Change of Basis for Vectors and Matrices

A change of basis rewrites the same vector or linear operator using a different coordinate system. The vector does not move; its coordinate column changes because the basis used to describe it has changed.

This page fixes the finite-dimensional linear-algebra convention used by the Toolkit. For the physics-facing workflow with state coefficients, observables, and probabilities, see Change of Basis.

Helpful background. bases and coordinates for expanding a vector in a chosen basis; matrix representations for matrix multiplication and operator action.

Basis vectors and coordinate columns

Let $V$ be an $n$-dimensional vector space over $\mathbb F$, where $\mathbb F$ is usually $\mathbb R$ or $\mathbb C$. Let

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

be an old ordered basis and let

$$\mathcal F=(f_1,\ldots,f_n)$$

be a new ordered basis.

Each new basis vector can be expanded in the old basis:

$$f_a = \sum_i P^i{}_a e_i.$$

The change-of-coordinates matrix $P$ is defined by putting these old-basis coordinate columns side by side:

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

With this convention, the columns of $P$ answer the question: how do the new basis vectors look in the old basis?

Because $\mathcal F$ is a basis, $P$ is invertible. If $P$ were not invertible, the proposed new basis vectors would be linearly dependent or would fail to span $V$.

Vector Coordinates

Write the same vector $v\in V$ in the two bases:

$$v = \sum_i c^i e_i = \sum_a d^a f_a.$$

Substituting $f_a=\sum_i P^i{}_a e_i$ gives

$$v = \sum_a d^a \left( \sum_i P^i{}_a e_i \right) = \sum_i \left( \sum_a P^i{}_a d^a \right) e_i.$$

By uniqueness of coordinates in the basis $\mathcal B$,

$$c^i = \sum_a P^i{}_a d^a.$$

In matrix form,

$$[v]_{\mathcal B} = P[v]_{\mathcal F}, \qquad [v]_{\mathcal F} = P^{-1}[v]_{\mathcal B}.$$

This is a passive basis change. The abstract vector $v$ is unchanged. Only the coordinate map has changed.

Operator Matrices

Let $A:V\to V$ be a linear operator. Its matrix in the old basis is $A_{\mathcal B}$, defined by

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

Its matrix in the new basis is $A_{\mathcal F}$, defined by

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

Use the coordinate relation twice:

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

But also

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

Since this holds for every coordinate column $[v]_{\mathcal F}$,

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

and therefore

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

This is a similarity transformation. It changes the matrix representation of $A$, not the abstract operator.

For a map $T:V\to W$ with a basis change in both domain and codomain, the corresponding formula is

$$[T]_{\mathcal G\leftarrow\mathcal F} = Q^{-1} [T]_{\mathcal C\leftarrow\mathcal B} P,$$

where $P$ changes domain coordinates from $\mathcal F$ to $\mathcal B$, and $Q$ changes codomain coordinates from $\mathcal G$ to $\mathcal C$.

Orthonormal Bases and Unitary Matrices

In a finite-dimensional complex inner-product space, quantum mechanics usually uses orthonormal bases. Let

$$\mathcal B=\{\lvert e_i\rangle\}, \qquad \mathcal F=\{\lvert f_a\rangle\}$$

be two orthonormal bases. Define the overlap matrix

$$S_{ai} = \langle f_a\vert e_i\rangle.$$

If

$$\lvert\psi\rangle = \sum_i c_i\lvert e_i\rangle = \sum_a d_a\lvert f_a\rangle,$$

then the new coefficients are

$$d_a = \langle f_a\vert\psi\rangle = \sum_i S_{ai}c_i.$$

Thus

$$d = S c.$$

The overlap matrix is unitary:

$$\sum_i S_{ai}S_{bi}^* = \sum_i \langle f_a\vert e_i\rangle \langle e_i\vert f_b\rangle = \langle f_a\vert f_b\rangle = \delta_{ab}.$$

This says $SS^\dagger=I$, and finite dimensionality then gives $S^\dagger S=I$ as well.

The earlier matrix $P$ has columns $[\lvert f_a\rangle]_{\mathcal B}$. Its entries are

$$P_{ia} = \langle e_i\vert f_a\rangle = S_{ai}^*,$$

so

$$P=S^\dagger, \qquad P^{-1}=S.$$

The operator matrix formula becomes

$$A_{\mathcal F} = S A_{\mathcal B}S^\dagger.$$

This is the form most often seen in finite-dimensional quantum mechanics: state coefficients transform by $d=Sc$, while operator matrices transform by $A_{\mathcal F}=S A_{\mathcal B}S^\dagger$.

Active Versus Passive Transformations

A passive basis change rewrites the same vector in a new coordinate system:

$$\lvert\psi\rangle \quad\text{unchanged}, \qquad c\mapsto d=Sc.$$

An active transformation applies a linear operator to produce a new vector:

$$\lvert\psi\rangle \mapsto U\lvert\psi\rangle.$$

Both discussions may use unitary matrices, so the notation alone is not enough. Identify what the matrix acts on: an active transformation changes the vector, while a passive transformation changes its coordinates. A rotation of the axes and a rotation of the vector in fixed axes therefore use inverse transformations.

