Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given $B^T=B$ and if


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 $X$

share|improve this question
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
show 1 more comment

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 $X^t=Y^t$

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.