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I'm having a bit of trouble with the following question.

Suppose $A$ is a square matrix.

a) Show that the matrix $B = A+A^T$ is symmetric.

Not sure how to do this. But here is my attempt. Well, let $A$ have a size of $n\times n$. That means $A^T$ must also have a size of $n$ by $n$ by the transpose property number of columns and rows are swapped. Therefore $B$ must also have a size $n$ by $n$. That's all I have at the moment.

b) Show that $C=AA^T$ is symmetric

I also have approached this like a) But not sure.

c) A matrix $M$ has a property that $M^T = - M$ (skew symmetry). Show that $D = (A - A^T)$ is a skew symmetric matrix.

d) Can you show how to write any square matrix as the sum of a symmetry and skew symmetric matrix.

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3 Answers 3

$$B^T = (A+A^T)^T=A^T+(A^T)^T=A+A^T=B$$$$C^T=(AA^T)^T=(A^T)^TA^T=AA^T$$$$D ^T= (A - A^T)^T=A^T-(A^T)^T=A^T-(-A)^T=-A+A^T=-D$$$$\forall B :B=\frac{B+B^T}{2}+\frac{B-B^T}{2}$$

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Thanks though, but I like to have an attempt. I was looking for a hint. –  Bobby Jun 16 '13 at 15:35
    
No worries thanks :-) –  Bobby Jun 16 '13 at 15:52
    
$\Huge\color{red}{+}$ –  Software Jun 17 '13 at 7:51

Hints: use the definitions.

  1. What does it mean for $B$ to be symmetric? It means $B^T=B$. Prove it.
  2. Similar to 1.
  3. Similar to 1.
  4. Given a square matrix $\Omega$, note that $\Omega=\dfrac{1}{2}\left(\Omega+\Omega^T+\Omega-\Omega^T\right)$. Try to extract the relevant matrices from this equality. (Use 1. and 3.)
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+1 for not giving the answer outright –  psoft Jun 16 '13 at 15:34
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Thanks for the hint. I'll give it a crack now. –  Bobby Jun 16 '13 at 15:34
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1. $B^T = (A+A^T)^T$. Note I used this: $(\lambda A + \mu A^T)^T = (\lambda A^T + \mu (A^T)^T).$ Now the transpose of that a transpose is just the matrix itself. $= A^T + A.$ By the commutative law of matrix addition. We arrive $B^T = A + A^T = B.$ Hence symmetry occurs. b) $C^T = (AA^T)^T=(A^T)^TA^T=AA^T=C.$ Is this right so far? –  Bobby Jun 16 '13 at 15:42
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$D^T=(A-A^T)^T=A^T-A=-(A-A^T)=-D.$ Hence skew symmetry. –  Bobby Jun 16 '13 at 15:45
    
@Bobby Nice job. –  Git Gud Jun 16 '13 at 15:47

$$B^T=(A+A^T)^T=A^T+A=B$$ $$C^T=(AA^T)^T=(A^T)^TA^T=AA^T=C$$ Also, $D^T=-D^T$ as previous equalities. Finally, $$A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$$.

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$A=A_1+A_2$, and as in previous inequalities, it is easy to check that $A_1$ is symmetric and $A_2$ is skew-symmetric. –  alans Jun 16 '13 at 15:39
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Thanks guys. Please I just want to have a crack for myself! :-) I just wanted a hint. Haha./ –  Bobby Jun 16 '13 at 15:44

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