Linear Maps, Matrices, and Change of Basis
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You should be comfortable with the definition of a vector space, the concept of a basis, and the fact that the dimension of a finite-dimensional vector space is invariant (see D-074). You should also recall how to express a vector as a linear combination of basis vectors.
Why this matters
Linear maps are the physical symmetries and evolutions of our systems, and matrices are merely their coordinate shadows.
In quantum mechanics, classical mechanics, and signal processing, the fundamental objects are linear transformations between vector spaces. While we often calculate with matrices, the matrix is not the map itself; it is a representation of the map relative to a specific choice of basis. Understanding how these representations change when we switch bases is crucial for identifying invariant properties, such as eigenvalues, and for simplifying complex systems into their normal modes. This lecture bridges the abstract definition of a linear map with its concrete matrix implementation.
The Abstract Map
Imagine a vector space \(V\) as a set of physical states. A linear map \(T: V \to W\) is a rule that takes every state in \(V\) and assigns a unique state in \(W\), preserving the structure of superposition. If you double a vector, the map doubles the result; if you add two vectors, the map adds the results. This linearity is the key property. Crucially, the map \(T\) exists independently of any coordinate system. It is a geometric relationship between the spaces. We do not need to pick a basis to define \(T\); we only need to know how it acts on vectors. The matrix is a tool we introduce later to perform calculations, but the map is the primary entity.
Matrix Representation
To compute with \(T\), we must choose a basis \(\mathcal{B} = \{e_1, \dots, e_n\}\) for \(V\) and a basis \(\mathcal{B}' = \{f_1, \dots, f_m\}\) for \(W\). The matrix of \(T\) with respect to these bases, denoted \([T]_{\mathcal{B} \to \mathcal{B}'}\), is constructed by applying \(T\) to each basis vector of \(V\) and expressing the result in terms of the basis of \(W\).
Specifically, the \(j\)-th column of the matrix contains the coordinates of \(T(e_j)\) in the basis \(\mathcal{B}'\). If \(T(e_j) = \sum_{i=1}^m a_{ij} f_i\), then the entry in row \(i\) and column \(j\) is \(a_{ij}\). This construction relies on the invariance of dimension (D-074), ensuring that the number of columns matches the dimension of the domain and the number of rows matches the dimension of the codomain.
Once this matrix is fixed, the action of \(T\) on any vector \(v \in V\) is simply matrix multiplication: if \(v\) has coordinate vector \([v]_{\mathcal{B}}\), then \([T(v)]_{\mathcal{B}'} = [T]_{\mathcal{B} \to \mathcal{B}'} [v]_{\mathcal{B}}\).
Change of Basis
Suppose we have the matrix of \(T\) relative to basis \(\mathcal{B}\), but we want the matrix relative to a new basis \(\mathcal{C}\). Let \(P\) be the change-of-basis matrix from \(\mathcal{C}\) to \(\mathcal{B}\). The columns of \(P\) are the coordinates of the new basis vectors \(c_j\) expressed in the old basis \(\mathcal{B}\). That is, \([c_j]_{\mathcal{B}}\) is the \(j\)-th column of \(P\).
The relationship between the coordinate vectors in the two bases is \([v]_{\mathcal{B}} = P [v]_{\mathcal{C}}\). To find the new matrix \([T]_{\mathcal{C}}\), we use the fact that the map \(T\) is invariant. We have \([T(v)]_{\mathcal{B}} = [T]_{\mathcal{B}} [v]_{\mathcal{B}}\). Substituting the change of basis, we get \([T(v)]_{\mathcal{B}} = [T]_{\mathcal{B}} P [v]_{\mathcal{C}}\).
We also know that \([T(v)]_{\mathcal{B}} = P [T(v)]_{\mathcal{C}}\). Equating the two expressions for \([T(v)]_{\mathcal{B}}\), we obtain \(P [T(v)]_{\mathcal{C}} = [T]_{\mathcal{B}} P [v]_{\mathcal{C}}\). Since this holds for all \(v\), we can cancel \([v]_{\mathcal{C}}\) (assuming \(v\) is arbitrary) and multiply by \(P^{-1}\) on the left:
This is the similarity transformation. It shows that matrices representing the same linear map under different bases are similar. The eigenvalues, trace, and determinant are invariant under this transformation, which is why they are physically meaningful quantities.
