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Linear Algebra Quick True/False Watch

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    (c) To prove that A<G, I can say that:

    A (2n+1) \times (2n+1) matrix is invertible if and only if it has non-zero determinant so A \subset G .

    Furthermore, A is non-empty since I_{2n+1} \in A since \text{det}(I_{2n+1})=1 .

    Let C,D \in A. Then \text{det}(C)=\text{det}(D)=1 . Now by the properties of determinants,

    \displaystyle \text{det}(CD^{-1}) = \text{det}(C)\text{det}(D^{-1}) =  \frac{\text{det}(C)}{\text{det}(  D)} = \frac{1}{1} = 1 .

    So CD^{-1} \in A and A<G.

    Now suppose P \in G and Q \in A

    Then \text{det}(PQP^{-1}) = \text{det}(P)\text{det}(Q)\text{  det}(P^{-1}) =  \text{det}(P)\text{det}(Q)\text{  det}(P)^{-1} = \text{det}(Q) = 1 .

    Therefore A \triangleleft G .

    Now B \neq \emptyset since I \in B (set u=1) .

    If U = \begin{bmatrix}u &  0  & \ldots & 0 \\ 0  &  u & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0  & 0 &\ldots & u\end{bmatrix} \in B and if V = \begin{bmatrix}v &  0  & \ldots & 0 \\ 0  &  v & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0  & 0 &\ldots & v\end{bmatrix} \in B

    Then UV^{-1} = \begin{bmatrix}uv^{-1} &  0  & \ldots & 0 \\ 0  &  uv^{-1} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0  & 0 &\ldots & uv^{-1}\end{bmatrix} \in B

    so that B < G .

    If P\in G is a (2n+1) \times (2n+1) matrix of the same size of U with arbitrary real entries

    i.e. P=\left[\begin{array}{cccc}

p_{1,1} & p_{1,2} & \hdots & p_{1,2n+1}\\

p_{2,1} & p_{2,2} & \hdots & p_{2,2n+1}\\

\vdots & \vdots & \ddots & \vdots\\

p_{2n+1,1} & p_{2n+1,2} & \hdots & p_{2n+1,2n+1}

\end{array}\right] for p_{i,j} \in \mathbb{R} for all 1 \leq i,j \leq 2n+1

    then UP=PU and it follows that PUP^{-1} = U (P\in GL_{2n+1}(\mathbb{R}) and hence having non-zero determinant guarantees it is invertible). Hence B \triangleleft G .

    Now since \text{det}(U) = u^{2n+1} and u\neq 0 it follows that if U \in A then u=1 .

    Therefore A \cap B = \{1\} .

    Let \text{det}(P) = r \neq 0 .

    Then \displaystyle P = \left[\begin{array}{cccc}

 \frac{p_{1,1}}{\sqrt[2n+1]{r}} & \frac{p_{1,2}}{\sqrt[2n+1]{r}} & \hdots & \frac{p_{1,2n+1}}{\sqrt[2n+1]{r}}\\

\frac{p_{2,1}}{\sqrt[2n+1]{r}} & \frac{p_{2,2}}{\sqrt[2n+1]{r}} & \hdots & \frac{p_{2,2n+1}}{\sqrt[2n+1]{r}}\\

\vdots & \vdots & \ddots & \vdots\\

\frac{p_{2n+1,1}}{\sqrt[2n+1]{r}} & \frac{p_{2n+1,2}}{\sqrt[2n+1]{r}} & \hdots & \frac{p_{2n+1,2n+1}}{\sqrt[2n+1]{r}}

\end{array}\right] \begin{bmatrix}\sqrt[2n+1]{r} &  0  & \ldots & 0 \\ 0  &  \sqrt[2n+1]{r} & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots\\0  & 0 &\ldots & \sqrt[2n+1]{r} \end{bmatrix} = QU (say)

    Note that the possibility that r<0 doesn't hurt since 2n+1 is odd for all n and consequently the 2n+1-th root is always real.

    We have Q \in A for \displaystyle \text{det}(Q) = \frac{\text{det}(P)}{(\sqrt[2n+1]{r})^{2n+1}} = \frac{r}{r} = 1 , and U \in B.

    Therefore G=AB and G is the internal direct product of A and B

    -------------------------------------------------------------------------------------------------------------------------------------

    Substituting \displaystyle y_1 = a_0 \sum_{n=0}^{\infty} \frac{1}{\prod_{i=0}^{n-1}( i - 1/2)( 2i + 2)} x^{ n - 1/2 }

    \displaystyle y_2 = a_0\sum_{n=0}^{\infty}  \frac{1}{\prod_{i=0}^{n-1}(i + 1)(2i + 5)} x^{ n + 1 }

    \displaystyle 2x^2 \sum_{n=0}^{\infty} (\sigma + n -1)(\sigma +n)a_nx^{n+\sigma -2}+ x \sum_{n=0}^{\infty} (n +\sigma) a_n x^{n+\sigma -1} - x \sum_{n=0}^{\infty} a_n x^{n + \sigma}

    \displaystyle - \sum_{n=0}^{\infty} a_n x^{n+\sigma} = 0

    \displaystyle \sum_{n=0}^{\infty} [2(\sigma +n-1)(\sigma +n) + \sigma + n -1 ]a_n x^{n+\sigma} - \sum_{n=0}^{\infty} a_n x^{n+\sigma + 1} = 0

    Setting n:=n+1 in the first sum

    \displaystyle \Rightarrow \sum_{n=-1}^{\infty} [2(\sigma + n)(\sigma +n+1)+\sigma + n]a_{n+1} x^{n+\sigma +1}

    \displaystyle ... = [2\sigma (\sigma -1)+\sigma -1]a_0 x^{\sigma} + \sum_{n=0}^{\infty} [2(\sigma +n)(\sigma +n+1)+\sigma +n]a_{n+1} x^{n+\sigma +1}

    Combining both sums:

    \displaystyle [2\sigma(\sigma - 1) +\sigma -1 ]a_0 x^{\sigma} + \sum_{n=0}^{\infty} \left[ (2(\sigma + n)(\sigma + n+1) + \sigma + n ) a_{n+1} - a_n \right] x^{n+\sigma + 1} = 0.

