Linear Operators and Matrix Representations

A linear operator is the correct notion of a structure-preserving map between vector spaces: a function $T:V\to W$ that sends sums to sums and scalar multiples to scalar multiples, $T(\alpha u+\beta v)=\alpha T(u)+\beta T(v)$ . That single algebraic requirement — additivity together with homogeneity — is deceptively powerful, because in finite dimensions it forces the map to be rigid. A linear operator cannot do anything surprising once you know what it does to a basis; every other value is fixed by linearity. Specify the $n$ image vectors $T(e_1),\dots,T(e_n)$ and the operator is completely determined. The payoff of that rigidity is the matrix representation. Expand each image $T(e_j)$ in a basis of the target space and the coefficients assemble into a rectangular array $A$ whose $j$ -th column is the coordinate vector of $T(e_j)$ . Abstract operators thereby acquire concrete coordinates, and every operator statement becomes a matrix statement: applying the map becomes a matrix–vector product $\mathbf y=A\mathbf x$ , composing two maps becomes matrix multiplication $BA$ , the identity operator becomes the identity matrix, and invertibility becomes nonsingularity. Heisenberg's matrix mechanics, a rotation of the sky, the derivative acting on polynomials, and a finite-difference Laplacian are all one idea wearing different clothes. The subtlety that organizes everything downstream is that the matrix depends on the chosen bases while the operator does not. Change the basis and the same operator is represented by a different matrix, related to the old one by a similarity transformation $A'=P^{-1}AP$ . The features that survive that conjugation — rank, trace, determinant, the characteristic polynomial, the spectrum — are the operator's genuine content, its physics; the individual entries are just one coordinate photograph. Diagonalization, the spectral theorem, and the Jordan form are all attempts to choose the basis in which that photograph is as simple as it can possibly be.

Key takeaways

A linear operator $T:V\to W$ satisfies $T(\alpha u+\beta v)=\alpha T(u)+\beta T(v)$ ; in finite dimensions it is fixed completely by its action on a basis.
Stacking the images of the basis vectors as columns produces the matrix $A$ , so the abstract map becomes the concrete product $\mathbf y=A\mathbf x$ .
Composition of operators is matrix multiplication — the row-by-column rule is forced, not arbitrary — and the identity operator is the identity matrix $I$ .
The matrix is basis-dependent: a change of basis conjugates it, $A'=P^{-1}AP$ (similarity) for an operator on one space, $A'=Q^{-1}AP$ (equivalence) for a map between two.
The set $\mathcal L(V,W)$ of linear maps is itself a vector space, of dimension $(\dim V)(\dim W)$ .
Rank, trace, determinant, and eigenvalues are invariants of the operator, unchanged by the representing basis.
Key idea: a linear map is fixed by its action on a basis; writing those images in coordinates gives a matrix, so abstract operator algebra (composition, inverse, identity) becomes concrete matrix algebra (multiplication, inverse, identity) — yet the matrix is basis-dependent while the operator is not.

The core insight

A linear map is fixed by its action on a basis; writing those images in coordinates gives a matrix, so abstract operator algebra (composition, inverse, identity) becomes concrete matrix algebra (multiplication, inverse, identity) — but the matrix is basis-dependent while the operator is not.

Where this lives today

The operator-and-representation dichotomy is not a historical curiosity — it is the working grammar of computational physics. When a quantum chemist solves for the electronic structure of a molecule, the Hamiltonian is an abstract operator on an infinite-dimensional Hilbert space; the practical calculation begins by choosing a finite basis of atomic orbitals, whereupon the operator becomes a finite Hermitian matrix and the energy levels become its eigenvalues, computed by exactly the diagonalization of Worked Example 2 scaled to thousands of dimensions. The same move drives lattice quantum chromodynamics, where the Dirac operator on a discretized spacetime is a sparse matrix whose spectrum encodes hadron masses, and it drives every finite-element and finite-difference PDE solver, where a differential operator like the Laplacian becomes the sparse matrix of Worked Example 1 generalized to a grid. In quantum computing, a circuit is literally a product of unitary matrices — composition as multiplication, equation (4), executed on hardware — and the entire enterprise of choosing a good basis to make a matrix sparse or block-diagonal is what modern numerical linear algebra, from Krylov solvers to tensor-network methods, exists to automate. Google's PageRank, a search for the dominant eigenvector of a web-link operator, is the same mathematics wearing a commercial suit.

