# What are the solutions to $X$ for $X^{T} A X = A$? Knowing that, what are the solutions to Y for $Y= (I−X)(I+X)^{-1}$ (with $\det(I+X) \neq 0$)?

All matrices are square matrices of real numbers. The goal is to use these properties to show that $$AY+Y^{T}A= 0$$

What do the equations in the title reveal about the properties of $$A$$, $$X$$, and $$Y$$? Can $$X$$ only be the identity matrix?

Thanks!

loup blanc's edit: This question is about showing that the solutions of $$X^TAX=A$$ are in isomorphism with the solutions of $$AY+Y^TA=0$$. Recent research (2011) has made it possible to know the dimension of the solutions of the second equation, then the dimension of the solutions of the first one.

• If $A$ is the identity matrix, then $X$ can be any orthogonal matrix. An analogous result applies whenever $A$ is symmetric and positive definite Commented Aug 31, 2019 at 18:20
• Welcome to MSE. In order to get responses that suit your needs, please include in the body of the question your own thoughts, the effort made so far, and the specific difficulties that got you stuck. Commented Sep 1, 2019 at 3:38
• For those who are interested, one can find the dimension of the algebraic set of the solutions of $X^TAX=0$ when $A$ is symmetric invertible or (more difficult) when $A$ is generic. cf. my answer in math.stackexchange.com/questions/3105787/…
– user91684
Commented Sep 1, 2019 at 21:17

In view of the expression for your $$Y$$, here is a pertinent result:

Theorem. Suppose $$A$$ is symmetric and nonsingular over a field of characteristic $$\ne2$$. Then every solution to the equation $$X^TAX=A$$ with $$\det(I+X)\ne0$$ is in the form of $$X=(A+K)^{-1}(A-K)$$ for some skew-symmetric matrix $$K$$ such that $$A+K$$ is nonsingular. Such a matrix $$X$$ is called a cogredient automorph or congruent automorph of $$A$$.

For a proof of the above theorem, see Sam Perlis, Theory of Matrices, pp.104-105.

In the special case where $$A=I$$ over $$\mathbb R$$, the equation $$X^TAX=A$$ reduces to $$X^TX=I$$ and hence $$X$$ is a real orthogonal matrix, and the expression $$X=(A+K)^{-1}(A-K)$$ in the theorem above becomes $$X=(I+K)^{-1}(I-K)$$, which is the familiar Cayley transform.

Some more general results for a nonsingular but perhaps non-symmetric $$A$$ can be found in sec. 37 (pp.65-68), chapter V of Mac Duffee's detailed survey The Theory of Matrices.

• Well done, I didn't know this generalization. When $A$ is symmetric $>0$, it's easy to see that it works, because $X=A^{-1/2}QA^{1/2}$ where $Q\in O(n)$. Then, using the above formula giving $X$, we obtain $A^{1/2}XA^{-1/2}=(I+A^{-1/2}KA^{-1/2})^{-1}(I-A^{-1/2}KA^{-1/2})=(I+H)^{-1}(I-H)$ where $H$ is skew symmetric. Moreover, $\{X,X^TAX=A\}$ is an algebraic set of dimension $n(n-1)/2$ (independent on $A$).
– user91684
Commented Sep 1, 2019 at 15:05
• In fact, the OP's problem is equivalent to show the formula in Perlis' book. According to the @Omnomnomnom 's post, $AY+Y^TA=0$ that is $(I+X)^{-1}(I-X)=A^{-1}K$ where $K$ is skew symmetric, that is, $X=(A+K)^{-1}(A-K)$.
– user91684
Commented Sep 1, 2019 at 15:32

Without characterizing the solutions to your equations, we can say the following:

Suppose that $$X$$ is a solution to $$X^TAX = A$$, and that $$Y = (I - X)(I + X)^{-1}$$. Because $$\det(I+X) \neq 0$$, it suffices to show that $$(I + X)^T[AY + Y^TA](I + X) = 0.$$ With that in mind, we note that we can expand \begin{align*} (I + X)^T[AY + Y^TA]&(I + X) = (I+X)^TA(I-X) + (I-X)^TA(I + X) \\ &= A - AX + X^TA - X^TAX + A - X^TA + AX - X^TAX\\ &= A - AX + X^TA - A + A - X^TA + AX - A = 0. \end{align*} The desired conclusion follows.

• Could you explain how you arrived at $(I + X)^T[AY + Y^TA](I + X) = 0$ ? Commented Aug 31, 2019 at 20:40
• I wanted to get rid of the $(I+X)^{-1}$ matrices; this was a convenient way to do so Commented Aug 31, 2019 at 21:06
• But what properties allows you to say that $(I + X)^T[AY + Y^TA](I + X) = [AY + Y^TA]$? It seems like you are using the given $X^{T}AX=A$ but instead you are using $(I+X)$ somehow. You also can't just assume they are equivalent because $AY+Y^TA=0$, because that's what you're trying to prove. Commented Aug 31, 2019 at 21:41
• @JungleFungus let $M = AY + Y^TA$, we're trying to show that $M=0$. If we know that $(I+X)^TM(I+X) = 0$, then we can say that $$(I+X)^TM(I+X) = 0 \implies\\ [(I+X)^TM(I+X)](I + X)^{-1} = 0(I+X)^{-1} \implies\\ (I+X)^TM = 0 \implies\\ (I+X)^{-T}(I+X)M = (I + X)^{-T}0 \implies\\ M = 0$$ The point was not that $(I + X)^TM(I + X) = M$. Commented Sep 1, 2019 at 9:28
• I'm still not sure this makes sense... you are saying that if we know $(I+X)^TM(I+X) = 0$ then we can show that it does equal zero. Isn't that circular reasoning? The point is that we don't know that $M = 0$, so we have to assume that it could be anything, then prove that it indeed is zero. Commented Sep 3, 2019 at 13:55