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Problem 1 from Axler's Linear Algebra Done Right, 3rd ed, page 160:

Suppose $T \in \mathcal{L}(V)$ is diagonalizable. Prove that $V = null (T) \oplus range (T)$.

The case for $V$ finite-dimensional is already done here: If $T\in\mathcal{L}(V)$ is diagonalizable then $V = \mathrm{null}\; T \oplus \mathrm{range}\; T$

But for the case $V$ infinite dimensional I can't figure out, then if you could prove or give a counter-example, I would appreciate.

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    $\begingroup$ For any linear operator $T:V\to V$, $\text{ker}(T)$ is just the eigenspace with eigenvalue $0$. You only need to prove that, if $T$ is diagonalizable, $\text{im}(T)$ is the (direct) sum of eigenspaces with nonzero eigenvalues. $\endgroup$ Feb 7, 2017 at 14:58
  • $\begingroup$ @Batominovski But, to exist a matrix for $T$, it is necessary exist a list of vectors $v_1,v_2,v_3,...$ that spans $V$. Is that true for any infinite dimensional vector space ? $\endgroup$ Feb 7, 2017 at 15:25
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    $\begingroup$ In infinite-dimensional cases, it doesn't really make sense to talk about the matrix of a linear operator. (Well, you can---sort of, by writing $T$ as a $2$-dimensional infinitely long table $\left[t_{x,y}\right]_{x,y\in B}$ of numbers, where $B$ is a basis of $V$.) Also, I think you need to find out what "diagonalizable" means in your book. The general notion is that a linear operator $T:V\to V$ is diagonalizable iff $V=\bigoplus_{\lambda}\,V_\lambda$, where $\lambda$ runs over all eigenvalues of $T$ and $V_\lambda$ denotes the eigenspace of eigenvalue $\lambda$. $\endgroup$ Feb 7, 2017 at 15:35
  • $\begingroup$ @Batominovski The Axler's definition for $T$ be diagonalizable is "An operator $T \in \mathcal{}L(V) $ is called diagonalizable if the operator has a diagonal matrix with respect to some basis of $V$". Do you think that Axler was just think in the finite dimensional case? $\endgroup$ Feb 7, 2017 at 15:39
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    $\begingroup$ @RafaelDeiga It seems safe to assume that. That particular definition isn't great for an operator on an infinite dimensional vector space. $\endgroup$
    – Ken Duna
    Feb 7, 2017 at 15:51

1 Answer 1

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Hints:

As $\;V\;$ contains a basis of eigenvectors of $\;T\;$ ,$\;V\;$ is the direct sum of the corresponding eigenspaces. But $\;\ker T=\text{null}\,T\;$ is just the eigenspace corresponding to the eigenvalue zero (which, btw, is the zero space if $\;T\;$ is invertible) ...

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