The physical picture

A linear operator is a transformation that keeps grid lines straight, parallel, and evenly spaced while pinning the origin in place. Squares become parallelograms, never curved cells; the tiling stays a tiling. That rigidity is the whole reason a matrix suffices to describe the map: knowing where the edge vectors of one cell go tells you where every point goes, because the rest of the plane is just copies of that cell laid end to end. The first figure makes this concrete for a specific operator — the shear-and-stretch with columns (2,0) and (1,1) — and reads off the one number that measures how the operator distorts area.

Figure

A linear operator keeps the grid straight and parallel: the shear-scale map with columns (2,0) and (1,1) carries the unit square to a parallelogram, and the factor by which every area is scaled is exactly det A = 2

The matrix is nothing more than a bookkeeping device for that picture: stack the image of each basis vector as a column, and multiplying the matrix by a coordinate column computes the image of the corresponding vector. Two features of the operator refuse to depend on which basis you chose to build the matrix. One is the area (or volume) scaling factor, the determinant, visible above as det A = 2. The other is the trace, and both are best seen for a rotation, where the operator is manifestly a geometric act — spinning the plane — that has nothing to do with coordinates. The second figure shows a 30° rotation and reads off its two invariants directly.

Figure

A rotation by 30° is a coordinate-free geometric act: it preserves length so det R = 1, and its trace 2cos30° = 1.732 is basis-independent — the same number in every coordinate system, while the eigenvalues are the unit-modulus pair e to the plus or minus i30°

The core equations

Everything begins with the defining property. A linear operator is exactly a map that respects the two operations a vector space is built from — addition and scalar multiplication — and a single equation encodes both at once.

T(\alpha u+\beta v)=\alpha\,T(u)+\beta\,T(v) \qquad (1)

where: $T$ — linear operator from $V$ to $W$ ; $u,v$ — vectors in $V$ ; $\alpha,\beta$ — scalars from the field $\mathbb F$ (typically $\mathbb R$ or $\mathbb C$ ). Reading it: whatever combination you form before applying $T$ , you get the same combination of the images afterward — so $T$ never bends or curves the space, and in particular $T(0)=0$ . Once bases $\{e_j\}$ of $V$ and $\{f_i\}$ of $W$ are fixed, the operator is pinned down by the coordinates of the images of the basis vectors. Those coordinates, arranged in columns, are the matrix.

T(e_j)=\sum_{i=1}^{m} A_{ij}\,f_i \qquad (2)

where: $e_j$ — $j$ -th basis vector of $V$ (dimension $n$ ); $f_i$ — $i$ -th basis vector of $W$ (dimension $m$ ); $A_{ij}$ — entry in row $i$ , column $j$ . Reading it: the $j$ -th column of $A$ is literally the image of the $j$ -th basis vector, written in the target basis. Transpose this convention by accident and every subsequent formula silently transposes with it. Because the map is linear and the matrix stores the basis images, applying $T$ to any vector reduces to a matrix–vector product on coordinates.

y_i=\sum_{j=1}^{n}A_{ij}\,x_j,\qquad \mathbf y=A\mathbf x \qquad (3)

where: $x_j$ — coordinates of $v\in V$ in $\{e_j\}$ ; $y_i$ — coordinates of $T(v)\in W$ in $\{f_i\}$ ; $A$ — the $m\times n$ matrix representing $T$ . Reading it: coordinates in, coordinates out, with the abstract operator replaced by an entirely mechanical row-times-column computation. Composition of operators corresponds to matrix multiplication — and this correspondence is what defines the multiplication rule, rather than the other way around.

