In Axler's linear algebra chapter 5, he explicitly shows that for a complex vector space $V$ and a linear operator $T$, there exists a choice of basis such that the matrix of the operator with respect to it is upper triangular. Here's the proof sketch:

Use induction on the dimension of $V$, for $T\in L(V)$, let $\lambda$ be an eigenvalue (which must exist for a complex vector space), let $U=\mathrm{im}(T-\lambda I)$. He showed that $U$ is a proper subspace of $V$, and it is invariant under $T$. So $T$ restricted to $U$ is an operator on $U$, by induction hypothesis, there exists basis $(u_1,\dots ,u_m)$ such that the matrix for $T$ restricted to $U$ is upper triangular.

Then extend $U$'s basis to $(u_1,\dots u_m, v_1, \dots, v_n)$ a basis for $V$.

$$Tv_k=(T-\lambda I)v_k + \lambda v_k$$

The expression shows that $Tv_k \in \mathrm{span} (u_1, \dots, u_m, v_1, \dots ,v_k)$ for each $k$. By a theorem proven before, this is equivalent to the matrix of $T$ with respect to this basis being upper triangular.

My question: The above proof only uses the fact that a complex vector space must have an eigenvalue, for a real vector space, if it is given that it has an eigenvalue, would the above construction to show that there exists a basis such that the matrix is upper triangular make sense? If so, does it mean that a real vector space have invariant subspace (of $\mathrm{im}(T)$ ) of different dimensions under $T$? (Since there exists a basis $(v_1,\dots , v_n)$ such that $Tv_1,\dots , Tv_k \in \mathrm{span}(v_1,\dots , v_k)$ for each $k$)


1 Answer 1


In a real vector space, we can't generally guarantee that $T$ will have $n$ eigenvalues. The real power of the complex case is that as you restrict $T$ to smaller subspaces, we can always find an eigenvalue/eigenvector within that space.

It is notable, however, that real operators can be "almost" upper triangularized (see p. 39 here). In particular, we note that a real linear operator in $L(V)$ always has an invariant subspace of dimension at most $2$.

  • $\begingroup$ Sorry, I am not getting this one. The proof only requires the existence of a single eigenvalue, so wouldnt it work if there is at least one real eigenvalue for a real vector space? Using similar construction, like finding a basis for $\mathrm{im}(T-\lambda I)$, etc. What makes this construction invalid? $\endgroup$
    – lEm
    Commented Aug 17, 2016 at 15:52
  • $\begingroup$ The key to the proof is that when you take the restriction of $T$ to an invariant subspace, that transformation also has an eigenvalue. It's not enough to simply say that $T$ itself will have an eigenvalue. $\endgroup$ Commented Aug 17, 2016 at 15:54
  • $\begingroup$ Also, did you mean that any invariant subspace of a real vector space at most has dimension $2$? That is actually very surprising for me. However I dont see the connection between this fact and the decomposition. $\endgroup$
    – lEm
    Commented Aug 17, 2016 at 15:56
  • $\begingroup$ That is not what I meant. I meant what I said, and I said what I meant. We can't guarantee a one dimensional invariant subsapce (i.e. an eigenvector and its span), but we can guarantee a two dimensional invariant subspace. $\endgroup$ Commented Aug 17, 2016 at 16:00
  • $\begingroup$ Sorry I misinterpreted what you meant. Now its clear. Thanks a lot for the help :) $\endgroup$
    – lEm
    Commented Aug 17, 2016 at 16:03

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