# Transpose theorem proof

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

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

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

PS: A and B are matrices.
• March 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 $(ab)^t=a^t b^t$? Are $a,b$ supposed to be matrices? If so, your formula is false, in general. You really want $(ab)^t=b^ta^t$. Also, what are $Aaxb$ and $Bbxc$?
• March 23rd 2011, 12:34 PM
Capes
Ahh, sorry about that, I meant $(A*B)^t = B^T * A^T$, yes, where both of them are matrices. $Aaxb$ means a general matrix A with a rows and b columns.
• March 23rd 2011, 02:13 PM
TheEmptySet
Let $\mathbf{r}_i$ be the ith row of $A$.
Let $\mathbf{c}_j$ be the jth column of $B$.

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

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

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

$B^TA^T$

Now this exchanges the rows and columns of A and B

So

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

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

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

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