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TsAmE
 Post subject: Matrix multiplication problem
PostPosted: Fri, 8 Oct 2010 22:45:25 UTC 
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Find all matrices X satisfying AX = XA, where A = \begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}.

\begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}\begin{pmatrix}x1 & y1\\ x2& y2\end{pmatrix} = \begin{pmatrix}x1 & y1\\ x2& y2\end{pmatrix}\begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}
\begin{pmatrix}x1 & y1\\ 2x1& 2y1\end{pmatrix}=\begin{pmatrix}x1 & 2y1\\ x2& 2y2\end{pmatrix}

I dont know what more to do


Last edited by TsAmE on Sat, 9 Oct 2010 14:53:19 UTC, edited 5 times in total.

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Matt
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PostPosted: Fri, 8 Oct 2010 23:00:00 UTC 
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Double check your work. The matrices weren't multiplied correctly.
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TsAmE
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PostPosted: Sat, 9 Oct 2010 13:46:10 UTC 
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I forgot the equals sign for the last step. I have edited the post. Otherwise I check my multiplication and didnt see anything wrong.
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outermeasure
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PostPosted: Sat, 9 Oct 2010 13:57:48 UTC 
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TsAmE wrote:
I forgot the equals sign for the last step. I have edited the post. Otherwise I check my multiplication and didnt see anything wrong.


Are you sure you copied A correctly?

TsAmE wrote:
Find all matrices X satisfying AX = XA, where A = \begin{pmatrix}1 & 2\\ 2& 0\end{pmatrix}.

\begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}\begin{pmatrix}x1 & y1\\ x2& y2\end{pmatrix} = \begin{pmatrix}x1 & y1\\ x2& y2\end{pmatrix}\begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}
\begin{pmatrix}x1 & y1\\ 2x1& 2y1\end{pmatrix}=\begin{pmatrix}x1 & 2y1\\ x2& 2y2\end{pmatrix}

I dont know what more to do

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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TsAmE
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PostPosted: Sat, 9 Oct 2010 14:56:12 UTC 
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Sorry made a mistake A is suppose to = \begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}.

I found my mistake. Here is my correction:

\begin{pmatrix}x1 &y1 \\ 2x1& 2y1\end{pmatrix} = \begin{pmatrix}x1 + 2y1 &0 \\ x2 + 2y2 & 0\end{pmatrix}

Im not sure what to do from here.
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helmut
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PostPosted: Sat, 9 Oct 2010 16:21:50 UTC 
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Two matrices are equal if their respective entries are equal. That gives you four equations...

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TsAmE
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PostPosted: Sat, 9 Oct 2010 19:20:50 UTC 
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helmut wrote:
Two matrices are equal if their respective entries are equal. That gives you four equations...


I then get the equations:

x1 = x1 + 2y1
y1 = 0
2x1 = x2 + 2y2
2y1 = 0

From this info I found that:

y1 = 0
x1 = x1
Cant solve for x2 :|
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Soroban
 Post subject: Re: Matrix multiplication problem
PostPosted: Sun, 10 Oct 2010 15:02:47 UTC 
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Hello, TsAmE!

Quote:
Find all matrices X satisfying AX = XA, where A \:=\:\begin{pmatrix}1 & 0\\ 2& 0\end{pmatrix}.

Let: .X \:=\:\begin{pmatrix}a&b\\c&d\end{pmatrix}

Then we have: .\begin{pmatrix}1&0\\2&0\end{pmatrix} \begin{pmatrix}a&b\\c&d\end{pmatrix} \;=\; \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\2&0\end{pmatrix}

. . . . . . . . . . . . . . . . \begin{pmatrix}a & b \\ 2a & 2b\end{pmatrix} \;=\;\begin{pmatrix}a+2b & 0 \\ c + 2d & 0 \end{pmatrix}


Hence, we have four equations: . \begin{array}{cccccccccc}a &=& a+2b & [1] && b&=& 0 & [2] \\ 2a &=& c+2d & [3] && 2b &=& 0 & [4] \end{array}


From [3] and [4] we have: .b \:=\: 0

. . . . . Then [1] becomes: .a \:=\: a

. . .And from [3] we have: .d \:=\:\dfrac{2a-c}{2}


Therefore: .X \;=\; \begin{pmatrix}a & 0 \\ c & \frac{2a-c}{2} \end{pmatrix} . for any values of a and c.
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