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Rotate the standard basis through $30^{\circ }$ using the rotation matrix whose columns are the new basis vectors:Here $u\cdot v=0$ and $u\cdot u=v\cdot v=1$, so $Q$ is an orthogonal matrix with $Q^{T}Q=I$, where $T$ denotes transpose and $I$ is the identity matrix. For the same vector $p=(2,1{)}^T$, its coordinate vector in this Cartesian basis isThus $\alpha \approx 2.232$ and $\beta \approx -0.134$: travel forwards along $u$ and a little backwards along $v$ to reconstruct $p=\alpha u+\beta v$. The orthogonal projections onto the two axes give these signed coordinates because the measuring vectors are both perpendicular and of unit length.

The squared norm remainsThis is useful when choosing axes aligned with a geometric feature: the coordinates change but distances and dot products retain the same sum-of-products formula. The two directions of the change of basis must be kept straight: $Q$ reconstructs the vector in old coordinates, $p=Q[p{]}_{\mathcal{U}}$, and $Q^T$ extracts its new coordinates. Applying $Q$ to the old coordinate vector $p$ instead would actively rotate the vector.

A short proof explains both formulas. If $p=Qc$, multiplication by $Q^T$ gives $c=Q^{T}p$. Then $\Vert p{\Vert }^2=c^T{Q}^{T}Qc=c^{T}c$. Perpendicular unit-length columns make the matrix inverse equal the transpose; this is the computational convenience supplied by a Cartesian basis.

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]

A circulant matrix applies the same rule at every position of a finite ring: moving both the input and output indices one step leaves its entries unchanged. Its first column determines the whole matrix. For a length-$N$ vector $c$, index rows and columns by $0,\dots ,N-1$ and defineThe indices use modular arithmetic. Thus each column is the previous column shifted down one place, with the last entry wrapping around to the top. Multiplication is exactly circular convolution:This makes a circulant matrix a natural model for a periodic finite difference rule or a filter acting on a periodic record. Strang's MIT lecture develops this connection between circulant matrices, circular convolution and the discrete Fourier transform.^[ref]

For example, the first column $c=(2,-1,0,-1{)}^T$ givesHere the indices wrap around modulo four. Each output compares a value with its two neighbors; this is the Graph Laplacian of a four-vertex cycle graph. A constant input gives zero, while an alternating input gives a large response. The Fourier diagonalization of a circulant matrix makes this distinction exact: its four eigenvalues are $0,2,4,2$.

The plotted discrete Fourier modes are complex vectors, shown through their real and imaginary parts. Each entry has modulus one; the Euclidean norm of each length-four vector is two. In particular, modes $k=1$ and $k=3$ have the same real part but opposite imaginary parts, so they are distinct eigenvectors with the same eigenvalue.

The useful surprise is that every length-$N$ circulant matrix shares the same basis of eigenvectors, whatever its first column. One change of basis therefore simplifies an entire family of linear maps, instead of requiring a separate eigenvalue computation for each one. Bamieh's tutorial develops the discrete Fourier transform from precisely this simultaneous diagonalization problem.^[ref]

For an ordered basis $\mathcal{B}=(b_1,\dots ,b_n)$, the coordinate vector $[v{]}_{\mathcal{B}}=(a_1,\dots ,a_n{)}^T$ records the unique coefficients satisfying $v={\sum }_i{a}_i{b}_i$. Change of basis changes this representation while preserving the underlying vector.

For $b_1=(1,0{)}^T$, $b_2=(1,1{)}^T$, the vector $v=(3,1{)}^T$ has coordinate vector $(2,1{)}^T$, since $2b_1+b_2=v$. These are the same coefficients extracted by the dual basis in its worked example. The vector is independent of the chosen basis; its list of coordinates is not. In an orthonormal eigenbasis, the coordinates additionally separate the action of a self-adjoint operator into independent scalar multiplications.

A covector is an element of the dual space, hence a linear functional on a vector space. In finite-dimensional coordinates, it acts as a row multiplied by a coordinate vector column. The row $\left(\begin{array}{cc}2& -1\end{array}\right)$ acts on $\left(\begin{array}{c}3\\ 1\end{array}\right)$ to give five. The terminology emphasizes its role as a scalar-valued measurement and its behavior under a change of basis.

For example, the total derivative of $f(x,y)=x^2+y^2$ at $(1,2)$ sends a small displacement $h$ to the linear change $df(h)=2h_x+4h_y$. This covector describes the first-order response. The gradient $\nabla f=(2,4)$ represents the same rule through the chosen Euclidean inner product. In differential geometry, the cotangent space collects these scalar-valued measurements of tangent vectors.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

For the basis $v_j=e_1+\cdots +e_j$ obtained from a basis $e_1,\dots ,e_n$, the dual basis is obtained by successive differences of the original dual basis. Indeed, if $x={\sum }_j{c}_j{v}_j$, its original coordinates are $x_i={\sum }_{j\ge i}{c}_j$. Hence $c_i=x_i-x_{i+1}$ for $i<n$, and $c_n=x_n$. These are precisely the new coordinate linear functionals. This gives an elementary example of the inverse relation between change of basis and covector transformation under a change of basis.