(S\circ T)\ \longleftrightarrow\ BA,\qquad (BA)_{ik}=\sum_{j} B_{ij}A_{jk} \qquad (4)

where: $T$ — the map applied first, matrix $A$ ; $S$ — the map applied second, matrix $B$ ; $BA$ — the single matrix representing the composite. Reading it: "do $T$ , then $S$ " is one operator, so it must have one matrix; the row-by-column formula is the unique rule that makes $BA$ that matrix. The matrix is not intrinsic to the operator — it changes when you change bases. For a map between two different spaces the transformation law is an equivalence; for an operator from a space to itself it specializes to a similarity.

A'=Q^{-1}AP\ \ \xrightarrow{\ V=W,\ Q=P\ }\ \ A'=P^{-1}AP \qquad (5)

where: $P$ — change-of-basis matrix in $V$ (its columns are the new basis vectors written in old coordinates); $Q$ — change-of-basis matrix in $W$ ; $A,A'$ — the old and new matrices of the same operator. Reading it: conjugation by $P$ is what "looking at the same operator from a different basis" does to its matrix. The linear maps between two fixed spaces do not merely exist one at a time — they can be added and scaled, so they form a vector space of their own, whose dimension counts the free entries of a matrix.

\dim\mathcal L(V,W)=(\dim V)(\dim W)=mn \qquad (6)

where: $\mathcal L(V,W)$ — the set of all linear maps $V\to W$ ; $m=\dim W$ , $n=\dim V$ . Reading it: an $m\times n$ matrix has $mn$ independently choosable entries, and each choice is a distinct map, so the space of maps is $mn$ -dimensional. Finally, the quantities that a change of basis leaves alone are the operator's true invariants — the numbers a physicist can quote without first agreeing on coordinates.

\operatorname{tr}A=\sum_i A_{ii},\qquad \det A,\qquad \operatorname{rank}A \qquad (7)

where: $\operatorname{tr}A$ — the trace, sum of diagonal entries; $\det A$ — the determinant, the signed volume-scaling factor; $\operatorname{rank}A$ — the dimension of the image. Reading it: each of these is unchanged under $A\mapsto P^{-1}AP$ , so each is a property of $T$ itself; the eigenvalues and the whole characteristic polynomial join them.

Derivation AThe similarity transformation law

The goal is to derive precisely how the matrix of an operator $T:V\to V$ changes when we keep the operator fixed but swap the basis from $\{e_j\}$ to a new basis $\{e_j'\}$ . This is the single most important computation in the subject, because it draws the line between what belongs to the operator and what belongs to a mere coordinate choice; every invariant we later trust — trace, determinant, spectrum — is trustworthy exactly because it survives this transformation. We define the change-of-basis matrix, track a vector's coordinates through old and new descriptions, and read off the new matrix.

Step 1. Define $P$ by writing each new basis vector in the old basis, $e_j'=\sum_i P_{ij}e_i$ ; the columns of $P$ are the new basis vectors in old coordinates. A vector with new coordinates $\mathbf x'$ then has old coordinates $\mathbf x=P\mathbf x'$ . Why: $P$ is exactly the dictionary that translates a coordinate list from the new language back into the old one.

Step 2. In the old basis the operator acts by its known matrix, $\mathbf y=A\mathbf x$ , where $\mathbf y$ is the image's old coordinates. Why: this is just the defining action of the representing matrix, equation (3), applied in the basis we already understand.

Step 3. Translate the image back into the new basis with the inverse dictionary, $\mathbf y'=P^{-1}\mathbf y$ , and substitute the previous two steps: $\mathbf y'=P^{-1}A\mathbf x=P^{-1}AP\,\mathbf x'$ . Why: we want the action expressed entirely in new coordinates, so both the input and the output must be converted through $P$ .

