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I need to find $\det X$ where $$8GX=XX^T,\quad G=\left(\begin{matrix}5 & 4\\3 & 2\\\end{matrix}\right).$$

My answer is that the determinant of $X$ is $-128$ and that is correct but there is one more value of $\det X$ that can solve the equation.

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    $\begingroup$ Huh? The equation looks like you're looking for a 2x2 matrix $X$, yet you state that the answer is $-128$... $\endgroup$ – fgp Apr 30 '13 at 12:19
  • $\begingroup$ If $X$ is a $2\times n$ matrix, then the left side is $2\times n$ while the right is $2\times2$, so $n$ must be $2$. So, how can the answer be a number? $\endgroup$ – Gerry Myerson Apr 30 '13 at 12:20
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    $\begingroup$ I think closing this question is a pointless exercise. The OP is confused and so the question makes no sense., but what good does closing it do? At least give the OP a bit more time to work out what the question should be... $\endgroup$ – user1729 Apr 30 '13 at 14:32
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    $\begingroup$ The OP can edit a closed question. $\endgroup$ – Emily Apr 30 '13 at 15:31
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    $\begingroup$ @Arkamis Sure, of course they can. But 1) They do not know that, and 2) They then have to garner the support of $5$ people to reopen it. Which is all a bit too much effort, considering it would have been harmless enough to have left it for, say, a day before closing it! $\endgroup$ – user1729 Apr 30 '13 at 16:22
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This question had been heavily downvoted and was once closed, but frankly I don't think this is an easy question and I don't understand the reasons behind the downvotes.

Let $A=8G=\pmatrix{40&32\\ 24&16}$. If $AX=XX^T$, then $\det(A)\det(X)=\det(X)^2$. So, the only possible values for $\det(X)$ are $0$ and $\det(A)=-128$. Clearly, the equation $AX=XX^T$ with $\det(X)=0$ is solvable --- $X=0$ is a trivial solution in the first place, and it is easy to show that all singular solutions are given by $X=0$ or $X=\lambda uu^T$, where $(\lambda,u)$ is an eigenpair of $A$ with $\|u\|=1$.

The real question, though, is whether the determinant $-128$ is attainable, i.e. whether there exists a nonsingular matrix $X$ such that $AX=XX^T$. This is equivalent to $A=XX^TX^{-1}$ and it apparently is a very strong requirement --- not only must $A$ be similar to $X^T$, the change-of-basis matrix must also be equal to $X$. Perhaps I am missing something here, but I honestly don't see any obvious answer to this problem.

Edit: For those who are interested, if we put $Y=AX$, then $AYA^T=AXX^TA^T=YY^T$. So, if $X$ is nonsingular, then $Y$ is symmetric positive definite and $\det(Y)=128^2=d$ (say). So, if we let $t=\operatorname{trace}(Y)$ and write $Y=\pmatrix{x&y\\ y&z}$, then by Cayley-Hamilton theorem, we get $AYA^T=YY^T=Y^2=tY-dI$ and this equation can be rewritten as $$ \pmatrix{x\\ y\\ z}= -256\left[\pmatrix{ 25&40&16\\ 15&22& 8\\ 9&12& 4} -sI\right]^{-1} \pmatrix{1\\ 0\\ 1},\tag{1} $$ where $s=t/64$. Since $t=64s=x+z$, if we left-multiply $(1)$ by $(1,0,1)$ on both sides, we will eventually obtain $p(s)=(s-2)(s^3 - 49s^2 - 208s - 216)=0$. It turns out that $p$ has two real roots. The root $s=2$ does not give rise to a valid $X$, but the other root, $s\approx53.00132225047258$ gives the solution $$ X=\pmatrix{ 43.30336531969982 &26.18058434772915\\ 25.88968225037343 &12.69663468030015 }. $$

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    $\begingroup$ The question was closed, not because it was thought to be easy, but because it was so badly stated as to be incomprehensible. It was re-opened after it was clarified that OP was after $\det X$ --- and that is an easy question. But I appreciate your efforts to answer the more interesting question that could have been asked. $\endgroup$ – Gerry Myerson May 3 '13 at 9:11
  • $\begingroup$ @GerryMyerson I see. Thanks for the explanation. $\endgroup$ – user1551 May 3 '13 at 13:26

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