    Equating the series to 0 term-by-term gives the indicial equation

    2\sigma (\sigma -1) + \sigma -1 = 0 \Rightarrow (2\sigma +1)(\sigma -1) = 0 \Rightarrow \sigma = -\frac{1}{2},\; \sigma = 1.

    We also get the recurrence relation

    \displaystyle a_{n+1} = \frac{a_n}{2(\sigma + n)(\sigma + n +1) + \sigma + n} = \frac{a_n}{(\sigma + i)(2 \sigma + 2i + 3)}

    In general,

    \displaystyle a_n = a_0 \prod_{i=0}^{n-1} \frac{1}{2(\sigma+i)(\sigma+i+1) + \sigma+i} = a_0 \prod_{i=0}^{n-1} \frac{1}{(\sigma + i)(2 \sigma + 2i + 3)}

    which give 2 independent solutions for \sigma = -\frac{1}{2},\; \sigma = 1.

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    \displaystyle k_2 = f(x,y) + \frac{\partial f}{\partial (x,y)} \begin{pmatrix} \frac{1}{2}h \\ \frac{1}{2}hf(x,y) \end{pmatrix} + R

    k_1

    k_2

    t= w^{-2}

    \displaystyle \frac{dt}{dw} = -2w^{-3}

    \displaystyle dw = -\frac{dt}{2w^{-3}}

    So the limits of the integral change from \int^{\infty}_1 to \int^0_1 = -\int^1_0

    \displaystyle \int^{\infty}_1 (w^2 -1)^{\frac{p-1}{2}} w^{-q} \; dw = \int^1_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{q}{2}} \frac{dt}{2w^{-3}}

    \displaystyle = \frac{1}{2} \int^1_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt

    we have that t^{-1} - 1 = \frac{1-t}{t}.

    Hence we have \displaystyle  \frac{1}{2} \int^1_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt = \frac{1}{2} \int^1_0 \left( \frac{1-t}{t} \right) ^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt

    \displaystyle = \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} \frac{ t^{\frac{1}{2} (q-3)} }{t^{ \frac{p-1}{2} }} \;dt

    \displaystyle = \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} t^{\frac{q-p}{2}-1} \;dt

    \displaystyle = \frac{1}{2} \int^1_0 (1-t)^{\frac{p+1}{2} - 1} t^{\frac{q-p}{2}-1} \;dt

    \displaystyle \int^{\infty}_0 \frac{\sinh^{p}x}{\cosh^{q}x} \;dx = \frac{1}{2} B \left (\frac{q-p}{2} , \frac{p+1}{2} \right) = \frac{1}{2} B \left (\frac{p+1}{2} , \frac{q-p}{2} \right)

    \displaystyle -\frac{1}{2} \int^1_0 (1-t)^{\frac{q-p}{2}-1} t^{\frac{p+1}{2} - 1} \;dt

    \displaystyle \delta ^{m+1} [x,y] = \sum^{m+1}_{k=0} \binom{m+1}{k} [ \delta ^k x , \delta ^{m-k+1} y]

    \text{orb}_G \left (  \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right ) = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right \}

    (\mathfrak{sl}_3 \mathbb{C}) ^\prime \equiv [\mathfrak{sl}_3 \mathbb{C} , \mathfrak{sl}_3 \mathbb{C} ] = \mathfrak{sl}_3 \mathbb{C}

    X,Y \in \mathfrak{sl}_3 \mathbb{C}

    \displaystyle \binom{m}{k-1} + \binom{m}{k} = \frac{m!}{(k-1)! (m-k+1)!} + \frac{m!}{k! (m-k)!}

    \displaystyle = \frac{(k)m!}{k! (m-k+1)!} + \frac{(m-k+1) m!}{k!(m-k+1)!}

    \displaystyle = \frac{(m+1)!}{k! (m-k+1)!}

    \displaystyle = \binom{m+1}{k}

    \displaystyle \delta ^{m+1} [x,y] = [x , \delta ^{m+1} y] + \sum^m_{k=1} \binom{m+1}{k} [ \delta ^k x , \delta ^{m-k+1} y] + [\delta ^{m+1} x , y]

    \displaystyle 



[tex][X,Y] = XY-YX

    |X|=16

    \frac{\pi}{2}

    \pi

    \frac{3\pi}{2}

    G = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} , \begin{bmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} , \begin{bmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix} , \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} , \begin{bmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} \end{bmatrix} , \begin{bmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{bmatrix} \right \}

    \text{stab}_G \left ( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right ) = \left \{ g\in G : g\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right \} = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right \}
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    Where have you said what A, B and G are?

    Whatever G is, it looks like it might be Abelian so there's no need to check that subgroups are normal.
 
 
 
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