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Linear Algebra

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Is my proof correct?

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  • $\begingroup$ What do "from $(A^TA^{-1})^{-1}$, $(A^{-1})^{-1}(A^T)^{-1}$" and "Then, $(A)(A^T)^{-1}$" and similar constructs mean? If there is no equality, what is your statement? I think I know what you want to say, but you have to form proper statements in order to have a proper proof. $\endgroup$ – Vedran Šego Mar 22 '15 at 0:44
  • $\begingroup$ (A<sup></sup>TA<sup></sup>-1)<sup></sup>-1 = (A^−1)^−1(A^T)−^1 Yes, thank you, I make error in my typing. So, with the correction will my proof hold. $\endgroup$ – Redneck Blue State Mar 22 '15 at 1:49
  • $\begingroup$ (A<sup>T</sup> A<sup>-1</sup>)<sup>-1</sup> = (A<sup>-1</sup>)<sup>-1</sup> (A<sup>T</sup>)<sup>-1</sup> Yes, thank you, I make error in my typing. So, with the correction will my proof hold. $\endgroup$ – Redneck Blue State Mar 22 '15 at 1:56
  • $\begingroup$ Sorry about my previous comment, new with markdown. Trying to put superscript but it didn't work. (A^T A^−1)^−1 = (A^−1)^−1 (A^T)^−1 $\endgroup$ – Redneck Blue State Mar 22 '15 at 1:58
  • $\begingroup$ Those steps would be better, yes. As for writing formulas on this site, check this out. $\endgroup$ – Vedran Šego Mar 23 '15 at 0:12
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The line that mentions commutativity seems to be unjustified. Other than that, the steps seem to be correct, but you should try to write them down more carefully.

Here is an alternative proof. We want to show that $$ \left(\mathbf{A}^{T}\mathbf{A}^{-1}\right)^{T} = \left(\mathbf{A}^{T}\mathbf{A}^{-1}\right)^{-1}. $$ Let $\mathbf{C} = \mathbf{A}^{T}\mathbf{A}^{-1}$. Essentially, we want to show that $\mathbf{C}^{T} = \mathbf{C}^{-1}$. It suffices to show that $\mathbf{C}^{T}\mathbf{C} = \mathbf{I}$. We have \begin{align} \mathbf{C}^{T}\mathbf{C} &= \left(\mathbf{A}^{T}\mathbf{A}^{-1}\right)^{T}\left(\mathbf{A}^{T}\mathbf{A}^{-1}\right)\\ &= \left(\mathbf{A}^{-1}\right)^{T}\left(\mathbf{A}^{T}\right)^{T}\left(\mathbf{A}^{T}\mathbf{A}^{-1}\right)\\ &= \left(\mathbf{A}^{T}\right)^{-1} \mathbf{A}\mathbf{A}^{T}\mathbf{A}^{-1}\\ &= \left(\mathbf{A}^{T}\right)^{-1} \mathbf{A}^{T}\mathbf{A}\mathbf{A}^{-1}\\ &= \mathbf{I}\cdot \mathbf{I} = \mathbf{I}. \end{align}

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  • $\begingroup$ Thank you for your suggestion. I will rewrite and correct my proof. And I can't help to notice how precise, concise and quite nice your proof is. I have a very long way to go. Math is definitely not my forte. $\endgroup$ – Redneck Blue State Mar 22 '15 at 15:21

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