Dan's Thoughts » qmismf http://danboykis.com Thinking somewhat carefully Tue, 18 Jan 2011 00:46:24 +0000 en hourly 1 http://wordpress.org/?v=3.1.4 QMISMF Chap 3 http://danboykis.com/2009/10/qmismf-chap-3/ http://danboykis.com/2009/10/qmismf-chap-3/#comments Wed, 21 Oct 2009 16:13:04 +0000 dan http://danboykis.com/?p=1989 3-1
\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}+\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}=\begin{pmatrix} 1 & -i \\ i & -1 \end{pmatrix}
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1-i \\ 1+i & 0 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+\frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}
3-2
\begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 3 \end{pmatrix} \begin{pmatrix} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 3 \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
3-3
\frac{1}{4}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}=\frac{1}{4}\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=\frac{1}{4}\begin{pmatrix} 2 & -2 \\ -2 & 2 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=\frac{1}{4}\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}=\frac{1}{4}\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
3-4
\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix} \begin{pmatrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{pmatrix} =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}
3-5
Assume: AM^{-1}=M^{-1}A \iff AM=MA
AM^{-1}=M^{-1}A \implies AM=MA
AM^{-1}=M^{-1}A
AM^{-1}M=M^{-1}AM
MAI=MM^{-1}AM
MA=IAM
MA=AM

AM=MA \implies AM^{-1}=M^{-1}A
AM=MA
M^{-1}AM=A
M^{-1}A=AM^{-1}
3-6
Assume: AM=MA and BM=MB
(A+B)M=M(A+B)
AM+BM=MA+MB
AM+BM-MA-MB=0
(AM-MA)+(BM-MB)=0
0+0=0 (AB)M=M(AB)
A(BM)=(MA)B
A(MB)=(AM)B
(AM)B=(AM)B (zB)M=M(zB)
z(BM)=(MB)z
z(BM)=(BM)z

3-7
\begin{pmatrix} 2 & 2-i \\ 2+i & -2 \end{pmatrix} \frac{1}{9}\begin{pmatrix} 2 & 2-i \\ 2+i & -2 \end{pmatrix}=\frac{1}{9}\begin{pmatrix} 9 & 0 \\ 0 & 9 \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

]]> http://danboykis.com/2009/10/qmismf-chap-3/feed/ 0 QMISMF: Chapter 2 http://danboykis.com/2009/10/qmismf-chapter-2/ http://danboykis.com/2009/10/qmismf-chapter-2/#comments Tue, 06 Oct 2009 00:13:02 +0000 dan http://danboykis.com/?p=1929 (+ 3-i 2+4i) 5+3i > (+ 1+3i 2) 3+3i > (- -5+2i 2+2i) -7 > (+ -2+i 2+2i) +3i > (* 3-i 2+4i) 10+10i > (* 1+3i 2) 2+6i > (* 0+i 1+3i) -3+i > (* -5+2i 2+3i) -16-11i > (* 2+3i -2+3i) -13 > (* 2+3i 3+2i) +13i 2-2: 2-3: 2-4: 2-5: [...]]]> 2-1:

> (+ 3-i 2+4i)
5+3i
> (+ 1+3i 2)
3+3i
> (- -5+2i 2+2i)
-7
> (+ -2+i 2+2i)
+3i
> (* 3-i 2+4i)
10+10i
> (* 1+3i 2)
2+6i
> (* 0+i 1+3i)
-3+i
> (* -5+2i 2+3i)
-16-11i
> (* 2+3i -2+3i)
-13
> (* 2+3i 3+2i)
+13i

2-2:

z^{-1}=\frac{1}{x^2+y^2}(x+iy)
\left(\frac{1}{i}\right)^{-1} \rightarrow 0-i=\frac{1}{1}(i)=i
i \cdot \frac{1}{i}= 1
\left(\frac{1}{-i}\right)^{-1} \rightarrow \frac{-i}{-1} = i = \frac{1}{1}(-i)=-i
-i \cdot \frac{1}{-i}= 1

2-3:

i^*i = -i \cdot i = 1
(-i)^*(-i)= i \cdot -i = 1
|i|^2=(i^*i)=1
|-i|^2=\left( (-i)^*(-i) \right)=1
|i| = \sqrt{i^*i} = \sqrt{-i \cdot i} = 1
|-i| = \sqrt{(-i)^*(-i)}= \sqrt{i \cdot -i} = 1

2-4:

z=3+4i
z^* = 3-4i
z^*z = (3-4i)(3+4i) = 25
|z| = \sqrt{z^*z} = \sqrt{25} = 5
z^2 = z \cdot z = (3+4i)(3+4i) = -7+24i
\frac{1}{z} = \frac{1}{z} \cdot \frac{z^*}{z^*} = \frac{z^*}{zz^*} = \frac{3-4i}{25}

2-5:

|w+z| < |w| + |z|
|3+4i| < |3| + |4i|
5 < 7

2-6:

\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2}}{2}} \cdot \left( \sqrt{1 + \frac{x}{\sqrt{x^2+y^2}}} + i\sqrt{1 - \frac{x}{\sqrt{x^2+y^2}}} \right)
for y \ge 0
\sqrt{z} = \sqrt{\frac{\sqrt{x^2+y^2}}{2}} \cdot \left( -\sqrt{1 + \frac{x}{\sqrt{x^2+y^2}}} + i\sqrt{1 - \frac{x}{\sqrt{x^2+y^2}}} \right)
for y \le 0 \sqrt{i} = \sqrt{0+1i} = \sqrt{\frac{\sqrt{0^2+1^2}}{2}} \cdot \left( \sqrt{1 + \frac{0}{\sqrt{0^2+1^2}}} + i\sqrt{1 - \frac{0}{\sqrt{0^2+1^2}}} \right)
\sqrt{i} = \sqrt{0+1i} = \sqrt{\frac{1}{2}} \cdot (1+i) = \sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}
\left(\sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}\right)^2 = \frac{1}{2} + i - \frac{1}{2} = i \sqrt{-i} = \sqrt{0-1i} = \sqrt{\frac{\sqrt{0^2+(-1)^2}}{2}} \cdot \left( -\sqrt{1 + \frac{0}{\sqrt{0^2+(-1)^2}}} + i\sqrt{1 - \frac{0}{\sqrt{0^2+(-1)^2}}} \right)
\sqrt{-i} = \sqrt{0-1i} = \sqrt{\frac{1}{2}} \cdot (-1+i) = -\sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}
\left(- \sqrt{\frac{1}{2}}+i \sqrt{\frac{1}{2}}\right) \cdot \left(- \sqrt{\frac{1}{2}}+i \sqrt{\frac{1}{2}}\right)=\frac{1}{2}-i-\frac{1}{2}=-i

2-7:

z=-\frac{b}{2a} \pm \sqrt{\left(\frac{b}{2a}\right)^2 - \frac{c}{a}}
z=-\frac{b}{2a} \pm \sqrt{\frac{b^2-4ac}{2^2a^2}}
z=-\frac{b}{2a} \pm \frac{1}{2a}\sqrt{b^2-4ac}
z=\frac{-b \pm \sqrt{b^2-4ac}}{2a} since the above is now in the usual form of the quadratic formula it is clear z is a solution.

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