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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 X 1 or 2 X 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.

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  • $\begingroup$ Any matrix is an upper triangular block diagonal matrix if you only consider one block. Here you might want to add that the diagonal blocks have size $1$ or $2$. $\endgroup$ – Julien Mar 26 '13 at 14:33
  • $\begingroup$ @julien done. thanks for the correction. Any direction on this ? $\endgroup$ – MathMan Mar 26 '13 at 14:43
  • $\begingroup$ ok , i think i got something. let U be the one dimension or 2 dimensional subspace such that T is invariant on U. Now , let Y be a subspace of V such that Y = Direct sum of U and W . => y = u + w => Ty = Tu+Tw => since Tu belongs to U ; the additional components must be common to both Ty and Tw. => T is invariant on U and W. $\endgroup$ – MathMan Mar 26 '13 at 15:05
  • $\begingroup$ If you really don't want to use induction on some $P_n$, you can use a proof by contradiction by supposing that $\exists n, \lnot P_n$... $\endgroup$ – xavierm02 Mar 26 '13 at 15:18
  • $\begingroup$ You can't always find an invariant $W$. Otherwise, this would prove that every real matrix has a Jordan block diagonal form. And there can be $1$'s above the blocks. But of course, you can find $W$ such that $V\oplus W$ is invariant. Just complete to the whole space...So I'm not sure what your strategy really is. $\endgroup$ – Julien Mar 26 '13 at 15:32
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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.

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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.

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