# Applied Linear Algebra

Linear Algebra

Is my proof correct?

• 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. – Vedran Šego Mar 22 '15 at 0:44
• (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. – Redneck Blue State Mar 22 '15 at 1:49
• (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. – Redneck Blue State Mar 22 '15 at 1:56
• 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 – Redneck Blue State Mar 22 '15 at 1:58
• Those steps would be better, yes. As for writing formulas on this site, check this out. – Vedran Šego Mar 23 '15 at 0:12

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}