Proof of Transpose Property of Matrix

**razorfever** · Jan 18th 2009, 11:54 AM

prove (AB)^T = (B^T)*(A^T)
using summation notation

i know how to prove it without summation notation but with summation notation i'm not getting anywhere
if someone can prove it and then explain the summation steps involved that would be of great help

**Constatine11** · Jan 18th 2009, 10:16 PM

Originally Posted by razorfever

prove (AB)^T = (B^T)*(A^T)
using summation notation

i know how to prove it without summation notation but with summation notation i'm not getting anywhere
if someone can prove it and then explain the summation steps involved that would be of great help

$(AB)^T_{i,j}=(AB)_{j,i}=\sum_k A_{j,k}B_{k,i}=\sum_k B_{k,i}A_{j,k}=\sum_k B^T_{i,k}A^T_{k,j}=(B^TA^T)_{i,j}$

.

**Isomorphism** · Jan 18th 2009, 11:38 PM

If transpose was defined to you as $[A^T]_{ij} = [A]_{ji}$ , then "summation sign" method as explained in the prev. post is most likely the only thing you could do.

Ignore whatever's below, I replied in a haste.
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However if you are doing linear algebra, and you are familiar with the notion of a linear functional $[x,y]$ (Halmos' notation), then one could define $A'$ , adjoint of linear transformation A as $[Ax,y] = [x,A'y]$ .If $[A]_{\cal{X}}$ is the matrix of linear transformation A w.r.t the basis $\cal{X}$ , then its easy to verify that the matrix of linear transformation A' is $[A]^T_{\cal{X}}$ .

Now with all this setup, its extremely easy to prove $(AB)^T = B^TA^T$ . Note that it suffices to prove $(AB)' = B'A'$ .

But $\forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]$

Thus $(AB)' = B'A'$

**Constatine11** · Jan 19th 2009, 03:52 AM

Originally Posted by Isomorphism

If transpose was defined to you as $[A^T]_{ij} = [A]_{ji}$ , then "summation sign" method as explained in the prev. post is most likely the only thing you could do.

However if you are doing linear algebra, and you are familiar with the notion of a linear functional $[x,y]$ (Halmos' notation), then one could define $A'$ , adjoint of linear transformation A as $[Ax,y] = [x,A'y]$ .If $[A]_{\cal{X}}$ is the matrix of linear transformation A w.r.t the basis $\cal{X}$ , then its easy to verify that the matrix of linear transformation A' is $[A]^T_{\cal{X}}$ .

Now with all this setup, its extremely easy to prove $(AB)^T = B^TA^T$ . Note that it suffices to prove $(AB)' = B'A'$ .

But $\forall x,y : [x,(AB)'y] = [ABx,y] = [A(Bx),y] = [Bx, A'y] = [x, B'A'y]$

Thus $(AB)' = B'A'$

The question explicity asks for a demonstration using summation notation.

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