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28. Eigenvalues and Eigenvectors

Dated: 12-06-2025

Fixed Points

A fixed point of an \(n \times n\) matrix1 is a vector2 \(\vec x\) in \(\mathbb R^n\) such that \(A \vec x = \vec x\)
Every square matrix has at least one fixed point namely \(\vec x = 0\).

General procedure to find fixed points of a matrix1 \(A\) is to rewrite the equation \(A \vec x = \vec x\) as \(A \vec x = I \vec x\) or as

\[(I - A) \vec x = \vec 0\]

Theorem

If \(A\) is an \(n \times n\) matrix1 then following are equivalent

  • \(A\) has non trivial fixed points
  • \(I - A\) is a singular matrix1
  • \(\det(I - A) = 0\)

Example

\[ A = \begin{bmatrix} 3 & 6 \\ 1 & 2 \end{bmatrix} \]
\[(I - A)\vec x = \vec 0\]
\[ \left( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 6 \\ 1 & 2 \end{bmatrix} \right) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]
\[ \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \]

General solution of this system is

\[x = 2t, \, y = t\]

These are parametric equations of the line3 \(y = \frac x 2\)

Eigenvalues and Eigenvectors

In the following equation

\[A \vec x = \lambda \vec x\]

If there is a non zero vector2 which satisfies this equation then it is called the eigenvector and \(\lambda\) is a scalar, called the eigenvalue.

Eigen Space

The set of all solutions of \((A - \lambda I) \vec x = \vec 0\) is just the null space4 of the matrix1 \(A - \lambda I\). So this set5 is a subspace6 of \(\mathbb R^n\) and is called eigenspace of \(A\) corresponding to \(\lambda\).

Example

\[A = \begin{bmatrix} 4 & -1 & 6 \\ 2 & 1 & 6 \\ 2 & -1 & 8 \end{bmatrix}\]

Find a basis for corresponding eigen space where eigen value of a matrix1 is \(2\).

Solution

\[ A - 2 I = \begin{bmatrix} 4 & -1 & 6 \\ 2 & 1 & 6 \\ 2 & -1 & 8 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -1 & 6 \\ 2 & -1 & 6 \\ 2 & -1 & 6 \end{bmatrix} \]

Now we will row reduce7 the augmented matrix.

\[ \left[ \begin{array}{ccc|c} 2 & -1 & 6 & 0 \\ 2 & -1 & 6 & 0 \\ 2 & -1 & 6 & 0 \\ \end{array} \right] \]
\[\Bigg\downarrow R_2 - R_1\]
\[\Bigg\downarrow R_3 - R_1\]
\[ \left[ \begin{array}{ccc|c} 2 & -1 & 6 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right] \]

The general solution is

\[2x_1 - x_2 + 6 x_3 = 0\]
\[x_2 = t, \, x_3 = s\]
\[2x_1 = t - 6s\]
\[x_1 = \frac t 2 - 3s\]
\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} t/2 - 3s \\ t \\ s \end{bmatrix} = \begin{bmatrix} t/2 \\ t \\ 0 \end{bmatrix} + \begin{bmatrix} -3s \\ 0 \\ s \end{bmatrix} = t \begin{bmatrix} 1/2 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \]
\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = x_2 \begin{bmatrix} 1/2 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \]
\[ \left\{ \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix} , \begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} \right\} \text{ is a basis} \]

The most direct way to find eigen values is to rewrite \(A \vec x = \lambda \vec x\) as \(A \vec x = \lambda I \vec x\).

\[\implies (\lambda I - A) \vec x = \vec 0\]

The above equation has non trivial solutions if

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

This equation is called the characteristic equation.

Theorem

If \(\lambda\) is a scalar and \(A\) is a \(n \times n\) matrix1 then the following statements are true

  • \(\lambda\) is an eigenvalue of \(A\).
  • \(\lambda\) is the solution of the equation \(\det(\lambda I - A) = \vec 0\)
  • The linear system8 \((\lambda I - A) \vec x = 0\) has non trivial solutions.

Eigenvalues of Triangular Matrices

The characteristic polynomial for a triangular matrix9 is

\[\det(\lambda I - A) = (\lambda - a_{11})(\lambda - a_{22})(\lambda - a_{33})\ldots(\lambda - a_{nn})\]

Which implies that eigen values are

\[\lambda_1 = a_{11}, \, \lambda_2 = a_{22}, \, \ldots, \, \lambda_n = a_{nn}\]

Eigenvalues of Powers of a Matrix

\[A^2 \vec x = A (A \vec x) = A( \lambda \vec x) = \lambda (A \vec x) = \lambda (\lambda \vec x) = \lambda^2 \vec x\]

Theorem

If \(\vec x\) is the corresponding eigen vector of \(A\) then eigen value of \(A^k\) is \(\lambda^k\).

Theorem

If \(A\) is an \(n \times n\) matrix,1 then the following statements are equivalent:

  • The reduced row echelon form7 of \(A\) is \(I_n\).
  • \(A\) is expressible as a product of elementary matrices.10
  • \(A\) is invertible.
  • \(Ax = 0\) has only the trivial solution.
  • \(Ax = b\) is consistent for every vector2 \(b\) in \(\mathbb{R}^n\).
  • \(Ax = b\) has exactly one solution for every vector2 \(b\) in \(\mathbb{R}^n\).
  • The column vectors of \(A\) are linearly independent.11
  • The row vectors of \(A\) are linearly independent.11
  • \(\det(A) \ne 0\).
  • \(\lambda = 0\) is not an eigenvalue of \(A\).

References

Read more about notations and symbols.

  1. Read more about matrices. ↩↩↩↩↩↩↩↩

  2. Read more about vectors. ↩↩↩↩

  3. Read more about lines. ↩

  4. Read more about null space. ↩

  5. Read more about sets. ↩

  6. Read more about vector spaces. ↩

  7. Read more about row reduction. ↩↩

  8. Read more about linear systems. ↩

  9. Read more about triangular matrices. ↩

  10. Read more about elementary matrices. ↩

  11. Read more about linear independence. ↩↩