1
$\begingroup$

I am struggling with the following question:

When is it true that I can use the (right, eventually) eigenvectors of a finite dimensional matrix as a basis and thus write down an eigenvector expansion of a given vector?

I am sure about the fact that an answer to my question is the spectral theorem: for symmetric matrices, I can. Moreover, in this case the eigenvectors are orthogonal.

My problem is more with non-symmetric matrices. Let me explain that:

  • First of all, a "logic" consideration: if we can always use the right eigenvectors of a non-symmetric matrix as a basis, I think it would be a well know result (in fact, the spectral theorem is well known).
  • Looking on books by notable authors, I get confused. Everybody says "if the matrix is symmetric then the eigenvectors are a basis". However, I have never read a statement about non-symmetric matrices. Despite it, some authors just use the right eigenvectors as a basis. Some other, instead, they say that they assume they can use the right eigenvectors as a basis.
  • Finally, the power method and the Jordan form: the power method is based on the fact that a generic vector can be written as a linear combination of eigenvectors of the matrix. It seems to me that the only condition required for this to be possible is for the Jordan form of the matrix to exist. On the other hand, it seems to me that the Jordan form of a matrix always exists. Thus I would conclude that I can always use the right eigenvectors of a non symmetric matrix. But it seems crazy to me that nobody ever says that.

Can somebody explain to me what is the state of the art about eigenvectors and their being a basis?

$\endgroup$
1
$\begingroup$

Eigenvectors for a nonsymmetric matrix need not form a basis. For example, the only eigenvector of the matrix $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ is $(1 \ \ 0)^T$. The issue is that zero is an eigenvalue of algebraic multiplicity $2$, but geometric multiplicity $1$.

There are many of well known statements which are equivalent to the existence of a basis of eigenvectors. A short list of equivalent conditions is

  1. The matrix is diagonalizable.

  2. The Jordan blocks of the matrix are all $1 \times 1$.

  3. The eigenvalues of the matrix have algebraic multiplicity equal to geometric multiplicity.

  4. The minimal polynomial of the matrix has the form $p (x) = \Pi_{i=1}^m (x-\lambda_i)$ where the $\lambda_i$ are distinct.

An easy sufficient condition for existence of a basis of eigenvectors is that the matrix has distinct eigenvalues counting algebraic multiplicity. In some applications authors may assume diagonalizability since a random matrix has distinct eigenvalues with probability 1.

One may define generalized eigenvectors for a specified eigenvalue. Given a matrix $A$, say $v$ is a generalized eigenvector for the eigenvalue $\lambda$ if there is some integer $k$ such that $$ (A-\lambda I)^k v =0. $$ The generalized eigenvectors of a matrix always do form a basis, although I personally do not know where they are used.

$\endgroup$
  • $\begingroup$ Thanks! Could you give a reference about the statement that "random matrices have distinct eigenvalue with probability 1" (hence the Jordan blocks have all size one and the eigenvectors form a basis)? $\endgroup$ – GRquanti Oct 24 at 18:54
  • 1
    $\begingroup$ I don't actually know a common reference for this. I think it is well known enough that people just use it without reference. A rough argument for why it holds: if $M$ is a random matrix sampled from a continuous distribution, then its characteristic polynomial is a random polynomial from some continous distribution, hence the roots are random variables in some continuous distributions, so with probability 1 you wont have repeated roots. $\endgroup$ – Eric Oct 25 at 10:57
  • $\begingroup$ Also if $M$ is $2 \times 2$ and has repeated roots then the coefficents of the characteristic polynomial must lie in the zero set of some system of polynomials (namely the discriminant must be zero). I think this generalizes without much trouble for n=3,n=4, but for $n\geq 5$ I guess there may be difficulties since there isn't a closed form for the roots of such a polynomial. $\endgroup$ – Eric Oct 25 at 10:59
  • $\begingroup$ What if M is a random matrix sampled from a discrete distribution (say a matrix with only 0-1 entries, independent and identically distributed)? The argument about the polynomial doesn't seem to hold anymore. Thank you by the way. $\endgroup$ – GRquanti Oct 25 at 15:13
  • $\begingroup$ If you randomly sample from a discrete distribution all bets are off. For example you distribution could only be supported on matrices which are not diagonalizable if it is discret. Nondiagonalizable matrices occur with positive probability if you sample only matrices with 0-1 entries. (e.g the original example I gave occurs with positive probability for 0-1 matrices of size $2 \times 2$.) $\endgroup$ – Eric Oct 26 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.