Matrices as Linear Maps

A matrix is a coordinate representation of a linear map after bases have been chosen. The linear map is the object; the matrix is how that object acts on coordinate columns.

This page is the bridge between abstract linear maps and the matrices used in finite-dimensional quantum mechanics, spin systems, numerical diagonalization, and operator representations.

Matrix of a Linear Map

Let

$$T:V\to W$$

be a linear map. Choose a basis

$$\mathcal B=(e_1,\ldots,e_n)$$

for $V$ and a basis

$$\mathcal C=(f_1,\ldots,f_m)$$

for $W$.

The matrix of $T$ in these bases is written

$$[T]_{\mathcal C\leftarrow\mathcal B}.$$

Its $j$th column is the coordinate column of $T(e_j)$ in the basis $\mathcal C$:

$$T(e_j) = \sum_{i=1}^m T^i{}_j f_i.$$

Thus the matrix entries $T^i{}_j$ record what the map does to each basis vector.

For the index conventions behind expressions such as $T^i{}_j$, see Index Notation and Summation Conventions.

Acting on Coordinates

If

$$v=\sum_j c^j e_j,$$

then linearity gives

$$T(v) = \sum_j c^jT(e_j) = \sum_{i,j}T^i{}_j c^j f_i.$$

In coordinate-column notation,

$$[T(v)]_{\mathcal C} = [T]_{\mathcal C\leftarrow\mathcal B} [v]_{\mathcal B}.$$

This is the precise meaning of multiplying a matrix by a column vector: it computes the coordinates of the output vector in the chosen output basis.

Matrix Multiplication Is Composition

Let

$$T:U\to V, \qquad S:V\to W.$$

Choose bases $\mathcal A$ for $U$, $\mathcal B$ for $V$, and $\mathcal C$ for $W$. Then

$$[S\circ T]_{\mathcal C\leftarrow\mathcal A} = [S]_{\mathcal C\leftarrow\mathcal B} [T]_{\mathcal B\leftarrow\mathcal A}.$$

Matrix multiplication appears because composition appears. The rightmost matrix acts first on coordinate columns.

This is also why matrix multiplication is generally not commutative. The maps $S\circ T$ and $T\circ S$ may have different meanings, different domains, or different results.

Operators and Square Matrices

When $T:V\to V$ maps a vector space to itself, it is represented by a square matrix after one basis of $V$ is chosen:

$$[T]_{\mathcal B\leftarrow\mathcal B}.$$

Finite-dimensional quantum operators are often introduced this way. For example, a two-level Hamiltonian, a spin component, or a quantum gate can be represented by a $2\times2$ matrix after a basis is chosen.

The matrix is basis-dependent. The operator is not.

Change of Basis and Similarity

Let $A:V\to V$ be a linear operator. Suppose $\mathcal B$ is an old basis and $\mathcal F=(f_1,\ldots,f_n)$ is a new basis.

Use the convention that $P$ has as its columns the $\mathcal B$-coordinates of the new basis vectors:

$$P = \begin{pmatrix} [f_1]_{\mathcal B} & \cdots & [f_n]_{\mathcal B} \end{pmatrix}.$$

Then coordinate columns satisfy

$$[v]_{\mathcal B} = P[v]_{\mathcal F}.$$

If $A_{\mathcal B}$ is the old matrix and $A_{\mathcal F}$ is the new matrix, consistency requires

$$A_{\mathcal F} = P^{-1}A_{\mathcal B}P.$$

This is a similarity transformation. It changes the matrix representation but not basis-independent information such as eigenvalues, trace, determinant, and the abstract operator itself.

The canonical mathematical convention for this transformation is developed in Change of Basis. Quantum pages often use unitary overlap matrices for orthonormal bases; the physics-facing passive convention is explained in Change of Basis.

Worked Example

Define $T:\mathbb R^2\to\mathbb R^2$ by

$$T \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} x+y\\ 2y \end{pmatrix}.$$

In the standard basis $e_1=(1,0)^T$, $e_2=(0,1)^T$,

$$T(e_1) = \begin{pmatrix} 1\\ 0 \end{pmatrix}, \qquad T(e_2) = \begin{pmatrix} 1\\ 2 \end{pmatrix}.$$

These output coordinate columns become the matrix columns:

$$[T] = \begin{pmatrix} 1 & 1\\ 0 & 2 \end{pmatrix}.$$

Then

$$[T] \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} x+y\\ 2y \end{pmatrix},$$

as required.

Why Quantum Mechanics Needs This

Matrix mechanics represents states and operators in chosen bases. A Hamiltonian matrix is not merely an array of numbers; it is the coordinate representation of a linear operator. A gate matrix represents a unitary map in a computational basis. A change to an energy basis, spin basis, or position-momentum representation changes the concrete form while preserving the underlying linear map.

This is why the same operator can appear as a Pauli matrix, a diagonal matrix, a differential expression, or an integral kernel depending on the representation.

Common Mistakes

• Treating matrices as primary and linear maps as optional interpretation.
• Forgetting to specify the input and output bases for a matrix.
• Multiplying matrices in the wrong order because composition order was ignored.
• Changing a state-coordinate column without changing the operator matrix consistently.
• Applying a function to matrix entries when the intended operation is a function of the linear operator.
• Thinking similar matrices are different operators rather than different representations of the same operator.

Cross-Links

• Linear Maps
• Bases and Coordinates
• Index Notation and Summation Conventions
• Change of Basis
• Diagonalization
• Eigenvalues and Eigenvectors
• Spectral Decomposition
• Matrix Diagonalization
• Operator Representations
• Change of Basis

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

1. Let $T:\mathbb R^2\to\mathbb R^2$ be $T(x,y)=(2x-y,x+y)$. Find the matrix of $T$ in the standard basis.

Solution

Compute the images of the standard basis vectors:

$$T(e_1)=T(1,0)=(2,1), \qquad T(e_2)=T(0,1)=(-1,1).$$

These become the columns:

$$[T] = \begin{pmatrix} 2 & -1\\ 1 & 1 \end{pmatrix}.$$

2. If $[v]_{\mathcal B}=P[v]_{\mathcal F}$, why is $A_{\mathcal F}=P^{-1}A_{\mathcal B}P$ the consistent matrix for the same operator in the $\mathcal F$ basis?

Solution

Start with

$$[Av]_{\mathcal B} = A_{\mathcal B}[v]_{\mathcal B} = A_{\mathcal B}P[v]_{\mathcal F}.$$

But also $[Av]_{\mathcal B}=P[Av]_{\mathcal F}=P A_{\mathcal F}[v]_{\mathcal F}$. Hence

$$P A_{\mathcal F} = A_{\mathcal B}P,$$

so

$$A_{\mathcal F} = P^{-1}A_{\mathcal B}P.$$

3. Explain why matrix multiplication is not just a computational rule but a representation of composition.

Solution

If $T$ is applied first and $S$ second, then the output is $S(T(u))$. In coordinates, applying $T$ gives one matrix multiplication, and applying $S$ gives the next. The combined matrix is therefore the product representing $S\circ T$.