Let the Klein four-group have generators $X,Y$ and $Z=XY$, with $X^2=Y^2=I$ and $XY=YX$. For every real $p$, the matricesdefine a group representation. Indeed, $D(Y{)}^2=I_2$, and $D(X)$ commutes with every matrix, so the group relations hold.

Set $S=\left(\begin{array}{cc}1& -p/2\\ 0& 1\end{array}\right)$. A direct change of basis givesConsequently all these representations are isomorphic representations, with character values $(2,-2,0,0)$ on $(I,X,Y,Z)$. Each is a direct sum of the two one-dimensional representations whose values on $X$ are $-1$ and whose values on $Y$ are $+1$ and $-1$. A group representation need not be orthogonal; nonzero $p$ changes the basis rather than the represented group.

On $K^4$, use the standard basis $\mathcal{U}=(u_1,u_2,u_3,u_4)$ and the alternating bilinear form with matrix $J$ above for $n=2$. A second symplectic basis isFor instance, $\omega (f_1,f_2)=1-1=0$, while $\omega (e_1,f_1)=\omega (e_2,f_2)=1$ and the cross-pairings vanish. The change of basis matrix, whose columns are these new basis vectors in $\mathcal{U}$, isThe symmetric matrix $C$ makes $S^{T}JS=J$, so $S$ is a symplectic matrix. Explicitly, multiplying the blocks gives a lower-right block $C^T-C=0$; the other blocks agree with $J$. For a fixed vector, change of basis means $[v{]}_{\mathcal{U}}=S[v{]}_{\mathcal{B}}$, henceTheir symplectic form gives $3\cdot 0+2\cdot 1-1\cdot 1-1\cdot 0=1$ in $\mathcal{U}$, and $1\cdot 0+1\cdot 1-1\cdot 0-1\cdot 0=1$ in $\mathcal{B}$. The separate coordinate-plane contributions change, but their sum does not.

The same matrix also represents an active symplectic linear map: keeping one basis fixed, it sends $(q,p)$ to $(q+Cp,p)$. The following diagram shows this active interpretation over the real numbers. Its preserved total pairing is precisely the test used for linear canonical transformations.

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]

Two square matrices $A,C$ are congruent when $C=P^{T}AP$ for an invertible matrix $P$. This is the transformation of the matrix of a bilinear form under change of basis, and also transforms a quadratic form under an invertible linear map on its coordinates. The matrices need not be symmetric. It differs from matrix similarity, which uses $P^{-1}AP$ for a linear operator. For real symmetric matrices, completing the square diagonalizes the corresponding quadratic form by matrix congruence.

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]

Choose a basis and let $A$ be the matrix of $T$. Use the monic conventionUnder a change of basis, $A$ becomes $B=P^{-1}AP$ for an invertible matrix $P$. Thenso multiplicativity of the determinant gives $\det (xI-B)=\det (xI-A)$. Therefore the characteristic polynomial is independent of the chosen basis and is well-defined for the linear transformation itself.

Use the $m$-dimensional space $V_m$ for this change of basis. With the columns of $P$ recording the new basis in the old one, bilinearity givesThus the matrices of the bilinear form satisfyThis is matrix congruence, rather than the similarity rule for an endomorphism: both arguments of the bilinear form change basis.

Use the standard structural fact that $\mathfrak{sp}(V)$ is a semisimple Lie algebra over an algebraically closed field of characteristic zero. If $\mathfrak{h}$ is a Cartan subalgebra, part (a)(ii) gives $C_{\mathfrak{sp}(V)}(\mathfrak{h})=\mathfrak{h}$, and $\mathfrak{h}$ is an abelian Lie algebra. Part (c) therefore shows that its weight spaces occur as $n$ pairs of one-dimensional spaces, with no zero weight space.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_n$ of its nonzero weights of a representation. Pick $p_i\ne 0$ in $V_{{\lambda }_i}$ and $q_i\ne 0$ in $V_{-{\lambda }_i}$, scaling $q_i$ so that $\omega (p_i,q_i)=1$. Part (b) shows that all other pairings among different pairs are zero. Each individual weight space is one-dimensional and the bilinear form is alternating, so the ordered basisis a symplectic basis. Every $h\in \mathfrak{h}$ is a diagonal matrix in this basis, with entries ${\lambda }_1(h),\dots ,{\lambda }_n(h),-{\lambda }_1(h),\dots ,-{\lambda }_n(h)$.