Invariants

Similarity transformations preserve the basis-independent information of an operator. In finite dimension,

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

has the same eigenvalues, trace, determinant, rank, and characteristic polynomial as $A_{\mathcal B}$.

Expectation values are also preserved when states and operators are transformed consistently. In the orthonormal convention above,

$$d=Sc, \qquad A_{\mathcal F}=SA_{\mathcal B}S^\dagger.$$

Then

$$d^\dagger A_{\mathcal F}d = c^\dagger S^\dagger \left( SA_{\mathcal B}S^\dagger \right) Sc = c^\dagger A_{\mathcal B}c.$$

What is invariant is the physical scalar or abstract object, not the individual matrix entries.

Nonorthogonal bases and the inner-product matrix

For general complex bases, the coordinate rule $c=Pd$ still holds, but the inner product is not usually the Euclidean pairing of columns. Let $G_{\mathcal B}$ and $G_{\mathcal F}$ be the Gram matrices of the two bases. Substitution into the squared norm gives

$$c^\dagger G_{\mathcal B}c =d^\dagger P^\dagger G_{\mathcal B}Pd, \qquad G_{\mathcal F}=P^\dagger G_{\mathcal B}P.$$

The Gram matrix transforms by congruence, while the coordinate matrix of an operator transforms by similarity, $A_{\mathcal F}=P^{-1}A_{\mathcal B}P$. Together these rules preserve the quadratic form:

$$d^\dagger G_{\mathcal F}A_{\mathcal F}d =d^\dagger P^\dagger G_{\mathcal B}A_{\mathcal B}Pd =c^\dagger G_{\mathcal B}A_{\mathcal B}c.$$

For a nonzero state, divide this expression by the corresponding squared norm to obtain its expectation value.

There are two different arrays often called the operator’s “matrix.” The coordinate matrix $A_{\mathcal F}$ maps the coefficients of $v$ to those of $Av$. The bra–ket matrix

$$(K_{\mathcal F})_{ab}=\langle f_a|A|f_b\rangle \quad\text{satisfies}\quad K_{\mathcal F}=G_{\mathcal F}A_{\mathcal F}.$$

The last identity follows by expanding $A|f_b\rangle$ in the $f$ basis and taking its inner product with $f_a$. These arrays coincide in an orthonormal basis. For a Hermitian operator, $K_{\mathcal F}$ is Hermitian, whereas $A_{\mathcal F}$ need only satisfy

$$A_{\mathcal F}^\dagger G_{\mathcal F} =G_{\mathcal F}A_{\mathcal F}.$$

Here daggers on arrays mean ordinary conjugate transpose; Hermitian Operators explains the adjoint with a Gram matrix. A nonsymmetric coordinate matrix therefore need not signal a non-Hermitian physical operator.

The same distinction explains the generalized eigenvalue equation:

$$A_{\mathcal F}d=\lambda d \quad\Longleftrightarrow\quad K_{\mathcal F}d=\lambda G_{\mathcal F}d.$$

Diagonalizing $K_{\mathcal F}$ alone would solve a different problem. For Hermitian $A$, the right-hand problem has a Hermitian pair and positive-definite $G_{\mathcal F}$, as in the LAPACK treatment. Matrix Diagonalization discusses structure-preserving numerical methods; this coordinate derivation does not recommend explicitly forming $G_{\mathcal F}^{-1}K_{\mathcal F}$.

Worked Example

Let $\mathcal B$ be the standard basis of $\mathbb R^2$ and let

$$f_1 = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}, \qquad f_2 = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ -1 \end{pmatrix}.$$

The change-of-coordinates matrix is

$$P = \frac{1}{\sqrt2} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}.$$

Here $P^{-1}=P^T=P$. For

$$v = \begin{pmatrix} 1\\ 0 \end{pmatrix}_{\mathcal B},$$

the new coordinate column is

$$[v]_{\mathcal F} = P^{-1}[v]_{\mathcal B} = \frac{1}{\sqrt2} \begin{pmatrix} 1\\ 1 \end{pmatrix}.$$

Now let the old matrix of an operator be

$$A_{\mathcal B} = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}.$$

The new matrix is

$$A_{\mathcal F} = P^{-1}A_{\mathcal B}P = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}.$$

The matrix changed from diagonal to off-diagonal, but the operator did not change. The expectation value is the same in both descriptions:

$$[v]_{\mathcal B}^T A_{\mathcal B} [v]_{\mathcal B} = 1, \qquad [v]_{\mathcal F}^T A_{\mathcal F} [v]_{\mathcal F} = 1.$$

This is the real two-dimensional version of the familiar conversion between the $z$ and $x$ spin bases.