Step 4. By definition the new matrix $A'$ is whatever satisfies $\mathbf y'=A'\mathbf x'$ for every $\mathbf x'$ , so comparison gives $A'=P^{-1}AP$ . Why: two matrices that agree on every input coordinate vector are equal, which forces the boxed identity.

A'=P^{-1}AP

The general two-space version is $A'=Q^{-1}AP$ .

Go deeper · Derivation B · 1 minThe trace is a similarity invariant

Derivation BThe trace is a similarity invariant

The goal is to prove that the trace depends only on the operator, never on the basis used to represent it, so that the diagonal sum $\sum_i A_{ii}$ — which looks hopelessly basis-specific — is in fact a genuine attribute of $T$ . This matters physically because the trace is the sum of eigenvalues and, in quantum mechanics, the tool behind expectation values and partition functions; if it changed with coordinates it would be useless. The engine is a single algebraic fact, the cyclic property of the trace, applied to the similarity law we just derived.

Step 1. Establish cyclicity: $\operatorname{tr}(XY)=\sum_i\sum_j X_{ij}Y_{ji}=\sum_j\sum_i Y_{ji}X_{ij}=\operatorname{tr}(YX)$ . Why: the double sum is symmetric under relabelling, so a product's trace is unchanged when the two factors swap places — the one identity the whole proof rests on.

Step 2. Apply it to the similarity form $\operatorname{tr}(P^{-1}AP)$ by grouping $X=P^{-1}A$ and $Y=P$ , then cycling: $\operatorname{tr}\big((P^{-1}A)P\big)=\operatorname{tr}\big(P(P^{-1}A)\big)$ . Why: cyclicity lets us move the rightmost factor $P$ around to the front, where it will meet its own inverse.

Step 3. Collapse the product: $\operatorname{tr}\big((PP^{-1})A\big)=\operatorname{tr}(IA)=\operatorname{tr}(A)$ . Why: $PP^{-1}=I$ , so the change of basis leaves no trace of itself — the diagonal sum is identical in both bases.

\operatorname{tr}(P^{-1}AP)=\operatorname{tr}(A)

The determinant is invariant by a multiplicative route instead: $\det(P^{-1}AP)=\det(P^{-1})\det(A)\det(P)=\tfrac{1}{\det P}\det(A)\det(P)=\det A$ .

Go deeper · Derivation C · 1 minThe space of linear maps has dimension m times n

Derivation CThe space of linear maps has dimension m times n

The goal is to show that $\mathcal L(V,W)$ is not just a set of maps but a vector space, and to compute its dimension by building an explicit basis for it out of the simplest possible operators. The result, $\dim\mathcal L(V,W)=mn$ , is the precise statement that the correspondence "operator to matrix" is a bijection onto the $mn$ -dimensional space of $m\times n$ arrays — the deepest justification for why matrices are the right bookkeeping device at all. We fix bases, define single-transition elementary maps, and check they span and are independent.

Step 1. Fix bases $\{e_1,\dots,e_n\}$ of $V$ and $\{f_1,\dots,f_m\}$ of $W$ , and define elementary maps by $E_{ij}(e_k)=\delta_{jk}f_i$ — the map that sends $e_j$ to $f_i$ and kills every other basis vector. Why: there is one such map for each row–column slot, the operator analogue of a matrix with a single 1.

Step 2. Show spanning: any $T$ equals $\sum_{i=1}^{m}\sum_{j=1}^{n}A_{ij}E_{ij}$ , where $A_{ij}$ are its matrix entries, because both sides send each $e_k$ to $\sum_i A_{ik}f_i=T(e_k)$ . Why: two linear maps that agree on a basis agree everywhere, so matching the basis action is enough to prove equality.