Let $D_{\mathcal{B}}$ be the Lie subalgebra consisting of all diagonal matricesEach such matrix lies in the symplectic Lie algebra and commutes with $\mathfrak{h}$. Henceand $\mathfrak{h}=D_{\mathcal{B}}$. In particular, every Cartan subalgebra has dimension $n$; the ${\lambda }_i$ are a basis of ${\mathfrak{h}}^{\ast }$.

Carry out the same construction for ${\mathfrak{h}}^{\prime }$, obtaining a symplectic basis ${\mathcal{B}}^{\prime }=(p_1^{\prime },\dots ,p_n^{\prime },q_1^{\prime },\dots ,q_n^{\prime })$ with ${\mathfrak{h}}^{\prime }=D_{{\mathcal{B}}^{\prime }}$. The unique linear map $g$ sending $p_i$ to $p_i^{\prime }$ and $q_i$ to $q_i^{\prime }$ is invertible and preserves all pairings of basis vectors. By bilinearity, it preserves $\omega$ on all of $V$, so $g$ belongs to the symplectic group $\mathrm{Sp}(V)$. The change of basis givesThus conjugacy of Cartan subalgebras of the symplectic Lie algebra follows by explicitly matching their symplectic bases. If $n=0$, the Lie algebra is zero and the same conclusion holds with $g$ the identity on the zero vector space.

Take the inner product with $u_i$ in $a={\sum }_j{a}_j^{\prime }{w}_j$. The reciprocal basis condition gives $a_i^{\prime }=⟨u_i,a⟩$. Expanding instead in the $u$ basis proves the reciprocal-basis coordinate transformationThus $G$ is the Gram matrix of the original basis. It is a positive-definite matrix, so this change of basis is invertible.

In the specified two-dimensional Cartesian frame,Solving the four dot product conditions givesThey are perpendicular to $u_2$ and $u_1$, respectively, with lengths adjusted so the other dot product is one. Equating the Cartesian components of $a=a_1{u}_1+a_2{u}_2=a_1^{\prime }{w}_1+a_2^{\prime }{w}_2$ givesConsequentlywhich agrees entry by entry with the Gram matrix computed directly from the two unit basis vectors.

The $j$th column of the matrix representation of a linear map is the coordinate vector of the image of the $j$th basis vector:Define $C$ by $w_j={\sum }_i{C}_{ij}{u}_i$, so its columns express the new basis vectors in the old basis. For every vector, $[v{]}_u=C[v{]}_w$. Applying $\varphi$ in either coordinate system gives $AC[v{]}_w=C{A}^{\prime }[v{]}_w$, proving the change of basis formulaTo make $A^{\prime }$ diagonal, choose the columns of $C$ to be linearly independent eigenvectors of $A$, expressed in the $u$ basis, and put their corresponding eigenvalues on the diagonal in the same order. This is possible precisely when $A$ is a diagonalizable matrix over the stated field; an eigenbasis need not exist over an arbitrary field.

The linear system of differential equations has matrixThe columns of $C$ are eigenvectors of $M$. The change of basis $v^{\prime }=C^{-1}v$ therefore gives $v_1^{\prime }=(v_1-i{v}_2)/2$, $v_2^{\prime }=(v_1+i{v}_2)/2$, with independent first-order linear differential equationsThe initial velocity gives $K_{+}=-i{v}_0/2$, $K_{-}=i{v}_0/2$. Transforming back, $v=C{v}^{\prime }$, yields the real velocity vector componentsThese also cover $\alpha =0$. Complex coordinates have decoupled a real rotation without making the physical velocity complex.

The characteristic polynomial isThus the eigenvalues are $i$ and $-i$. There is no real diagonalization, because a real diagonal matrix would have real eigenvalues. Over the complex numbers, the two distinct eigenvalues give linearly independent eigenvectors, so the matrix is diagonalizable.

For $\lambda =i$, the first row gives $(1-i)x+2y=0$. Choose $x=2$, giving $y=-1+i$. Its standard complex norm is $\sqrt{|2{|}^2+|-1+i{|}^2}=\sqrt{6}$; the conjugate eigenvector belongs to $-i$. Hence normalized eigenvectors and a suitable change of basis areDirectly, $AP=P\mathrm{diag}(i,-i)$ and $\det P=-2i/3\ne 0$, soNormalization does not assert that these eigenvectors are orthogonal.

For an invertible matrix $P$ whose columns give the new basis in old coordinates, the new matrix is $B=P^{-1}AP$. Multiplicativity of the determinant givessince $\det (P^{-1})\det P=1$.

To justify the cyclic property of the trace directly, for two square matrices $R,S$,Apply it with $R=P^{-1}A$ and $S=P$:Thus both the determinant and trace are invariant under change of basis.