## linear algebra – What are the solutions for \$ X \$ for \$ X ^ {T} A X = A \$? If you know that, what is the solution for Y for \$ Y = (I – X) (I + X) ^ {- 1} \$ (with \$ det (I + X) neq 0 \$)?

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

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

Thanks a lot!

## Complexity Theory – Would \$ Sigma_i ^ P neq Pi_i ^ P \$ mean that the polynomial hierarchy can not collapse to the \$ i \$ -th level?

If $$Sigma_i ^ P = Pi_i ^ P$$, then it follows that the polynomial hierarchy collapses to that $$i$$-th level.

What about the case? $$Sigma_i ^ P neq Pi_i ^ P$$?
For example, consider the case of $$NP neq coNP$$, As far as I know, this would mean that the polynomial hierarchy can not collapse to the first level, because if $$PH = NP$$then in particular $$coNP subseteq NP$$, which does ______________ mean $$NP = coNP$$, Can we extend this idea to prove the general case:
$$Sigma_i ^ P neq Pi_i ^ P$$ implied $$PH$$ can not collapse $$i$$-th level?

## cv.complex variables – Why is \$ (1- | Cu-D | ^ {- 1}) operatorname {Im} (C (Cu-D)) ^ 2 = operatorname {Re} (C (Cu-D) ) ^ 2 \$ impossible for \$ C neq 0 \$

From the system of differential equations
$$pmatrix {g & f \ – fg & 1 + f ^ 2 \ – f & g \ 1 + g ^ 2 & -fg} pmatrix {f & # 39; & gt; \ g & # 39 ;; = pmatrix {6f? g? \ – 3gf & # 39; ^ 2 + 3ff? G & # 39; – 3f & # 39; ^ 2 + 3g & gt; 2 & gt; 3gf & gt; & gt; -3fg & gt; ^ 2}$$

The first and third equations can be combined
$$pmatrix {f & -g \ g & f} pmatrix {f & # 39; & gt; \ g & # 39;} = 3 pmatrix {f & gt; 2-g & gt; ^ 2 \ 2f & ggr; & gt;} \ text {or} \ (f + ig) (f & + + ig &)) = 3 (f + + ig)) ^ 2$$
this is through unique integration $$f & # 39; + ig & # 39; = C (f + ig) ^ 3 tag {*},$$ and twice
$$(f + ig) ^ {- 2} = D-Cu implies f ^ 2 + g ^ 2 = | D-Cu | ^ {- 1}.$$

Multiply the 4th or 2nd equation with $$f,$$ or. $$g$$ and sum there
$$ff & # 39; & # 39; + gg & # 39; & gt; = – 3 (f ^ 2g? 2-2ff? Gg? + F? 2g ^ 2) = – 3 (fg? -F? G) ^ 2 = -3Im (((f-ig) (f + + ig)) ^ 2$$
But also
$$ff & # 39; & # 39; + gg & # 39; & gt; = Re ((f-ig) (f & + + ig &))) = 3 frac {Re Bigl ((f-ig) (f & # 39; + ig & # 39;) ^ 2 Bigr)} {f ^ 2 + g ^ 2}$$
If we insert the above differential equation of first order (*), we get the identity
$$– operatorname {Im} (C (f ^ 2 + g ^ 2) (f + ig) ^ 2) ^ 2 = frac { operatorname {Re} Bigl ((C (f ^ 2 + g ^ 2) (f + ig) ^ 2) ^ 2 Bigr}} {f ^ 2 + g ^ 2} \ iff \ – (f ^ 2 + g ^ 2) operatorname {Im} (C (f + ig) ^ 2) ^ 2 = operatorname {Re} (C ^ 2 (f + ig) ^ 4) = Re (C ( f + ig) ^ 2) ^ 2-Im (C (f + ig) ^ 2) ^ 2 \ iff \ (1- | Cu-D | ^ {- 1}) operatorname {Im} (C (Cu-D)) ^ 2 = operatorname {R} (C (Cu-D)) ^ 2$$
The solution says that the last should be impossible $$C neq 0$$, Can someone explain why this is impossible? It is clear if $$C$$ is real, but I think it can be complex. features $$f$$ and $$g$$ are real functions of the same real parameter.

## turing machines – A condition for \$ emptyset neq S subset RE \$, under the \$ L_S notin RE \$

I read some lecture notes on computational theory and after citing and proving the sentence: $$emptyset in S Rightarrow L_S = { langle M rangle: L (M) in S } notin RE$$ it stands that $$emptyset in S$$ is not a sufficient condition, i. $$L_S notin RE$$ does not give way $$emptyset in S$$by giving the counterexample $$L _ { Sigma ^ *} notin RE$$ , However, it means that there is a necessary and sufficient condition under which $$L_S notin RE$$, I searched for this condition in Sipser's book but did not find it. I would be very happy to receive a reference for this condition.