Step 3. Show independence: if $\sum_{ij}c_{ij}E_{ij}=0$ , evaluate on $e_k$ to get $\sum_i c_{ik}f_i=0$ , and since the $f_i$ are independent every $c_{ik}=0$ . Why: only the trivial combination gives the zero map, which is exactly what linear independence demands.

Step 4. Conclude that the $mn$ maps $\{E_{ij}\}$ form a basis, so $\dim\mathcal L(V,W)=mn$ . Why: a spanning, independent set is a basis, and its cardinality is the dimension.

\dim\mathcal L(V,W)=mn

Limiting cases

Special operators, or specially chosen bases, collapse the matrix to a recognizable canonical form. Each row below is a sanity check the general theory must reproduce.

Case	Result	Meaning
Identity operator	$T=\mathrm{id}\ \Rightarrow\ A=I_n$ in every basis	the unique operator whose matrix is basis-independent; $P^{-1}I_nP=I_n$ for all $P$
Diagonalizable operator, eigenbasis	$A=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$	when a full basis of eigenvectors exists, the eigenbasis makes the matrix diagonal with eigenvalues on the diagonal
Defective operator	$A=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$	too few eigenvectors to diagonalize; the simplest form is the Jordan block with off-diagonal 1's
Projection	$P^2=P\ \Rightarrow\ A\sim\operatorname{diag}(1,\dots,1,0,\dots,0)$	idempotent; splits the space into image and kernel, eigenvalues only 0 and 1
Nilpotent operator	$A^k=0$ for some $k$	e.g. the derivative on degree- $<k$ polynomials; strictly upper-triangular in a suitable basis, so trace and determinant vanish

Common trapConfusing the operator with its matrix

The single most common error is treating a property of one matrix as though it were a property of the operator. The matrix changes under a change of basis by $A'=P^{-1}AP$ , so anything tied to specific entries — the values in the array, whether it happens to be upper-triangular, whether it is symmetric — is a fact about the basis, not the operator. Two matrices that look completely different can be the very same operator seen through two windows. What genuinely belongs to the operator is precisely the list of quantities invariant under conjugation: rank, trace, determinant, characteristic polynomial, and eigenvalues. A second, quieter trap lives in the column convention: the $j$ -th column of $A$ is the image of the $j$ -th basis vector, expressed in the target basis. Build the matrix row-wise by habit and you have silently represented the transpose $T^{\mathsf T}$ (or the adjoint), and every composition and eigenvalue computation downstream inherits the mistake.

Worked examples

1 · The derivative as a matrix (fully worked)

Let $V$ be the space of real polynomials of degree at most 2, with the ordered basis $\{1,\,x,\,x^2\}$ . The differentiation operator $D=\mathrm d/\mathrm dx$ maps $V\to V$ . Build its matrix, read off its invariants, and use it to differentiate $p(x)=3+2x+5x^2$ .

Step 1. Apply $D$ to each basis vector and expand the result in the basis. $D(1)=0$ , $D(x)=1$ , $D(x^2)=2x$ . In coordinates $(c_0,c_1,c_2)$ for $c_0+c_1x+c_2x^2$ , these images are $(0,0,0)$ , $(1,0,0)$ , and $(0,2,0)$ .

Step 2. Stack the images as the columns of the matrix, in order — column $j$ is the image of the $j$ -th basis vector (equation 2):

A=\begin{pmatrix}0&1&0\\0&0&2\\0&0&0\end{pmatrix}

Step 3. Read the invariants. The matrix is strictly upper-triangular, so $\operatorname{tr}A=0$ and $\det A=0$ ; its two nonzero, independent columns give $\operatorname{rank}A=2$ (the image is the degree- $\le1$ polynomials, and the kernel is the constants). Cubing it gives $A^3=0$ : the operator is nilpotent, as it must be, since three derivatives annihilate any quadratic.

