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Given $B^T=B$ and if

$$A^{T}BA=0,$$

with $B\in{\rm{M}}_{2\times2}(\mathbb{C})$ and $A\in{\rm{M}}_{2\times2}(\mathbb{R})$

what values may $B$ take to satisfy this equation?

I think $B=0$ is one solution, any others?


more questions: just yes/ no answer is okay for these :)

if $ABC=0$ then does $(ABC)^t=0^t=0$

where $X^t$ is the transpose of $X$

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Please do not write multiple questions in one question. –  Phira Nov 1 '12 at 12:43
    
$\mathbb{C}^2$ is the set of all $2 \times 1$ column vectors with entries in $\mathbb{C}$. Do you mean $B$ is a $2 \times 2$ matrix with entries in $\mathbb{C}$? –  littleO Nov 1 '12 at 12:43
    
What is the meaning of $B^t=B$ for vectors in $\mathbb{C}^2$? –  Dennis Gulko Nov 1 '12 at 12:45
    
@littleO, yes sorry for the confusion B is a 2 by 2 matrix; I meant B may have complex entries,real or imaginary or both. –  laurie Nov 1 '12 at 12:48
    
@DDennis Gulko: B is a 2 by 2 matrix satisfying the condition $B^t=B$, where entries of B may take complex values. I'm somewhat unfamiliar with much mathematical jargon still... –  laurie Nov 1 '12 at 12:50
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1 Answer

up vote 1 down vote accepted

For the first part: If you mean matrices $2\times 2$, then you should just play with those: denote $$A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right),\hspace{10pt}B=\left(\begin{array}{cc}x&y\\y&z\end{array}\right),\hspace{10pt}$$ Calculate $A^tBA$ directly and compare to 0.
For the second part: if $X=Y$ then $X^t=Y^t$

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right thanks! ill get working on the solution... also does the condition $B^t=B$ add to finding the solution in anyway? –  laurie Nov 1 '12 at 12:53
    
ah never mind... I re-read your comment regarding B. –  laurie Nov 1 '12 at 12:54
    
Yes, it does: observe that in $B$ I denoted the entries by $x,y,z$ - three parameters instead of four, since if $B^t=B$ the of-diagonal entries are equal. –  Dennis Gulko Nov 1 '12 at 12:55
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