Introduction: Simplifying Operators
Diagonalization transforms a matrix into a diagonal matrix (all zeros except on the diagonal). In the eigenbasis, an operator becomes trivially simple—it just multiplies each component by the corresponding eigenvalue.
The Diagonalization Process
If \(A\) has \(n\) linearly independent eigenvectors, we can write:
\[A = PDP^{-1}\]where:
- \(P\) is the matrix whose columns are eigenvectors
- \(D\) is diagonal with eigenvalues on the diagonal
Equivalently: \(D = P^{-1}AP\)
Why Diagonalization is Powerful
- Powers: \(A^n = PD^nP^{-1}\) (just raise diagonal entries to \(n\))
- Functions: \(f(A) = Pf(D)P^{-1}\) (apply \(f\) to eigenvalues)
- Exponentials: \(e^A = Pe^DP^{-1}\)
Worked Examples
Example 1: Diagonalizing a 2×2 Matrix
Diagonalize \(A = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}\) (from previous lesson):
Eigenvalues: \(\lambda_1 = 5\), \(\lambda_2 = 2\)
Eigenvectors: \(\vec{v}_1 = (2, 1)\), \(\vec{v}_2 = (1, -1)\)
\[P = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}\]Verify: \(P^{-1} = \frac{1}{-3}\begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix}\)
Example 2: Computing A²
Using diagonalization:
\[A^2 = PD^2P^{-1} = P\begin{pmatrix} 25 & 0 \\ 0 & 4 \end{pmatrix}P^{-1}\]Much easier than multiplying \(A \times A\) directly!
Example 3: Diagonal Matrix in Eigenbasis
In its own eigenbasis, any diagonalizable operator looks like:
\[\hat{A} = \sum_n \lambda_n |n\rangle\langle n|\]This is the spectral decomposition.
The Quantum Connection
In quantum mechanics, choosing the eigenbasis of an observable diagonalizes it. The Hamiltonian in its eigenbasis is:
\[\hat{H} = \sum_n E_n |n\rangle\langle n|\]Time evolution becomes trivial: each energy eigenstate just picks up a phase \(e^{-iE_n t/\hbar}\). This is why finding energy eigenstates is the central problem of quantum mechanics.