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I need to prove that for every real linear operator $T:\mathbb{R}^n\longrightarrow \mathbb{R}^n$, there exists an orthonormal basis of $\mathbb{R}^n$ such that the corresponding matrix is block diagonal and each block is $1\times 1$ or $2\times 2$.

I've spent hours trying to prove that but I haven't a clue.

I'll be really grateful for any help.

Julien
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  • Are you sure your question is correct? –  Mar 31 '13 at 14:31
  • Yeah... I've just checked it:) –  Mar 31 '13 at 14:33
  • "Prove that for every square real matrix A of the dimension nxn, there exists an orthonormal basis B of $\mathbb{R}^n$ such that $\left[f_{A}\right]_{B}^{B}$ is block diagonal and each block is 1x1 or 2x2" –  Mar 31 '13 at 14:41
  • I'm totally confused:-/ Some hint please?:) –  Mar 31 '13 at 16:53
  • Maybe I am being stupid, but I still think the condition "orthonormal" is too strong. Hope someone can prove or disprove this claim. I would very much like to see the solution. –  Mar 31 '13 at 18:02
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    @TongZhang This is not even true with an arbitrary basis. A fortiori not for an orthonormal basis. – Julien Mar 31 '13 at 18:22
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    Where did you get this question? Are you sure there are no other assumptions? – Julien Mar 31 '13 at 18:32
  • It's a homework assignment... I guess it should be "block upper triangular and the diagonal blocks are 1x1 or 2x2". If so, real Schur decomposition would be the answer. –  Mar 31 '13 at 19:04
  • Yes, that's precisely the statement of real Schur decomposition. – Julien Mar 31 '13 at 19:12

1 Answers1

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You should stop trying because this is false. Otherwise, this would make the whole Jordan normal form theory almost pointless. Indeed, when there are $1$'s on the super diagonal of the Jordan normal form, we can't get rid of them like that.

Consider for instance the matrix $$ A=\left(\matrix{0&1&0\\0&0&1\\0&0&0} \right). $$ Note that the characteristic polynomial of $A$ is $x^3$. If the result was true, $A$ would be similar to a block diagonal matrix with blocks of size $1\times 1$ or $2\times 2$. Without loss of generality, we would get $$ PAP^{-1}=\left(\matrix{*&0&0\\0&*&*\\0&*&*} \right) $$ including the three $1\times 1$ blocks case. Given that the characteristic polynomial $x^3$ of $A$ is the product of the characteristic polynomials of these two diagonal blocks, we see that the $1\times 1$ block is $0$, and the $2\times 2$ block is triangularizable with $0$ diagonal, as its characteristic polynomial $x^2$ splits. So without loss of generality, $$ PAP^{-1}=\left(\matrix{0&0&0\\0&0&*\\0&0&0} \right). $$ On one hand, the rank of $A$ is $2$. On the other hand, the rank of $PAP^{-1}$ is $0$ or $1$. Contradiction.

Julien
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