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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.

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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.

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  • $\begingroup$ 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

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