What Are Eigenvalues and Eigenvectors?
For a square matrix \(A\), a nonzero vector \(\mathbf v\) is an eigenvector when multiplying by \(A\) changes only its scale or complex phase:
$$A\mathbf v=\lambda\mathbf v$$
The scalar \(\lambda\) is the corresponding eigenvalue. Eigenvalues are the roots of the characteristic equation \(\det(\lambda I-A)=0\). A repeated eigenvalue can have fewer independent eigenvectors than its algebraic multiplicity; that distinction determines whether the matrix is diagonalizable.
How This Calculator Works
The calculator accepts real square matrices from \(2\times2\) through \(6\times6\). It balances the matrix, reduces it to Hessenberg form, and applies shifted QR iterations. Real and complex eigenvectors are recovered from \((A-\lambda I)\mathbf v=0\), normalized, and checked using the residual
$$\frac{\lVert A\mathbf v-\lambda\mathbf v\rVert}{(\lVert A\rVert+|\lambda|)\lVert\mathbf v\rVert}.$$
Results are withheld if the iteration does not converge or the eigenpairs fail the residual check. Numerical coefficients may be rounded, but small legitimate values are retained and displayed in scientific notation.
Understanding the Results
- Characteristic polynomial: \(\det(\lambda I-A)\), whose roots are the eigenvalues.
- Multiplicity: The algebraic multiplicity of an eigenvalue in the characteristic polynomial.
- Eigenvector basis: Independent vectors spanning the eigenspace for that eigenvalue.
- Diagonalizable over ℝ: The matrix has a full basis of real eigenvectors.
- Diagonalizable over ℂ: The matrix has a full basis of possibly complex eigenvectors.
- Normalized residual: A scale-independent check of how closely each reported pair satisfies \(A\mathbf v=\lambda\mathbf v\).
Examples
Symmetric matrix
$$A=\begin{bmatrix}2&1\\1&2\end{bmatrix}$$
The characteristic polynomial is \(\lambda^2-4\lambda+3\). The eigenvalues are \(1\) and \(3\), with eigenvectors proportional to \((-1,1)^T\) and \((1,1)^T\). Two independent real eigenvectors make the matrix diagonalizable over both ℝ and ℂ.
Complex eigenvalues
$$A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$
The eigenvalues are \(i\) and \(-i\). The matrix has no real eigenvector basis, so it is not diagonalizable over ℝ, but its two complex eigenvectors make it diagonalizable over ℂ.
Defective repeated eigenvalue
$$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$
The eigenvalue \(1\) has algebraic multiplicity two but only one independent eigenvector. The matrix is therefore not diagonalizable over either ℝ or ℂ.
Limitations and Numerical Conditioning
Eigenvalue problems can be sensitive to rounding, especially for nearly defective or highly non-normal matrices. Inputs use standard double-precision arithmetic and are limited to 15 significant digits. A small residual confirms that a displayed pair fits the entered matrix, but it does not guarantee that a poorly conditioned eigenvalue would remain unchanged after a small perturbation to the data.
Repeated eigenvalues are grouped only when they are indistinguishable at the solver's scale-aware error threshold. Coefficients or determinants outside the ordinary floating-point range are shown in scaled scientific notation. Verify important academic, scientific, or engineering work independently.
How to Use This Calculator
- Choose a matrix size from \(2\times2\) through \(6\times6\).
- Enter a finite number in every cell, or paste a complete square matrix.
- Review the eigenvalues, multiplicities, eigenvector bases, trace, determinant, and diagonalizability over ℝ and ℂ.
- Check the convergence status and maximum normalized residual before using the result.
- Use the derivation cards and copy controls to inspect or reuse individual results.
Frequently Asked Questions
Why can an eigenvector be complex when the matrix is real?
A real matrix can have non-real eigenvalues in conjugate pairs. Their eigenvectors are generally complex as well.
Why are eigenvectors normalized?
Any nonzero multiple of an eigenvector represents the same eigendirection. Normalization gives a consistent length and makes results easier to compare.
Does a repeated eigenvalue always make a matrix non-diagonalizable?
No. It is diagonalizable when the total number of independent eigenvectors still equals the matrix size.
Sources and References
- Gene H. Golub and Charles F. Van Loan, Matrix Computations, 4th edition, Johns Hopkins University Press.
- LAPACK Users' Guide, “Nonsymmetric Eigenproblems.” Netlib LAPACK documentation.
- NIST Digital Library of Mathematical Functions, Linear Algebra.