Operator in a real vector space has an upper block triangular matrix

Question

I was trying to prove that an operator $T$ in a real vector space $V$ has an upper block triangular matrix with each block being $1 \times 1$ or $2 \times 2$ and without using induction.

The procedure which i followed was :

We already know that an operator in a real vector space has either a one dimensional invariant subspace or a 2 dimensional invariant subspace.

Whatever be the case now, lets begin with the vector(s) which span these subspaces.

Let U denote this subspace ----- (1)

Now, if i am able to prove that there exists an another subspace W such that T is an invariant operator on the direct sum of U and W , then we can prove that operator T in a real vector space V has an upper block triangular matrix .

I need a direction on proving the latter part.

I sincerely thank you for the help.

Answer 1

Write $V=U\oplus W$ for some $W$ and consider the projection $p:V\to W$ in the $W$ component of the sum. We get a new operator $\tilde T:W\to W$ by restriction of $p\circ T$ to $W$. Since $\dim W<\dim V$, we know that $\tilde T$ has a triangular block decomposition with blocks of size $1$ and $2$. Let $\mathcal B_1$ be a basis of $W$ given such decomposition and let $\mathcal B_2$ be any basis of $U$. It is easy to see that $\mathcal B:= \mathcal B_1 \cup \mathcal B_2$ is a basis of $V$ and that $T$ in this basis decomposes as you want.

EDIT: I miss the part of "without using induction" in the question but you can adapt my proof adding some "handwaving" to avoid the induction.

Answer 2

I don't know what you mean by not using induction; it seems that you're getting only one $1\times1$ or $2\times2$ block from your reasoning, and the other ones can follow only by using induction.

And the argument is simpler than you think. If you take your $d\in\{1,2\}$-dimensional invariant subspace, choose a basis of the whole space starting with $d$ vectors in that subspace, and express $T$ in this basis, then the first $d$ columns have nonzero entries only in the first $d$ rows, so your matrix is already block-upper-triangular with square diagonal blocks of size $d$ and $n-d$. Now apply induction for the operator corresponding to the $n-d$ square block (which is really an operator on the quotient space by the $d$-dimensional invariant subspace, but you do not really need this interpretation); this leads to a different basis of $\mathbf R^{n-d}$ in which the operator originally given by the square $n-d$-block has the desired form. The corresponding change of basis (affecting only the final $n-d$ basis vectors of the original basis) will bring your entire matrix into the correct shape.