05 Mar 2026

Eigenvalues and Eigenvectors of Matrices

Characteristic equation, eigenvectors, diagonal form, and physical meaning of matrix modes.

msc semester-i numerical-methods eigenvalues matrices

In many problems a matrix represents a linear operation: stretching a vector, coupling two oscillators, rotating axes, or acting with an operator on a state. Usually the direction of a vector changes after the operation. An eigenvector is a special vector whose direction is preserved.

For a square matrix $A$, a non-zero vector $\mathbf{x}$ is an eigenvector if multiplication by $A$ only changes its scale:

\[A\mathbf{x}=\lambda\mathbf{x}.\]

The number $\lambda$ is the corresponding eigenvalue.

Characteristic equation

Rearrange the eigenvalue equation:

\[(A-\lambda I)\mathbf{x}=0.\]

For a non-zero solution, the determinant must vanish:

\[\det(A-\lambda I)=0.\]

This is the characteristic equation.

The determinant condition is necessary because $(A-\lambda I)\mathbf{x}=0$ is a homogeneous system. A non-zero solution exists only when the coefficient matrix is singular.

Eigenvectors

After finding an eigenvalue $\lambda$, substitute it into

\[(A-\lambda I)\mathbf{x}=0\]

and solve for the components of $\mathbf{x}$.

Physical meaning

Eigenvalue problems appear throughout physics:

• normal modes of coupled oscillators,
• energy levels in quantum mechanics,
• principal axes of inertia,
• stability analysis near equilibrium.

Numerical viewpoint

For large matrices, one generally does not expand the determinant by hand. Numerical algorithms are used to compute eigenvalues and eigenvectors efficiently and accurately.

Two-by-two example

Find the eigenvalues and eigenvectors of

\[A= \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix}.\]

The characteristic equation is

\[\det(A-\lambda I)= \begin{vmatrix} 2-\lambda&1\\ 1&2-\lambda \end{vmatrix} =0.\]

Therefore

\[(2-\lambda)^2-1=0,\]

or

\[\lambda^2-4\lambda+3=0.\]

Hence

\[\lambda=3,\quad 1.\]

For $\lambda=3$,

\[(A-3I)\mathbf{x}=0 \quad\Rightarrow\quad -x+y=0,\]

so an eigenvector is

\[\mathbf{x}_1= \begin{pmatrix} 1\\ 1 \end{pmatrix}.\]

For $\lambda=1$,

\[(A-I)\mathbf{x}=0 \quad\Rightarrow\quad x+y=0,\]

so an eigenvector is

\[\mathbf{x}_2= \begin{pmatrix} 1\\ -1 \end{pmatrix}.\]

These two eigenvectors are perpendicular, as expected for a real symmetric matrix.

Key points

• Eigenvectors give special directions of a linear transformation.
• Eigenvalues give the scale factors along those directions.
• In symmetric or Hermitian problems, eigenvalues are real and eigenvectors have strong orthogonality properties.

Practice questions

1. Define eigenvalue and eigenvector of a matrix.
2. Derive the characteristic equation $\det(A-\lambda I)=0$.
3. Find the eigenvalues of $\begin{pmatrix}2&1\1&2\end{pmatrix}$.
4. Find an eigenvector corresponding to each eigenvalue in the previous question.
5. Mention two physical problems where eigenvalues occur.