of the same vector space. The vector stays fixed; its coordinate vector changes because the measuring directions change. Both bases must describe every vector uniquely, so a change of basis is reversible.

Start with the Cartesian basis $e_1=(1,0)$, $e_2=(0,1)$, and choose the new basis vectors $u=(1,1)$ and $v=(1,-1)$. For the vector $z=(2,0)$,The second coordinate vector says to travel once along $u$ and once along $v$. Their vertical contributions cancel, leaving exactly the original horizontal arrow. The diagram shows two descriptions of that same arrow, with the same origin and endpoint.

Choosing a useful basis can reveal simple structure hidden by the old coordinate vectors. Here the new basis describes common and opposing contributions. For two equal measurements $(5,5)$, the new coordinate vector is $(5,0)$: only the common contribution is present. In larger examples, a suitable basis can concentrate important information into a few coefficients. Strang's lecture on change of basis develops this principle through applications: the reversible change of basis preserves information, while subsequently discarding small coefficients loses information.^[ref]

Fix the convention: the columns of the change-of-basis matrix $P$ are the new basis vectors written in the old basis. Thus $P$ converts new coordinate vectors into old ones; its matrix inverse converts old coordinate vectors into new ones. For the preceding basis,Its determinant is $-2$, so $P$ is an invertible matrix. Checking which direction $P$ converts is essential: different books may call either $P$ or $P^{-1}$ the change-of-basis matrix.

For a linear operator, convert the input into the old basis, apply the old matrix, and convert the output back. Reading that sequence from right to left givesFor example, let the old diagonal matrix stretch horizontal components by $2$ and vertical components by $3$. ThenThe output represents $2u+2v=(4,0)$, just as $A_{\mathrm{old}}(2,0{)}^T=(4,0{)}^T$. The matrices look different because their bases differ, but they encode the same linear operator. This relation is called matrix similarity; it preserves the underlying action.^[ref]

The vector space ${\mathbb{R}}^3$ with the cross product as Lie bracket, $[x,y]=x\times y$. Bilinearity and $x\times x=0$ give an Alternating Lie bracket; the vector triple product identity proves the Jacobi identity because the three expressions $x\times (y\times z)=y(x\cdot z)-z(x\cdot y)$ cancel cyclically, using the symmetric dot product.

Its relation to the SO(3) Lie algebra is explicit. For a vector $x=(x_1,x_2,x_3)$, define the skew-symmetric matrixThen $[\hat{x},\hat{y}]=\widehat{x\times y}$: acting on $v$, the left side is $x\times (y\times v)-y\times (x\times v)=(x\times y)\times v$. Every real $3\times 3$ skew-symmetric matrix occurs uniquely, so this hat map is a Lie algebra isomorphism. With the standard basis vectors $e_1,e_2,e_3$,and their matrix commutators are $[X,Y]=Z$, $[Y,Z]=X$, $[Z,X]=Y$. With active rotation matrices acting on column vectors, $\exp (t\hat{x})$ rotates through angle $t\Vert x\Vert$ about the oriented axis $x$ when $x\ne 0$. In particular $\exp (tZ)e_1=(\cos t,\sin t,0)$, fixing the positive-angle convention used in the small rotation commutator.

The map $x\mapsto -(i/2)x\cdot \sigma$ is also a Lie algebra isomorphism to the compact real SU(2) Lie algebra, using the Pauli matrix commutator identity. These two identifications concern Lie algebras; they do not identify the corresponding Lie groups globally.

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.

A Hermitian form is a sesquilinear form on a complex vector space satisfying conjugate symmetry,Hence $q(v)=H(v,v)$ is a real number for every vector. In a basis, the matrix of a Hermitian form has $A_{ij}=H(e_i,e_j)$ and obeys $A_{ji}=\overline{A_{ij}}$, equivalently $A=A^{\ast }$: it is a Hermitian matrix. The coordinate rule remains $H(x,y)=x^{T}A\overline{y}$. Conjugate symmetry does not imply positivity.^[ref]

The diagonal contains more information than might be expected. For any complex sesquilinear form $B$, put $q(v)=B(v,v)$. Expanding four expressions gives the complex polarization identity in the linear-first convention:This is an algebraic identity, with no positivity assumption. If every $q(v)$ is real, interchanging $u,v$ preserves the first difference and negates the second: use $q(\alpha v)=|\alpha {|}^{2}q(v)$ for $\alpha =-1,i$. The identity then gives $B(v,u)=\overline{B(u,v)}$. Thus a complex sesquilinear form is a Hermitian form if and only if all its diagonal values are real. Testing only the basis vectors is insufficient: the off-diagonal example in sesquilinear form has zero diagonal entries in its matrix but a nonreal value at $(1,i)$.

