Greetings again,

I am wondering about the proof of a theorem. I suspect that the proof, as described in the textbook I'm reading (see the comments below for detailed description) is either imprecise (maybe intentionally - this is an introductory level book) or, what is infinitely more probable, my understanding of diagonalizable matrices, eigenvectors and eigenspaces has some flaws. The theorem concerns a method for approximating dominant eigenvalue. In the textbook it is called "power method" and It seems to me that the same thing is described in Wikipedia as "power iteration algorithm" or Von Mises Iteration.

Theorem: Let A be a $\displaystyle n\times n$ diagonalizable matrix with dominant eigenvalue $\displaystyle \lambda_{1}$. Then there exists a nonzero vector $\displaystyle x_{0}$ such, that the sequence of vectors $\displaystyle x_{k}$ defined by $\displaystyle x_{1}=Ax_{0}, x_{2}=Ax_{1},x_{3}=Ax_{2}, ... ,x_{k}=Ax_{k-1}, ...$ approaches a dominant eigenvector of A.

Proof: We may assume that the eigenvalues of A have been labeled so that:

$\displaystyle |\lambda_{1}| > |\lambda_{2}| \geq |\lambda_{3}| \geq ...\geq |\lambda_{n}|$

Let $\displaystyle v_{1}, v_{2}, v_{3}, ... , v_{n}$ be the corresponding eigenvectors. Since $\displaystyle v_{1}, v_{2}, v_{3}, ... , v_{n}$ are linearly independent, they form a basis for $\displaystyle \mathbb{R}^{n}$. Consequently we can write $\displaystyle x_{0}$ as linear combination of these eigenvectors. Say:

$\displaystyle x_{0}=c_{1}v_{1}+c_{2}v_{2}+ ... + c_{n}v_{n}$

We use that $\displaystyle x_{k}=A^{k}x_0$ and substitute the above equation into it to get:

$\displaystyle x_{k}=A^{k}x_0=c_{1} \lambda_{1}^{k}v_{1}+c_{2} \lambda_{2}^{k}v_{2}+ ... + c_{n} \lambda_{n}^k v_{n}$

We next factor out $\displaystyle \lambda_{1}^k$ to get

$\displaystyle x_{k} = \lambda_{1}^{k} ( c_{1}v_{1} + c_{2} (\frac{\lambda_{2}}{\lambda_{1}})^{k} + ... + c_{n} (\frac{\lambda_{n}}{\lambda_{1}})^{k} )$

Doing so we use the fact that $\displaystyle \lambda_{1}$ is not equal to zero.

So far so good. But I cannot get why the dominant eigenvalue cannot be zero. It is obvious that If there are more than one distinct eigenvalues, for $\displaystyle \lambda_{1} $ to be dominant, its absolute value must be greater than the absolute value of every other eigenvalue. So it cannot be zero. But what if the only eigenvalue is 0? It certainly is not wrong to have eigenvalue equals zero and it's certainly not wrong diagonalizable matrix to have less than n distinct eigenvalues. It seems to me at first look, that only a zero matrix can have zero as only eigenvalue (and will have $\displaystyle \mathbb{R}^n$ as eigenspace). But If we substitute zero matrix for A, things certainly won't behave as in the theorem? [0][0]...[0]x will converge to the zero vector which is not eigenvector at all.

Uhhh. Help???

The book in question is - David Poole's Linear Algebra (Second Edition). (Great book so far as a layman as me can say). This is theorem 4.28 on page 309.