# Transpose theorem proof

• Mar 23rd 2011, 11:48 AM
Capes
Transpose theorem proof
$\displaystyle (A*B)^t = B^t * A^t$

So, my problem is, how to proof this is true?

What I could do:
Lets Say $\displaystyle A axb$ and $\displaystyle B bxc$, so, in this case, $\displaystyle AB axc$, and obviously $\displaystyle AB^T cxa$. The only way of getting $\displaystyle AB^T cxa$ is multiplying $\displaystyle B^t cxb$ for $\displaystyle A^t bxa$, which would be false if we used $\displaystyle A^t bxa * B^t cxa$, since $\displaystyle a =/= c$. Even though this makes sense, is this a valid proof? And is there a cleaner one?

PS: A and B are matrices.
• Mar 23rd 2011, 12:19 PM
roninpro
To be honest, I have no idea what you are talking about, in the statement and your "proof". What do you mean by $\displaystyle (ab)^t=a^t b^t$? Are $\displaystyle a,b$ supposed to be matrices? If so, your formula is false, in general. You really want $\displaystyle (ab)^t=b^ta^t$. Also, what are $\displaystyle Aaxb$ and $\displaystyle Bbxc$?
• Mar 23rd 2011, 12:34 PM
Capes
Ahh, sorry about that, I meant $\displaystyle (A*B)^t = B^T * A^T$, yes, where both of them are matrices. $\displaystyle Aaxb$ means a general matrix A with a rows and b columns.
• Mar 23rd 2011, 02:13 PM
TheEmptySet
Let $\displaystyle \mathbf{r}_i$ be the ith row of $\displaystyle A$.
Let $\displaystyle \mathbf{c}_j$ be the jth column of $\displaystyle B$.

So the $\displaystyle d_{ij}$ i,jth entry $\displaystyle D$ of the product $\displaystyle AB=D$

$\displaystyle \mathbf{d}_{ij}=\mathbf{r}_i\cdot \mathbf{c}_j$

$\displaystyle (AB)^T=D^T$ Now lets compare with

$\displaystyle B^TA^T$

Now this exchanges the rows and columns of A and B

So

$\displaystyle \mathbf{d}_{ji}=\mathbf{c}_j\cdot \mathbf{r}_i$

Now what can we conclude?
• Mar 23rd 2011, 02:19 PM
roninpro
Quote:

Originally Posted by Capes
Ahh, sorry about that, I meant $\displaystyle (A*B)^t = B^T * A^T$, yes, where both of them are matrices. $\displaystyle Aaxb$ means a general matrix A with a rows and b columns.

No problem. You might want to try something like $\displaystyle A_{a\times b}$ (code: A_{a\times b}) next time.