Step 4. Differentiate $p=3+2x+5x^2$ by matrix–vector multiplication. Its coordinate vector is $(3,2,5)^{\mathsf T}$ , and

A\begin{pmatrix}3\\2\\5\end{pmatrix}=\begin{pmatrix}0\cdot3+1\cdot2+0\cdot5\\0\cdot3+0\cdot2+2\cdot5\\0\end{pmatrix}=\begin{pmatrix}2\\10\\0\end{pmatrix}

which is the polynomial $2+10x$ . And indeed $\dfrac{\mathrm d}{\mathrm dx}(3+2x+5x^2)=2+10x$ . The abstract operator "differentiate" is fully captured by one $3\times3$ matrix, and calculus becomes matrix arithmetic.

2 · Diagonalizing by a similarity transformation (fully worked)

The operator $T$ on $\mathbb R^2$ has, in the standard basis, the matrix $A=\begin{pmatrix}2&1\\0&1\end{pmatrix}$ — the very shear-scale of the first figure. Find a basis in which its matrix is diagonal, and confirm that trace and determinant are basis-independent.

Step 1. Find the eigenvalues from $\det(A-\lambda I)=0$ . Since $A$ is upper-triangular, $(2-\lambda)(1-\lambda)=0$ , so $\lambda_1=2$ and $\lambda_2=1$ . Two distinct eigenvalues in two dimensions guarantee a full eigenbasis, hence diagonalizability.

Step 2. Find an eigenvector for each. For $\lambda_1=2$ : $(A-2I)\mathbf v=0$ gives $\mathbf v_1=(1,0)^{\mathsf T}$ . For $\lambda_2=1$ : $(A-I)\mathbf v=0$ gives $\begin{pmatrix}1&1\\0&0\end{pmatrix}\mathbf v=0$ , so $\mathbf v_2=(-1,1)^{\mathsf T}$ .

Step 3. Assemble the change-of-basis matrix from the eigenvectors as columns, and invert it:

P=\begin{pmatrix}1&-1\\0&1\end{pmatrix},\qquad P^{-1}=\begin{pmatrix}1&1\\0&1\end{pmatrix}

Step 4. Apply the similarity law $A'=P^{-1}AP$ :

P^{-1}AP=\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}2&1\\0&1\end{pmatrix}\begin{pmatrix}1&-1\\0&1\end{pmatrix}=\begin{pmatrix}2&0\\0&1\end{pmatrix}

The matrix is now diagonal, with the eigenvalues on the diagonal — the eigenbasis is the coordinate system in which $T$ merely stretches by 2 along $\mathbf v_1$ and does nothing along $\mathbf v_2$ .

Step 5. Check the invariants against the two representations. Trace: $2+1=3$ for $A$ and $2+1=3$ for $A'$ . Determinant: $2\cdot1-1\cdot0=2$ for $A$ and $2\cdot1=2$ for $A'$ . Both are unchanged, exactly as the invariance derivations promised — the diagonal form is a simpler photograph of the same operator.

Physics of the portrait

Matrices reached the world late. For decades a "linear substitution" was written out as a system of equations, and its composition was a laborious substitution by hand; it took a mathematician's eye to see that the array of coefficients was an object in its own right, with an algebra obeying rules of its own.

