Linear Maps

A linear map is a function between vector spaces that preserves addition and scalar multiplication. In quantum mechanics, ordinary state-space operators are linear maps: their action on a superposition is determined by their action on the pieces.

Helpful background. Vector spaces introduce linear combinations. For the physical uses of linear maps, see Operators.

Linear maps and superposition

Let $V$ and $W$ be vector spaces over the same field, usually $\mathbb C$ in quantum mechanics. The complex-scalar case is developed in Complex Vector Spaces. A map

$$T:V\to W$$

is linear if, for all $v,w\in V$ and scalars $a,b$,

$$T(av+bw) = aT(v)+bT(w).$$

Equivalently, a linear map preserves every finite linear combination:

$$T\left(\sum_i a_i v_i\right) = \sum_i a_i T(v_i).$$

This is why knowing $T$ on a basis determines $T$ everywhere. A useful first check is $T(0)=0$: setting $a=b=0$ forces it. For example, $F(x)=2x+1$ fails this check. Passing it is necessary but not sufficient; $F(x)=x^2$ sends zero to zero but fails additivity.

Kernel and Image

The kernel of $T:V\to W$ is the set of vectors mapped to zero:

$$\ker T = \{v\in V:T(v)=0\}.$$

The image, or range, is the set of vectors in $W$ that are actually hit:

$$\operatorname{im}T = \{T(v)\in W:v\in V\}.$$

Both are subspaces: $\ker T$ is a subspace of $V$, and $\operatorname{im}T$ is a subspace of $W$.

In finite dimension, the rank-nullity theorem says

$$\dim V = \dim\ker T+\dim\operatorname{im}T.$$

It counts the independent input directions sent to zero and the independent output directions that remain.

Here is why no directions are missed or counted twice. Choose a basis $k_1,\ldots,k_r$ of $\ker T$ and extend it to a basis $k_1,\ldots,k_r,v_1,\ldots,v_s$ of $V$. Applying $T$ to an arbitrary basis expansion removes the $k_i$ terms, so $Tv_1,\ldots,Tv_s$ span the image. They are also independent: if $\sum_j a_jTv_j=0$, then $\sum_j a_jv_j$ belongs to the kernel and is therefore a combination of the $k_i$. Independence of the full input basis forces every $a_j=0$. Thus the image has dimension $s$, the kernel has dimension $r$, and $\dim V=r+s$. This argument uses the finite-dimensional fact that an independent list can be extended to a basis.

Composition

If

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

then the composition $S\circ T:U\to W$ is

$$(S\circ T)(u)=S(T(u)).$$

The composition of linear maps is linear:

$$(S\circ T)(au+bv) = a(S\circ T)(u)+b(S\circ T)(v).$$

Composition is usually not commutative. In quantum mechanics, this noncommutativity becomes the algebraic origin of commutators such as $[A,B]=AB-BA$.

Invertibility

A linear map $T:V\to W$ is invertible if there is a linear map

$$T^{-1}:W\to V$$

such that

$$T^{-1}T=I_V, \qquad TT^{-1}=I_W.$$

In finite dimension, a linear map between vector spaces of the same dimension is invertible exactly when its kernel is zero:

$$\ker T=\{0\}.$$

Equivalently, its image is all of $W$.

Matrices Represent Linear Maps

After choosing a basis of $V$ and a basis of $W$, a linear map $T:V\to W$ is represented by a matrix. If $V=W=\mathbb C^n$ with the standard basis, the formula is the familiar one:

$$(Tv)^i = \sum_j T^i{}_j v^j.$$

The matrix depends on the chosen bases. The linear map does not. This distinction is essential when translating between abstract operator equations and matrix calculations.

Operators as Endomorphisms

A linear map from a vector space to itself is often called an operator or endomorphism:

$$A:V\to V.$$

Quantum mechanics mainly uses operators on Hilbert spaces. In finite dimension, this is straightforward:

$$A:\mathcal H\to\mathcal H.$$

In infinite-dimensional Hilbert spaces, many important operators are not defined on all of $\mathcal H$. A differential operator may instead be a map

$$A:\mathcal D(A)\to\mathcal H, \qquad \mathcal D(A)\subseteq\mathcal H.$$

The domain $\mathcal D(A)$ is part of the operator. Suppressing it is often harmless in a first finite-dimensional example and dangerous in a rigorous wave-mechanics problem.

Worked Example

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

$$T \begin{pmatrix} x\\ y \end{pmatrix} = \begin{pmatrix} x+y\\ 0 \end{pmatrix}.$$

This map is linear because

$$T \left( a \begin{pmatrix} x_1\\ y_1 \end{pmatrix} + b \begin{pmatrix} x_2\\ y_2 \end{pmatrix} \right) = aT \begin{pmatrix} x_1\\ y_1 \end{pmatrix} + bT \begin{pmatrix} x_2\\ y_2 \end{pmatrix}.$$

Its kernel consists of vectors with $x+y=0$:

$$\ker T = \left\{ \begin{pmatrix} x\\ -x \end{pmatrix} :x\in\mathbb C \right\}.$$

Its image is the one-dimensional subspace

$$\operatorname{im}T = \left\{ \begin{pmatrix} z\\ 0 \end{pmatrix} :z\in\mathbb C \right\}.$$

Thus $\dim\ker T=1$ and $\dim\operatorname{im}T=1$, matching $\dim\mathbb C^2=2$.

Common Mistakes

• Calling a map linear because it is written with a formula.
• Forgetting that linearity requires preservation of both addition and scalar multiplication.
• Treating a matrix as the map itself rather than a representation in chosen bases.
• Assuming composition commutes.
• Ignoring the domain of an operator in infinite-dimensional examples.
• Confusing invertibility with having no zero entries in a matrix.

References

• S. Axler, Linear Algebra Done Right, 4th ed., Springer, 2024, §3B — null spaces, ranges, and the dimension theorem. Open text.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.

Exercises

1. Let $T:\mathbb C^2\to\mathbb C^2$ be

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

Show that $T$ is linear.

Solution

For vectors $u=(x_1,y_1)^T$ and $v=(x_2,y_2)^T$,

$$T(au+bv) = \begin{pmatrix} ax_1+bx_2\\ a(x_1+y_1)+b(x_2+y_2) \end{pmatrix} = aT(u)+bT(v).$$

Hence $T$ is linear.

2. For the map in the first exercise, find the kernel.

Solution

The equation $T(x,y)^T=0$ gives

$$x=0, \qquad x+y=0.$$

Thus $x=0$ and $y=0$, so

$$\ker T=\{0\}.$$

3. Give an example of a map $\mathbb C\to\mathbb C$ that is not linear.

Solution

The map $F(z)=\lvert z\rvert^2$ is not linear. For example,

$$F(iz)=\lvert iz\rvert^2=\lvert z\rvert^2,$$

but

$$iF(z)=i\lvert z\rvert^2,$$

which is not equal to $F(iz)$ unless $z=0$.