# Linear algebra proofs - traces, symmetricity and inversion

I have a few proofs I need some help with.

a) Prove that $AB-BA = I$ does not have any solutions for any $A,B$. All matrices are regular.

I based my proof on matrix traces. $tr(AB) = tr(BA)$. Since $tr(X+Y) = tr(X) + tr(Y)$, it holds that $tr(AB - BA) = tr(AB) - tr(BA) = 0$ and $tr(I) = m$ so the diagonal numbers cant be "önes". Is this proof correct or do I have to use some other method?

b) Prove that $(AB)^{-1} = B^{-1}A^{-1}$. I believe that I can prove this by simply writing all the matrices products down.. is there some simplier and more "elegant" way how to prove this?

c) Prove that $A + A^T$ is symetric for a square $A$. Not sure abotu this one...

Thanks for any help in advance!

EDIT: $A$ in c) is square, not rectangular!

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Your proof of (a) is correct and for (b), if you mean just computing $(B^{-1}A^{-1})(AB)$ and $(AB)(B^{-1}A^{-1})$, then that's the canonical proof. As for (c), what is the definition of a symmetric matrix? –  Branimir Ćaćić Jan 3 '13 at 10:31
a) and b) are good, for c), check first how transpose acts with sums. That is, how is $(A + B)^T$ related to $A^T$ and $B^T$. –  levap Jan 3 '13 at 10:32
Note that you don't need to compute both $(B^{-1}A^{-1})(AB)$ and $(AB)(B^{-1}A^{-1})$ if $A$ and $B$ are matrices (i.e. linear maps on a finite-dimensional vector space), as one of them will suffice in this case. –  fbg Jan 3 '13 at 10:54

a) The proof is fine

b) More "elegant" proof (to me) would be the argument "invertible matrices form a group with the group operation as matrix multiplication"

c) Are you sure $A$ is rectangular. Let us say $A$ is $2 \times 4$, so $A^T$ is $4 \times 2$, Now thinking about adding two matrices $A$ and $A^T$ which are of different size.

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edit! A is square... sorry for this –  Smajl Jan 3 '13 at 12:26

There are multiple other ways to do (a) as well.

For (b), you could reason straight from the definition of the inverse of a matrix as follows. Let $X=AB$ and $Y=B^{-1}A^{-1}$. Now $$XY=(AB)(B^{-1}A^{-1})=ABB^{-1}A^{-1}=AIA^{-1}=AA^{-1}=I,$$ and similarly $YX=I$. Therefore, $Y=X^{-1}$.

In (c), you must mean square rather than rectangular. If so, then $$(A+A^T)^T=A^T+(A^T)^T=A+A^T$$ and thus $A+A^T$ is symmetric.

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So a and b are correct. Thanks

As for c, I believe it goes like this

$A = \begin{bmatrix} a_{1,1} & \dots & a_{1,n} \\ \dots \\ a_{n,1} & \dots & a_{n,n} \end{bmatrix}$

$A^T = \begin{bmatrix} a_{1,1} & \dots & a_{n,1} \\ \dots \\ a_{1,n} & \dots & a_{n,n} \end{bmatrix}$

so $A + A^T = \begin{bmatrix} 2a_{1,1} & \dots & a_{1,n} +a_{n,1} \\ \dots \\ a_{n,1} + a_{1,n} & \dots & 2a_{n,n} \end{bmatrix}$ which is symetric.. correct?

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That's correct, but you don't even need to go into that detail, if you recall that $(A+B)^T = A^T + B^T$ and that $(A^T)^T = A$. –  Branimir Ćaćić Jan 3 '13 at 10:59
Perhaps you mean $A$ is square rather than rectangular. –  JohnD Jan 3 '13 at 12:23
Yes - my mistake.. I edited the post –  Smajl Jan 3 '13 at 12:27
Let C = A + $A^T$ then $C^T$ = C –  Ram Jan 3 '13 at 12:35