I am reading a proof on the diagonalization theorem (Linear Algebra: A Modern Introduction) and I'm struggling with understanding the first part. The theorem is as follows:

The Diagonalization Theorem

Let A be an n x n matrix whose distinct eigenvalues are $\lambda_{1}, \lambda_{2},...,\lambda_{k}$. The following statements are equivalent:

a. A is diagonalizable

b. The union $\beta$ of the bases of the eigenspaces of A (as in Theorem 4.24) contains n vectors.

c. The algebraic multiplicity of each eigenvalue equals its geometric multiplicity.

The proof of the first part is as follows:

Proof (a)=>(b) If A is diagonalizable, then it has n linearly independent eigenvectors, by Theorem 4.23. If $n_{i}$ of these eigenvectors correspond to the eigenvalue $\lambda_{i}$, then $\beta_{i}$ contains at least $n_{i}$ vectors. (We already know that these $n_{i}$ vectors are linearly independent; the only thing that might prevent them from being a basis for $E_{\lambda_{i}}$ is that they might not span it.) Thus $\beta$ contains at least n vectors. But, by theorem 4.24, $\beta$ is a linearly independent set in $\Re^{n}$; hence it contains exactly n vectors.

Why might the $n_{i}$ vectors not span $E_{\lambda_{i}}$? Shouldn't the eigenvectors of an eigenvalue always span the corresponding eigenspace as an eigenspace is defined as the collection of all eigenvectors corresponding to $\lambda$ together with the zero vector.


That the $n_i$ vectors span the eigenspace $E_{\lambda_i}$ does require a proof: After all, they aren't all the eigenvectors corresponding to $\lambda_i$, only a small subset of these eigenvectors.

We might throw a bit of light on the question by generalizing the setting a bit: Assume $A$ is not diagonalizable, and let $\beta$ be maximal set of eigenvectors of $A$. If $\beta_i$ are those vectors in $\beta$ corresponding to the eigenvalue $\lambda_i$, it is still true that $\beta_i$ spans $E_{\lambda_i}$. (Otherwise consider a vector in $E_{\lambda_i}$ not in the span of $\beta_i$, and show that this could be added to $\beta$, contradicting the maximality of $\beta$.)

What goes wrong with the proof, if $A$ is not diagonalizable, is that $\beta$ will have fewer than $n$ members. But that is a different issue.

  • $\begingroup$ So the book says that the eigenvectors of $\lambda_{i}$ might not span $E_{\lambda_{i}}$ simply because they haven't proved it yet. $\endgroup$ – Ruben23630 May 23 '17 at 20:39
  • $\begingroup$ That does indeed seem to be the case. $\endgroup$ – Harald Hanche-Olsen May 23 '17 at 20:50

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.