Origins · 1 minArthur Cayley and how this was discovered

Arthur Cayley

1821–1895 · the invention of matrix algebra

Arthur Cayley was a Cambridge-trained mathematician who spent fourteen years as a conveyancing barrister — publishing some 250 papers in his spare time — before returning to a professorship in 1863. His decisive contribution to this subject is the 1858 "A Memoir on the Theory of Matrices," which for the first time treated a rectangular array of numbers as a single algebraic quantity that could be added, multiplied, and inverted. Crucially, Cayley did not define matrix multiplication arbitrarily: he defined it so that the product of two matrices would represent the composition of the two linear substitutions they stood for — precisely the reasoning behind equation (4) here. He recognized that this multiplication is associative but not commutative, isolated the identity and zero matrices, and stated what is now the Cayley–Hamilton theorem, that every matrix satisfies its own characteristic equation (he verified it in the 2×2 and 3×3 cases and declared the general case needless to prove). The word "matrix" itself was coined a few years earlier, in 1850, by his lifelong friend and collaborator James Joseph Sylvester, who meant it as the "womb" from which determinants are born. Cayley's abstraction lay largely dormant until the twentieth century, when Werner Heisenberg — unknowingly reinventing matrix multiplication — found that the non-commuting arrays of his 1925 quantum mechanics were exactly Cayley's objects, and the theory of linear operators became the language of physics.

History

1843

— Hamilton's quaternions expose a useful non-commutative multiplication

1850

— Sylvester coins the term matrix for an array generating determinants

1858

— Cayley's Memoir on the Theory of Matrices founds matrix algebra

1878

— Frobenius develops rank, minimal polynomials, and the general Cayley–Hamilton theorem

1888

— Peano gives the modern axiomatic definition of a vector space and linear map

1925

— Heisenberg, Born, and Jordan build quantum matrix mechanics on operator algebra

Going further

Texts. Arfken & Weber, Mathematical Methods for Physicists, 7th ed., Ch. 3 (matrices and operators) · Hassani, Mathematical Physics, Ch. 3–4 (vectors, operators, matrices) · Stone & Goldbart, Mathematics for Physics, Ch. on linear operators · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 3–4 · Nakahara, Geometry, Topology and Physics, §2.2 (linear maps as a prelude to tensors) · Axler, Linear Algebra Done Right, Ch. 3 (the basis-free viewpoint) · Halmos, Finite-Dimensional Vector Spaces. The core results are: the column convention $T(e_j)=\sum_i A_{ij}f_i$ with induced action $\mathbf y=A\mathbf x$ ; the composition rule $(BA)_{ik}=\sum_j B_{ij}A_{jk}$ representing $S\circ T$ ; the similarity law $A'=P^{-1}AP$ with trace and determinant invariant by cyclicity; and the dimension count $\dim\mathcal L(V,W)=mn$ . The two worked examples are exact: the derivative matrix on $\{1,x,x^2\}$ is $A=\big(\begin{smallmatrix}0&1&0\\0&0&2\\0&0&0\end{smallmatrix}\big)$ with $A(3,2,5)^{\mathsf T}=(2,10,0)^{\mathsf T}$ ; and $P^{-1}AP=\operatorname{diag}(2,1)$ for $A=\big(\begin{smallmatrix}2&1\\0&1\end{smallmatrix}\big)$ , $P=\big(\begin{smallmatrix}1&-1\\0&1\end{smallmatrix}\big)$ , with trace 3 and determinant 2 in both bases. The figures assert only $\det\big(\begin{smallmatrix}2&1\\0&1\end{smallmatrix}\big)=2$ and, for $R(30°)$ , $\det R=1$ with $\operatorname{tr}R=2\cos30°=1.732$ .

Prerequisites. Vector Spaces and Subspaces · Linear Independence, Basis, and Dimension. Leads to. Change of Basis and Similarity Transformations · Rank, Nullity, and the Fundamental Theorem of Linear Maps · Eigenvalues, Eigenvectors, and the Characteristic Equation · Adjoint and Hermitian Operators.

Ask the tutor: ask me to build the matrix of any operator you name — a rotation, a projection, the derivative, a spin observable — and then diagonalize it and read off its invariants step by step.

Is there any difference between a linear operator and its matrix?

Yes, and keeping them distinct is the whole point of this article. The operator is the geometric or algebraic action itself — "rotate the plane," "differentiate the polynomial" — and it exists before any coordinates are chosen. The matrix is a numerical record of that action relative to a specific basis; change the basis and the record changes while the action does not. The relationship is like that between a physical rotation of an object and the list of numbers you would write down to describe it in a particular coordinate frame. This is why physicists care about invariants such as the trace, determinant, and spectrum: they are the features of the record that do not depend on the frame, so they describe the operator rather than your description of it.