Composition and Identity
Linear maps compose naturally. If \(S: U \to V\) and \(T: V \to W\), then the composition \(T \circ S: U \to W\) is also linear. The matrix of the composition is the product of the matrices, provided the bases are chosen consistently. If \([S]_{\mathcal{A} \to \mathcal{B}}\) and \([T]_{\mathcal{B} \to \mathcal{C}}\) are the matrices, then \([T \circ S]_{\mathcal{A} \to \mathcal{C}} = [T]_{\mathcal{B} \to \mathcal{C}} [S]_{\mathcal{A} \to \mathcal{B}}\).
Note the order: the rightmost matrix acts first. This is consistent with function notation. The identity map \(I: V \to V\) has the identity matrix \(I_n\) in any basis. A map \(T\) is invertible if and only if its matrix is invertible, and the inverse map \(T^{-1}\) has the matrix \([T]^{-1}\).
The derivation
Invariance of Dimension of a Vector Space
Let \(V\) be a vector space over a field \(\mathbb{F}\) that is spanned by a finite set. Then any two bases of \(V\) are finite and have exactly the same number of elements. Consequently the integer \(\dim V\), defined as the cardinality of any basis, is a well-defined invariant of \(V\) that does not depend on the choice of basis.
A common error is to treat the matrix as the map itself. Remember that the matrix depends on the basis. If you change the basis, the matrix changes, but the map does not. When computing \(P^{-1} [T]_{\mathcal{B}} P\), ensure that \(P\) is defined as the matrix whose columns are the new basis vectors in the old basis. If you define \(P\) the other way around, the formula becomes \(P [T]_{\mathcal{B}} P^{-1}\). Consistency in the definition of the change-of-basis matrix is critical. C
Worked example
Problem. Let \(V = \mathbb{R}^2\). Consider the linear map \(T\) defined by \(T(x, y) = (x + y, x - y)\). Let \(\mathcal{B} = \{(1,0), (0,1)\}\) be the standard basis. Let \(\mathcal{C} = \{(1,1), (1,-1)\}\) be a new basis. Find the matrix of \(T\) with respect to \(\mathcal{B}\), and then find the matrix of \(T\) with respect to \(\mathcal{C}\).
- Find \([T]_{\mathcal{B}}\). Apply \(T\) to the basis vectors of \(\mathcal{B}\): \(T(1,0) = (1,1)\) and \(T(0,1) = (1,-1)\). The columns of \([T]_{\mathcal{B}}\) are \((1,1)^T\) and \((1,-1)^T\). Thus, \([T]_{\mathcal{B}} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\).
- Find the change-of-basis matrix \(P\) from \(\mathcal{C}\) to \(\mathcal{B}\). The columns of \(P\) are the coordinates of the vectors in \(\mathcal{C}\) expressed in \(\mathcal{B}\). Since \(\mathcal{B}\) is the standard basis, the coordinates are just the components. So, \(P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\).
- Compute \(P^{-1}\). The determinant of \(P\) is \((1)(-1) - (1)(1) = -2\). The inverse is \(P^{-1} = \frac{1}{-2} \begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\).
- Compute \([T]_{\mathcal{C}} = P^{-1} [T]_{\mathcal{B}} P\). First, \([T]_{\mathcal{B}} P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\). Then \([T]_{\mathcal{C}} = P^{-1}(2I) = 2P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\). Check directly: \(T(1,1) = (2,0) = 1\cdot(1,1) + 1\cdot(1,-1)\), so the first column is \((1,1)^{\mathsf T}\); \(T(1,-1) = (0,2) = 1\cdot(1,1) - 1\cdot(1,-1)\), so the second column is \((1,-1)^{\mathsf T}\). (Here the new matrix happens to equal the old one because \([T]_{\mathcal{B}} = P\), so \(P^{-1}[T]_{\mathcal{B}}P = P\).)
Units check. The entries are dimensionless coefficients, as expected for a linear map on a vector space without physical units.
Three quick tests
- Is the trace of a matrix invariant under a change of basis?
Answer
Yes, the trace is invariant under similarity transformations. - If \(P\) is the change-of-basis matrix from \(\mathcal{C}\) to \(\mathcal{B}\), what is the matrix of the identity map in basis \(\mathcal{C}\)?
Answer
The identity matrix \(I\). - Can a singular matrix represent an invertible linear map?
Answer
No, a singular matrix represents a non-invertible map.
Problem set for PU-104 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.