# Thread: Properties of Transpose Proof!

1. ## Properties of Transpose Proof!

does anyone maybe have a link to prove the properties of the transpose?
Transpose - Wikipedia, the free encyclopedia

My teacher said we should know how to prove these...
In the website link, # 1 , 2 , 3 , 4 , and (A^r)^T = (A^T)^r for all nonnegative integers.

2. Originally Posted by aeubz
does anyone maybe have a link to prove the properties of the transpose?
Transpose - Wikipedia, the free encyclopedia

My teacher said we should know how to prove these...
In the website link, # 1 , 2 , 3 , 4 , and (A^r)^T = (A^T)^r for all nonnegative integers.
Hello

Some of these proofs aren't that bad

$\displaystyle \left(A^T\right)^T=A$

Let $\displaystyle A=\left[a_{ij}\right]$, $\displaystyle A^T=\left[a_{ji}\right]$ and $\displaystyle A^T=C=\left[c_{ij}\right]$

Thus, $\displaystyle c_{ij}^T=c_{ji}=a_{ji}^T=a_{ij}$

$\displaystyle \mathbb{Q.E.D.}$

-------------------------------------------------------------------------

$\displaystyle \left(A+B\right)^T=A^T+B^T$

Let $\displaystyle A=\left[a_{ij}\right]$, $\displaystyle B=\left[b_{ij}\right]$ and $\displaystyle A+B=C=\left[c_{ij}\right]$

Thus, $\displaystyle c_{ij}^T=c_{ji}=a_{ji}+b_{ji}=a_{ij}^T+b_{ij}^T$

$\displaystyle \mathbb{Q.E.D.}$

-------------------------------------------------------------------------

$\displaystyle \left(cA\right)^T=cA^T$

Let $\displaystyle A=\left[a_{ij}\right]$ and $\displaystyle cA=C=\left[c_{ij}\right]$

Thus, $\displaystyle c_{ij}^T=c_{ji}=ca_{ji}=ca_{ij}^T$

$\displaystyle \mathbb{Q.E.D.}$

-------------------------------------------------------------------------

Maybe other can elaborate on these...because I think I made it too short and sweet

--Chris

3. Originally Posted by aeubz
(A^r)^T = (A^T)^r for all nonnegative integers.
You need to prove $\displaystyle (AB)^T = B^T A^T$.

Then by induction $\displaystyle (A_1...A_k)^T = A_k^T ... A_1^T$.

Therefore, $\displaystyle (A^r)^T = (A...A)^T = (A^T) ... (A^T) = (A^T)^r$.

4. Originally Posted by ThePerfectHacker
Then by induction $\displaystyle (A_1...A_k)^T = A_k^T ... A_1^T$.
Hm, what if I need to prove this? By induction I have to assume this is true and then try to prove it's true for $\displaystyle A_n$ where $\displaystyle n=k+1$? I have to do it right now....I thought I would need to treat $\displaystyle (A_1...A_k)$ as a single matrix, correct? I either have to prove it this way or prove it for three, assuming it's true for two. But I wanted to use mathematical induction. The proof for two is given in the book, so I kinda extended the idea, treating this whole thing as a single $\displaystyle (A_1...A_k)$ matrix. Or there is a different way to prove it?