Edit: Given the answer from @dkaeae, I would like to know what the stronger feature is that can be derived from a non-trivial one $$S subset RE$$ in the case $$L_S notin RE$$,

## Field theory – show that \$ mathbb {Q} ( sqrt[n]{2}) neq mathbb {Q} ( sqrt[n]{3}) \$

I want to show that $$mathbb {Q} ( sqrt[n]{2}) neq mathbb {Q} ( sqrt[n]{3})$$ for a even $$n$$,

I was advised that I should use the following fact (which I have already proved):

if $$L / mathbb {Q}$$ finite field extension and $$A = L cap bar { mathbb {Z}}$$ (from where $$bar { mathbb {Z}}$$ is the algebraic integer ring or $$O_L$$) and $$B subseteq A$$ Side ring with $$Frac (B) = L$$ ($$Frac (B)$$ is the fraction field), then $$n ^ 2 cdot Delta_ {A / mathbb {Q}} = Delta_ {B / mathbb {Q}}$$ from where $$Delta_ {A / mathbb {Q}}$$ is the discriminant of $$A / mathbb {Q}$$ and $$n in mathbb {Z} _ {> 0}$$,

## Homology Cohomology – If \$ textrm {Tor} _n ^ A (-, A / radA) neq 0 \$? (\$ A \$ a finite-dimensional \$ K \$ algebra)

This question arises from a proof in paper form: Unlimited derived categories and finite dimension presumption – Jeremy Rickard, more precisely theorem 4.3.
The question is:

To let $$A$$ Be a finite dimensional algebra over a field $$K$$ and $$M$$ a right $$A$$Module with projective dimension $$d$$, So let it go $$P ^ { bullet}$$ Let be the minimum projective resolution of $$M$$ considered as complex. Then
$$textrm {Tor} _d ^ A (M, A / radA) neq 0$$
This is, $$P ^ { bullet}[-d] otimes_A (A / (radA))$$ has a non-zero cohomology in degrees zero. ($$P ^ { bullet}[-d]$$ the complex is moved $$d$$ times to the right)

I am thankful for every help.

## Linear Algebra – Exercise with \$ GL (V) \$, which shows that the group for \$ dim V geq 2 \$ and \$ F neq {0,1 } \$ is not commutative

In my lecture scripts we introduced the general linear group. We already know that $$L (V, V) = {f: V rightarrow V | f text {is linear} }$$ is a ring with unity and therefore $$GL (V) = {x in L (V, V) | x text {is invertible} }$$ is a group. I do not understand the proof of the above statement in the script:

I do not understand why $$A_2 ^ {- 1} = A_2$$, there $$A_2 (A_2 (v_1)) = A_2 (-v_1)$$

What I do not understand here is the definition of $$A_2$$ we have that $$v_1 neq-v_1$$ and therefore $$A_2 (-v_1) = – v_1$$, On the other hand, a linear map is uniquely determined by the values โโof the basis vectors and thus $$A_2$$ has to be linear and that means $$A_2 (-v_1) = – (A_2 (v_1)) = v_1$$,

But then we have $$v_1 = -v_1$$, This is not the case, for example $$V = mathbb {R} ^ 2$$

Can someone tell me where I am wrong?

## linear algebra – Show \$ || A || _2 = mathrm {sup} _ {x neq 0} frac {x ^ T A x} {x ^ T x} \$ where \$ A \$ is symmetric and positive definite

### problem

Show:
$$|| A || _2 = mathrm {sup} _ {0 neq x in mathbb {R}} frac {x ^ T A x} {x ^ T x}$$
from where $$A$$ : symmetrical and positive.

## To attempt

Since

begin {align} || A || _2 & = mathrm {sup} _ {0 neq x in mathbb {R}} frac {|| A x || _2} {|| x || _2} \ & = mathrm {sup} _ {0 neq x in mathbb {R}} frac {x ^ T A ^ T A x} {x ^ T x} end

I think the problem boils down to showing

$$mathrm {sup} _ {x neq0} x ^ T A x = mathrm {sup} _ {x neq0} x ^ T A ^ T A x$$

where I am stuck

Every help is appreciated.

## formal languages โโ- Construct a decidable set \$ B \$ such that \$ B neq A_w \$ stands for every \$ w in Sigma ^ star \$

I have been holding this problem for some time. All references would be grateful!

To let $$A subseteq Sigma ^ star$$ be decidable Given $$w in Sigma ^ star$$, define $$A_w = {x in Sigma ^ star : | : langle x, w rangle in A }.$$
Construct a decidable set $$B$$ so that $$B neq A_w$$ for each $$w in Sigma ^ star$$,

## Means \$ | (a, b) | = 1 (a, b) neq 0 \$?

So $$(a, b) in mathbb {R ^ 2}$$ and if $$sqrt {a ^ 2 + b ^ 2} = 1$$ Does that mean the vector $$(a, b) neq (0,0)$$?

Is this statement true in general?

$$(a, b) = (0,0) iff sqrt {a ^ 2 + b ^ 2} = 0?$$