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Given the $(4,4)$-matrix

$$Q = \begin{bmatrix}w&-x&-y&-z\\x&w&-z&y\\y&z&w&-x\\z&-y&x&w\end{bmatrix}$$

I am supposed to compute $Q$ multiplied by its transpose, then "from this computation, determine under which conditions on $x,y,w$,and $z$ is $Q$ invertible. If it exists, compute $Q$ inverse."

So I computed the product, which gives me a $4 \times 4$ diagonal matrix with $w^2+x^2+y^2+z^2$ down the diagonal. I have absolutely no idea how this helps me determine if $Q$ itself is invertible; I can easily get the inverse of $QQ^T$, (given the condition that $w^2+x^2+y^2+z^2$ does not equal zero) but what do I know about the inverse of $Q$? And how do I compute this inverse?

EDIT: I just found this statement in my book: "If A is an invertible matrix, then $AA^T$ and $A^TA$ are also invertible.". Is the converse of this true? ie. If $AA^T$ is invertible, then is A also invertible? Would this be a valid proof as to why the same condition on $w,x,y,z$ for $QQ^T$ to be invertible is also the condition for $Q$ to be invertible? Still have no clue how to compute the actual matrix from this though.

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  • $\begingroup$ math.stackexchange.com/questions/2642352/… $\endgroup$ – Andres Mejia Feb 15 '18 at 4:35
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    $\begingroup$ $QQ^T = kI$ where $k=x^2+y^2+z^2+w^2,$ according to what you have said. What happens if you divide both sides by $k$? $\endgroup$ – saulspatz Feb 15 '18 at 4:38
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    $\begingroup$ Do you know that $A$ is invertible if and only if $det(A) \neq 0$? $det(AA^{T}) =(det(A))^{2}$ ( assuming that you are considering real matrices) so it is obvious that $A$ is invertible if and only if $AA^{T}$ is invertible. $\endgroup$ – Kavi Rama Murthy Feb 15 '18 at 5:53
  • $\begingroup$ If the matrix is real, then $x^2+y^2+z^2+w^2>0$ unless $Q$ is the zero matrix. $\endgroup$ – Lord Shark the Unknown Feb 15 '18 at 6:11

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