Physical Interpretation

A basis choice is a representation choice. A state may be written in an energy basis, a spin-$z$ basis, a spin-$x$ basis, a position representation, or a momentum representation. The coordinate list changes, but the state does not.

Likewise, an observable may be diagonal in one basis and dense in another. Diagonalizing a Hamiltonian is not changing the Hamiltonian into a different physical operator. It is finding a basis in which its action is easiest to read.

For finite spin and qubit systems, the basis-change matrices are ordinary unitary matrices. For position and momentum, the same idea becomes a Fourier transform, and the matrix sums become integrals.

Common Mistakes

• Using $P$ where $P^{-1}$ is required.
• Transforming the state-coordinate column but leaving the operator matrix in the old basis.
• Assuming every basis change is unitary; nonorthonormal bases require a general invertible matrix.
• Forgetting that basis order matters.
• Comparing individual matrix entries from different bases as if they were basis-independent quantities.
• Treating a passive basis change as a physical time evolution or symmetry operation.
• Using an overlap matrix without tracking which basis labels its rows and columns.

References

• E. Anderson et al., LAPACK Users’ Guide, 3rd ed., SIAM, 1999, “Generalized Symmetric Definite Eigenproblems.”
• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• P. R. Halmos, Finite-Dimensional Vector Spaces, 2nd ed., Springer, 1974.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• G. Strang, Linear Algebra and Its Applications, 4th ed., Brooks/Cole, 2006.

Exercises

1. In $\mathbb R^2$, let $\mathcal B$ be the standard basis and let $\mathcal F=(f_1,f_2)$ with $f_1=(1,0)^T$ and $f_2=(1,1)^T$. Find $P$ and the $\mathcal F$-coordinates of $v=(3,2)^T$.

Solution

The columns of $P$ are the standard coordinates of $f_1$ and $f_2$:

$$P = \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$

Since

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

the new coordinate column is

$$[v]_{\mathcal F} = P^{-1}[v]_{\mathcal B} = \begin{pmatrix} 1\\ 2 \end{pmatrix}.$$

Indeed,

$$1f_1+2f_2 = \begin{pmatrix} 3\\ 2 \end{pmatrix}.$$

2. With the same bases as in the first exercise, let

$$A_{\mathcal B} = \begin{pmatrix} 2 & 0\\ 0 & 3 \end{pmatrix}.$$

Find $A_{\mathcal F}$, the Gram matrix $G_{\mathcal F}$, and the bra–ket matrix $K_{\mathcal F}$. Explain why the nonsymmetric $A_{\mathcal F}$ still represents a Hermitian operator. Using $v=(3,2)^T$ from Exercise 1, verify its normalized expectation value in both bases.

Solution

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

$$A_{\mathcal F} = \begin{pmatrix} 1 & -1\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0\\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -1\\ 0 & 3 \end{pmatrix}.$$

The off-diagonal entry appears because the new basis is not made of eigenvectors of the operator.

Since the old basis is orthonormal,

$$G_{\mathcal F}=P^\dagger P =\begin{pmatrix}1&1\\1&2\end{pmatrix}, \qquad K_{\mathcal F}=G_{\mathcal F}A_{\mathcal F} =\begin{pmatrix}2&2\\2&5\end{pmatrix}.$$

Thus $K_{\mathcal F}$ is Hermitian and $A_{\mathcal F}^\dagger G_{\mathcal F}=G_{\mathcal F}A_{\mathcal F}$, even though $A_{\mathcal F}^\dagger\ne A_{\mathcal F}$. With $c=(3,2)^T$ and $d=(1,2)^T$,

$$\frac{c^\dagger A_{\mathcal B}c}{c^\dagger c} =\frac{30}{13} =\frac{d^\dagger K_{\mathcal F}d}{d^\dagger G_{\mathcal F}d}.$$

Using $d^\dagger A_{\mathcal F}d/(d^\dagger d)$ instead would give $12/5$. That discrepancy comes from using the wrong inner product, not from changing the observable. Likewise, the correct eigenvalues $2,3$ follow from $\det(K_{\mathcal F}-\lambda G_{\mathcal F})=(2-\lambda)(3-\lambda)$; the ordinary eigenvalues $1,6$ of $K_{\mathcal F}$ are not the spectrum of $A$.

3. Let $\mathcal B=\{\lvert e_i\rangle\}$ and $\mathcal F=\{\lvert f_a\rangle\}$ be orthonormal bases, and let $S_{ai}=\langle f_a\vert e_i\rangle$. Prove that $S$ is unitary.

Solution

Use completeness of the old basis:

$$(SS^\dagger)_{ab} = \sum_i S_{ai}S_{bi}^* = \sum_i \langle f_a\vert e_i\rangle \langle e_i\vert f_b\rangle = \langle f_a\vert f_b\rangle = \delta_{ab}.$$

Thus $SS^\dagger=I$. Since $S$ is a square matrix, this also implies $S^\dagger S=I$.