# Theory for find the eigenvules of a matrix

1. Feb 24, 2014

### Jhenrique

Exist so much theory for find the eigenvules of a matrix (invariants, characteristic polynomials, algebraic formula with trace and determinant...), but don't exist none formula for find the eigenvectors of a matrix? I never saw none! Please, if such formula exist, give me it or tell when I can study this.

2. Feb 24, 2014

### pasmith

You need to find eigenvalues before you can find eigenvectors. Two matrices with the same eigenvalues can have entirely different eigenvectors, so there is no general formula for finding eigenvectors beyond the definition: $v \neq 0$ is an eigenvector of $M$ if and only if there exists a scalar $\lambda$ such that
$$Mv = \lambda v.$$
Having found the eigenvalues as the roots of the characteristic polynomial
$$\chi_M(z) = \det (M - zI)$$
you can then find the corresponding eigenvector(s) by using the definition above.

3. Feb 24, 2014

### Staff: Mentor

If you use similarity transformations of an $n\times n$ matrix $A$ to reduce it a diagonal form, i.e., repeat
\begin{align} A_{1} &= P_1^{-1} A P_1 \\ A_{2} &= P_2^{-1} P_1^{-1} A P_1 P_2 \\ \vdots \\ A_{k} &= P_k^{-1} \cdots P_1^{-1} A P_1 \cdots P_k \end{align}
until $A_{k}$ is diagonal, then
$$A_{k} = \mathrm{diag}( \lambda_1, \lambda_2, \ldots, \lambda_n)$$
with $\lambda_i$ the eigenvalues of $A$ and the columns of the matrix
$$X = P_1 P_2 \cdots P_k$$
are the corresponding eigenvectors.

The QR algorithm is such a method.

4. Feb 24, 2014

### pasmith

Not all matrices are diagonalizable, even over $\mathbb{C}$! For example
$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$

The best you can do is reduce a matrix to its Jordan normal form, which the above matrix is.

EDIT: QR will converge to the Schur form of $A$, which is upper triangular so the eigenvalues will appear on the diagonal, but is it necessarily the case that the columns of $P_1 \dots P_k$ are generalized eigenvectors of $A$?

Last edited: Feb 24, 2014
5. Feb 24, 2014

### Staff: Mentor

I never said they were.