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I'm trying to understand a proof about this theorem and I find myself stuck in a step.

Let's consider two quadratic forms $\langle q, Aq \rangle$ and $\langle q, Bq \rangle$, where the first one is positive definite. We can write $A$ as $UDU^{-1}$ ($U$ is orthogonal and $D$ is diagonal). If we define $\alpha=U\sqrt{D}U^{-1}$, where $\sqrt{D}$ is the diagonal matrix whose elements are $\sqrt{D_{ii}}$. It's then easy to show that $\alpha^2=A$. Then the proof goes on showing that:

$$\alpha^T=\left(U\sqrt{D}U^{-1}\right)^T=U\sqrt{D}U^{-1}=\alpha$$

and justifies this by saying that $U$ is orthogonal. I don't get how this works. If I compute this explicitly I get:

$$\alpha^T=\left(U\sqrt{D}U^{-1}\right)^T=U^T\sqrt{D}^T\left(U^T\right)^T=U^{-1}\sqrt{D}U.$$

Now, how is it possible that in the first case $U$ and $U^{-1}$ are inverted?

Michael Albanese
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Mr. Dee
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  • take a look at these four: http://math.stackexchange.com/questions/88022/congruence-and-diagonalizations http://math.stackexchange.com/questions/88026/is-there-a-simpler-way-to-falsify-this http://math.stackexchange.com/questions/1406359/simultaneously-diagonalise-two-real-quadratic-forms http://math.stackexchange.com/questions/1414002/simultaneous-orthogonal-diagonalization-of-two-matrices – Will Jagy Nov 21 '15 at 17:29

1 Answers1

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When you take the transpose of the product, you reverse the order, i.e. $(AB)^T = B^TA^T$. Therefore, the correct computation is

$$\alpha^T = (U\sqrt{D}U^{-1})^T = (U^{-1})^T\sqrt{D}^TU^T = U\sqrt{D}U^{-1} = \alpha.$$

Yai0Phah
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Michael Albanese
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