Consider the indefinite Hermitian formIts values at $e_1,e_2$ are $1,-1$. Moreover $u=(1,1)$ and $v=(1,-1)$ are nonzero isotropic vectors: $q_{-}(u)=q_{-}(v)=0$, yet $H_{-}(u,v)=2$. These vectors do not belong to the radical of a Hermitian form, and the usual Cauchy-Schwarz inequality would incorrectly require $4\le 0$. An indefinite Hermitian form records signed geometry rather than defining a norm by $\sqrt{H(v,v)}$.

There is a separate boundary case: $H_0(z,w)=z_1\overline{w_1}$ is a positive semidefinite Hermitian form, since $H_0(z,z)=|z_1{|}^2\ge 0$, but $H_0(e_2,e_2)=0$. Its radical of a Hermitian form is the entire second coordinate line. Positive definiteness is exactly the additional condition in a Hermitian inner product. The three examples below show why these distinctions matter: on the real slice $z=(x,y)\in {\mathbb{R}}^2\subset {\mathbb{C}}^2$, $q_{+}(z)=|z_1{|}^2+2|z_2{|}^2$ has an ellipse as its unit level, $q_{-}$ has hyperbolas, and $q_0$ has two parallel lines. These are slices of the complex examples, not drawings of their full four-dimensional domains.

A bounded operator is Hermitian, or self-adjoint, when $T=T^{\ast }$.

Apply the preceding Householder reflection to the columns ofThe first column becomes $(-5,0{)}^T$ and the second becomes $(-11/5,2/5{)}^T$:Thus an orthogonal matrix multiplying an upper triangular matrix reconstructs $A$. The zero below the diagonal came from a geometric operation that preserves Euclidean norms and inner products.

Our positive-diagonal convention for QR decomposition is obtained by a sign adjustment. With $D=\mathrm{diag}(-1,1)$, putThen $Q^{T}Q=I$ and $\boxed{A=Q{R}_{+}}$. The intermediate sign chooses an orientation for a basis vector; the adjustment makes the diagonal positive.

For a larger matrix, the next Householder reflection acts only on the lower rows of the remaining columns, retaining the zeros already created. Store its defining vector and apply $z\mapsto z-2u(u^{T}z)$ instead of forming a dense reflection matrix. This links a simple mirror operation to efficient QR decomposition, including least-squares solutions.

Lie theory studies continuous symmetries through Lie groups, Lie algebras, and their representations. A rotation can change by an arbitrarily small angle: its parameter varies continuously, yet composing two rotations obeys an exact group operation. The central question is how such finite transformations are controlled by their velocities at the identity element.

An infinite group need not require infinitely many independent parameters. The planar rotation group has infinitely many elements but only one local angle coordinate. More generally, an $n$-dimensional Lie group has an $n$-dimensional tangent space $\mathfrak{g}=T_{e}G$ at its identity element. A basis of this Lie algebra, together with the Lie brackets between its basis vectors, supplies finite data for infinitesimal composition. These brackets record the first correction to adding small transformation parameters. The Exponential map of a Lie group turns a chosen infinitesimal direction into an actual motion.

Finite tangent data consists of a finite basis and a finite table of Lie brackets. For planar rotation, every generator is $aJ$ with $a\in \mathbb{R}$, using the quarter-turn matrix $J$ computed below: one real coefficient specifies any angular velocity. This Lie algebra has infinitely many vectors, but its dimension is one. Linear combinations of the chosen basis vectors supply all the possible infinitesimal motions.

This passage between geometry and algebra is useful because linear maps and matrix commutators are easier to compute than an entire family of transformations. The local information does not determine every global identification: the additive real line and the circle group have isomorphic one-dimensional Lie algebras, but only the latter wraps around. For the local-versus-global viewpoint and the connection with differential equations, see Meinrenken, Lie groups and Lie algebras, introduction and Section 4.

Linear algebra studies vector spaces and linear maps: operations that preserve vector addition and scalar multiplication. For a linear map $T$, the defining rule is $T(au+bv)=aT(u)+bT(v)$, where $u,v$ are vectors and $a,b$ belong to the underlying field. The vectors might be arrows, lists of measurements, or polynomials. What matters is that their linear combinations make sense and that the same rule governs their transformation.

