Eigenvalues and Eigenvectors Calculator

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Interactive tool to compute eigenvalues and eigenvectors of any square matrix, along with detailed explanations adapted from linear algebra theory.

What are Eigenvalues & Eigenvectors?

For a square matrix \(A\) of size \(n \times n\), a scalar \(\lambda\) is called an eigenvalue, and a non-zero vector \(v\) is called an eigenvector associated with \(\lambda\) if:

\(A v = \lambda v\)

This implies that the vector \(v\) does not change direction under the linear transformation represented by \(A\); it is only scaled by \(\lambda\).

Characteristic Polynomial & How to Find Eigenvalues

Eigenvalues are found by solving the characteristic equation. The equation is derived from:

\(\det(A - \lambda I) = 0\)

Here, \(I\) is the identity matrix of size \(n \times n\). Expanding the determinant gives a polynomial in \(\lambda\) of degree \(n\). The roots of this polynomial are the eigenvalues of \(A\).

Important Properties

The trace of \(A\) (sum of its diagonal elements) equals the sum of its eigenvalues.
The determinant of \(A\) (product of diagonal elements if diagonalizable) equals the product of its eigenvalues.
If \(A\) is symmetric (or Hermitian), all eigenvalues are real and eigenvectors for distinct eigenvalues are orthogonal.
The matrix \(A\) is invertible if and only if none of its eigenvalues are zero.

Diagonalization & Eigenspaces

If \(A\) has \(n\) linearly independent eigenvectors \(v_1, \ldots, v_n\), we can form a matrix \(P\) whose columns are these eigenvectors, and a diagonal matrix \(\Lambda\) whose entries are the corresponding eigenvalues. Then:

\(A = P \Lambda P^{-1}\)

The eigenspace associated with an eigenvalue \(\lambda\) is the null space of \(A - \lambda I\). Its geometric multiplicity is the dimension of that space.

Example: 2×2 Matrix

Consider the matrix:

\(A = \begin{pmatrix}2 & 1\\1 & 2\end{pmatrix}\)

Compute the characteristic polynomial:

\(\det(A - \lambda I) = \begin{vmatrix}2-\lambda & 1\\1 & 2-\lambda\end{vmatrix} = (\;2-\lambda\;)(\;2-\lambda\;) - 1\cdot1 = \lambda^2 - 4\lambda + 3\)

Solve \(\lambda^2 - 4\lambda + 3 = 0\):

\(\lambda^2 - 4\lambda + 3 = (\lambda - 3)(\lambda - 1) = 0\)

Hence:

\(\lambda_1 = 3,\quad \lambda_2 = 1\)

Then for each \(\lambda\), solve \((A - \lambda I)v = 0\) to find the eigenvectors:

A - 3I = \begin{pmatrix}-1 & 1\\1 & -1\end{pmatrix},\quad v_1 = \begin{pmatrix}1\\1\end{pmatrix}

A - 1I = \begin{pmatrix}1 & 1\\1 & 1\end{pmatrix},\quad v_2 = \begin{pmatrix}1\\-1\end{pmatrix}

Frequently Asked Questions

Are eigenvalues unique?

Yes, the eigenvalues of a given matrix are uniquely determined, but they may repeat (have multiplicity).

Can a real matrix have complex eigenvalues?

Yes, non-symmetric real matrices may have complex eigenvalues. Symmetric real matrices always have real eigenvalues.

What if an eigenvalue has multiplicity greater than one?

Its geometric multiplicity (number of independent eigenvectors) may be less than its algebraic multiplicity, which means the matrix might not be diagonalizable.

What is an eigenspace?

The eigenspace corresponding to an eigenvalue \(\lambda\) is the set of all eigenvectors associated with \(\lambda\), plus the zero vector. It forms a vector subspace of the domain.

Why is the characteristic polynomial degree n for an n×n matrix?

Because \(\det(A - \lambda I)\) expands into a polynomial in \(\lambda\) of degree \(n\), where \(n\) is the size of the matrix. The highest power of \(\lambda\) comes from the diagonal terms when expanding the determinant.