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I got $A^t = -A$ (A antisymetric) and B symetric: $B^t = B$. I need to know if $(A + B)^2$ is symetric. I couldn't find a formula which describe it. In addition, I know that A and B are non zero and with order of 3x3.

The best I could find to try and prove the symetry:

$$ [(A+B)^2]_i,_j = \sum_1^3(\alpha_i,_k+\beta_i,_k)(\alpha_k,_j+\beta_k,_j) $$ $$ [(A+B)^2]_j,_i = \sum_1^3(\alpha_j,_k+\beta_j,_k)(\alpha_k,_i+\beta_k,_i)= \sum_1^k(-\alpha_i,_k+\beta_i,_k)(-\alpha_j,_k+\beta_j,_k) $$ Both of them are almost identical beside the fact that one got -alphas and the other got +alphas. Thus, i think there could be matrices A and B which it will be true and a pair which it won't be true.

How can I solve this question? Or how can I find those pairs of matrices?

MyNick
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1 Answers1

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Note that $(A+B)^2 = A^2 + AB + BA + B^2$. Now note $(AB)^T = B^T A^T = -BA$ and $(BA)^T = A^T B^T = -A B$. $(A^2)^T = (A A)T = A^T A^T = (-A)(-A) = A^2$ and similarly $(B^2)^T = B^T B^T = B B = B^2$.

So, $( (A+B)^2))^T = (A^2 + AB + BA + B^2)^T = (A^2)^T + (AB)^T + (BA)^T + (B^2)^T = A^2 - BA - AB + B^2$.

So, $(A+B)^2$ is symmetric if and only if $A^2 + AB + BA + B^2 = A^2 - BA - AB + B^2$ or $AB +BA = 0$.

Now check this condition.

Batman
  • 19,390
  • First of all thank you very much. I know that if AB=BA then the only solution is if B=cI (or A=cI). But for AB=-BA there is nothing much i can say. I think there is no such matrix B which it is true for her. But how can i prove it? – MyNick Feb 02 '16 at 20:26
  • An antisymmetric matrix has zeros on its diagonal. – Batman Feb 03 '16 at 02:47
  • Thats why A can't be of the form cI. But B can be cI. it still doesn't true. – MyNick Feb 03 '16 at 04:59
  • As far as i can see, there is no such matrix B which the statment is true for her. But how can i prove it? – MyNick Feb 04 '16 at 12:29