Eigenvalues and Eigenvectors

What are eigenvalues and eigenvectors?

A square matrix $A$ acts on column vectors by rotating and stretching them. Most vectors change direction when multiplied by $A$. An eigenvector is special: it is a non-zero vector $|v\rangle$ that $A$ only scales, never rotates:

The scalar $\lambda$ is the corresponding eigenvalue. The pair $(\lambda,\, |v\rangle)$ is called an eigenpair. Every eigenvector points along a direction that the matrix leaves invariant.

Finding eigenvalues: the characteristic equation

Rearranging the definition gives $(A - \lambda I)\,|v\rangle = 0$. A non-zero solution $|v\rangle$ exists only when the matrix $A - \lambda I$ is singular — that is, when its determinant is zero:

This is the characteristic equation (or characteristic polynomial) of $A$. For an $n \times n$ matrix it is a degree-$n$ polynomial in $\lambda$, so there are exactly $n$ eigenvalues (counted with multiplicity, possibly complex).

Worked example for 2x2 matrices

For a general $2 \times 2$ matrix

the characteristic equation is

Expanding:

The coefficient $a + d$ is the trace of $A$, denoted $\operatorname{tr}(A)$, and $ad - bc$ is the determinant $\det(A)$. So the characteristic equation always takes the form

The two eigenvalues are the roots of this quadratic.

Finding eigenvectors once you have the eigenvalues

Once $\lambda$ is known, plug it back into $(A - \lambda I)\,|v\rangle = 0$ and solve for $|v\rangle$. The system will always be underdetermined (because the matrix is singular by construction), leaving one free parameter — that is the eigenvector direction. Conventionally, eigenvectors are normalised so that $\langle v | v \rangle = 1$.

A concrete 2x2 example

Let

The trace is $3 + 3 = 6$ and the determinant is $3 \cdot 3 - 1 \cdot 1 = 8$, so the characteristic equation is

Factoring: $(\lambda - 4)(\lambda - 2) = 0$, giving eigenvalues $\lambda_1 = 4$ and $\lambda_2 = 2$.

Eigenvector for $\lambda_1 = 4$: Solve $(A - 4I)\,|v\rangle = 0$:

Normalising gives $|v_1\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$.

Eigenvector for $\lambda_2 = 2$: Solve $(A - 2I)\,|v\rangle = 0$:

Normalising gives $|v_2\rangle = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}$.

Notice that the two eigenvectors are orthogonal: $\langle v_1 | v_2 \rangle = \tfrac{1}{2}(1)(1) + \tfrac{1}{2}(1)(-1) = 0$. This is not a coincidence — the matrix $A$ above is symmetric (Hermitian), and Hermitian matrices always have real eigenvalues and orthogonal eigenvectors. That property is precisely why Hermitian matrices represent physical observables.

Why this matters for quantum mechanics

The energy levels of a quantum system are eigenvalues of its Hamiltonian operator $H$. The time-independent Schrödinger equation is exactly the eigenvalue equation

where $E$ is the energy eigenvalue and $|\psi\rangle$ is the corresponding energy eigenstate. Every time you solve for the allowed energies of a potential well, a harmonic oscillator, or a hydrogen atom, you are finding eigenvalues of a Hermitian matrix or operator.

Try it

This is a numerical exercise — your code should return a number. For the matrix $A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}$, find both eigenvalues by solving the characteristic equation $\det(A - \lambda I) = 0$, then return the larger eigenvalue.

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