Why exactly is matrix multiplication defined so strangely?

Because it has to represent composition of maps, and no other rule does. If $T$ is applied first (matrix $A$ ) and $S$ second (matrix $B$ ), the composite $S\circ T$ is a single linear operator and therefore has a single matrix. Tracking a basis vector through both maps, $(S\circ T)(e_k)=\sum_i\big(\sum_j B_{ij}A_{jk}\big)f_i$ , and the coefficient in the parentheses is forced to be the $(i,k)$ entry of the product $BA$ . The row-by-column rule is the unique definition consistent with "do one map, then the other." Entrywise multiplication, which looks more natural, corresponds to no operation on the underlying operators at all, which is why it plays no role in this theory.

What is the difference between similarity and equivalence?

Both describe how a matrix changes with a change of basis, but they apply to different situations. Equivalence, $A'=Q^{-1}AP$ , is the general law for a map $T:V\to W$ between two different spaces, where you may change the basis of the source ( $P$ ) and independently the basis of the target ( $Q$ ). Similarity, $A'=P^{-1}AP$ , is the special case where the map goes from a space to itself, $V=W$ , so there is only one basis to change and $Q=P$ . The distinction matters because only similarity preserves eigenvalues, trace, and determinant — those concepts require a single space where inputs and outputs can be compared. Under mere equivalence only the rank survives, which is why the rank is the complete invariant classifying maps between distinct spaces.

If two matrices share their trace, determinant, and eigenvalues, are they the same operator?

Not necessarily. Equal invariants are necessary for two matrices to represent the same operator, but they are not sufficient. The classic counterexample is a pair of $2\times2$ matrices both with eigenvalue $\lambda$ twice: the diagonal matrix $\lambda I$ and the Jordan block $\big(\begin{smallmatrix}\lambda&1\\0&\lambda\end{smallmatrix}\big)$ have the same trace $2\lambda$ , determinant $\lambda^2$ , and spectrum $\{\lambda,\lambda\}$ , yet they are genuinely different operators — one is diagonalizable and one is defective, and no invertible $P$ conjugates one into the other. The full invariant that decides similarity is the Jordan canonical form (equivalently, the list of elementary divisors), which records not just the eigenvalues but the sizes of the Jordan blocks.

How does an infinite-dimensional operator become a finite matrix in practice?

By truncation in a chosen basis. A differential operator on a function space is genuinely infinite-dimensional, but physics almost always proceeds by expanding functions in a finite set of basis functions — plane waves, atomic orbitals, finite-element bump functions, grid samples — and keeping only their coefficients. In that finite basis the operator's action is captured, approximately, by a finite matrix whose entries are the operator "sandwiched" between basis functions. The approximation improves as the basis grows, and the whole art of numerical physics is choosing a basis in which a modest matrix already captures the physics you care about. Worked Example 1 is the exact version of this: on the polynomials of degree at most two, the derivative is captured with no approximation at all by a $3\times3$ matrix.

Is the identity operator the only operator with a basis-independent matrix?

Essentially yes, up to scaling. The identity has matrix $I_n$ in every basis because $P^{-1}I_nP=I_n$ for all invertible $P$ . More generally, any scalar operator $\lambda\,\mathrm{id}$ has the same matrix $\lambda I_n$ in every basis, since $P^{-1}(\lambda I_n)P=\lambda I_n$ . These scalar multiples of the identity are exactly the operators that commute with everything, and they are the only ones whose matrix is a similarity invariant on the nose. Every other operator looks different in different bases — which is fortunate, because it is precisely by choosing a clever basis (the eigenbasis) that we simplify a non-scalar operator to diagonal form, as in Worked Example 2.