A concrete linear map is $T(x,y)=(x+y,x-y)$. Its two outputs record a sum and a difference. Since $T(1,0)=(1,1)$ and $T(0,1)=(1,-1)$, we immediately obtain $T(2,1)=2(1,1)+(1,-1)=(3,1)$. Knowing the outputs of the basis vectors determines the output of every vector; a matrix stores exactly that information in its columns.

The attraction of linear algebra is that one small collection of ideas answers several different questions. A system of linear equations asks which inputs produce a prescribed output. The image of a linear map describes the attainable outputs, while the kernel describes the invisible input changes. A basis gives every vector an address, and change of basis lets us choose addresses that make a linear map easier to understand. This shared language turns a collection of separate calculations into a picture of what is possible and how much freedom remains. Gilbert Strang's MIT course emphasizes these connections and their usefulness across other disciplines.^[ref]

Choose an ordered basis $(e_1,\dots ,e_n)$ of a finite-dimensional vector space. The matrix of a bilinear form records all pairings of basis vectors, with $A_{ij}=B(e_i,e_j)$. Writing $v={\sum }_i{v}_i{e}_i$ and $w={\sum }_j{w}_j{e}_j$, separate linearity giveswhere the last expression uses the coordinate vectors of $v,w$. Conversely, every matrix $A$ defines a bilinear form by this formula.^[ref]

For a concrete bilinear form on ${\mathbb{R}}^2$, takeThen $B((1,0),(0,1))=2$ but $B((0,1),(1,0))=0$: a bilinear form need not be a symmetric bilinear form. For $v=(2,1{)}^T$ and $w=(-1,3{)}^T$, first compute $Aw=(5,3{)}^T$, then $v^{T}Aw=2\cdot 5+1\cdot 3=13$. Reversing the inputs gives $Av=(4,1{)}^T$ and $w^{T}Av=-4+3=-1$.

Under change of basis, let the columns of an invertible matrix $P$ be the new basis vectors in old coordinates. Thus $v=P{v}^{\prime }$ and $w=P{w}^{\prime }$, soThis is matrix congruence, rather than the matrix similarity $P^{-1}AP$ for an endomorphism. Both inputs have changed coordinates, which explains the two factors of $P$. For the displayed $A$ and $P=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$, direct multiplication gives $A^{\prime }=\left(\begin{array}{cc}1& 3\\ 1& 4\end{array}\right)$; its $(1,2)$ entry is $B(e_1,e_1+e_2)=3$.^[ref]

The nullity of a linear map $T:V\to W$ is $n(T)=\dim \ker T$, the dimension of its kernel. It counts independent input directions that can change without changing the output. For a finite-dimensional vector space $V$, the rank-nullity theorem givesFor an $m\times n$ matrix $A$ of matrix rank $r$, this is $n-r$, using the number of columns, not the number of rows. This dimension identity is ^[ref].

The nullity answers a practical question about a linear system: once one input is known to produce the requested output, how much freedom remains? If $T(x_0)=b$, its fiber of a function over $b$ isIndeed, $T(x)=T(x_0)$ exactly when $T(x-x_0)=0$. Thus every nonempty fiber is an affine subspace of dimension $n(T)$, with the same direction space as the kernel. The kernel itself is the fiber over zero. Outputs outside the image of a linear map have empty fibers; the nullity does not guarantee existence, and we assign no solution-space dimension to an empty fiber here.

In the diagram, $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ has matrix $\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$, matrix rank one and nullity one. Its kernel has basis $(1,-1)$, and the full solution of $F(x,y)=(c,0)$ is $(x,y)=(c,0)+s(1,-1)$. One free scalar moves the solution along a straight line; the output $(c,1)$ is impossible.

The existing three-input example $T(x,y,z)=(x+y,z)$ instead has rank two and nullity one. Its image of a linear map is all of ${\mathbb{R}}^2$, andEvery output has a solution straight line. In contrast, $U(x,y,z)=(x+y,0)$ has rank one and nullity two, with kernel basis $(1,-1,0),(0,0,1)$. Its reachable output $(a,0)$ has the solution affine subspacea plane, while $(a,b)$ with $b\ne 0$ has no solution. These two linear maps have the same input and output dimensions but different freedom and reachability. Computing the kernel separates what an output determines from what an extra constraint would have to fix. For either example, a basis of the kernel gives one independent free scalar per basis vector, which is why nullity measures freedom rather than the number of solutions.

An ordered basis is a basis equipped with an ordering of its basis vectors. The order fixes which position of a coordinate vector corresponds to each basis vector.