Maybe this question is basic however I'm not familiar with operator theory. Does every operator have a matrix? I also would like to see some proof of this fact (if it's elementary) or at least get some good reference.
Does every operator have a matrix?
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0I'm looking for as general answer to this question as possible. – 2012-11-13
1 Answers
If you think about the relationship between an operator and its matrix, you can see that you can do it in the maximum possible generality: consider a linear map $T:V\to W$, where $V,W$ are vector spaces over some field $\mathbb F$. Let $\{x_\alpha\}_{\alpha\in A}$ be a basis for $V$, $\{y_\beta\}_{\beta\in B}$ a basis for $W$.
As $Tx_\alpha\in W$, there exist coefficients $t_{\alpha,\beta}$, $\beta\in B$, such that $Tx_\alpha=\sum_\beta t_{\alpha,\beta}y_\beta$ (only finitely many $t_{\alpha,\beta}$ are nonzero for fixed $\alpha$).
So we can associate, to the operator $T$, the "matrix" $\{t_{\alpha,\beta}\}_{\alpha\in A,\beta\in B}$.
Now, the fact that you can do this, doesn't really make it useful in infinite-dimensional cases (even when $\mathbb F=\mathbb C$ or $\mathbb R$). Because while in finite dimension you can get a fair amount of information about the operator from the matrix (i.e. eigenvalues, Jordan form) you cannot really do that in infinite dimension (at least for a general operator).
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1In infinite dimension it makes more sense to talk about the spectrum, and not just about eigenvalues. And, while the spectrum can be determined in many concrete examples, there is no general method. – 2012-11-13