I think this had something to do with the basis vectors e1, e2, e3 but I'm not sure how to go about it or if that was relevant.
If M is a mxn matrix, then prove that $(MN)^T=N^TM^T$
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linear-algebra
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1At what point exactly are you having trouble proving this? – 2017-02-08
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0Ok, I think I just figured it out. Thanks for the help. – 2017-02-08
2 Answers
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Hint: An efficient approach is as follows: begin by showing that $(Nx)^T = x^TN^T$ for any column vector $x$. It then follows that for the standard basis vectors $e_i$ and $e_j$, we have $$ e_i^T(MN)^Te_j = [(MN)e_i]^Te_j = [M(Ne_i)]^Te_j = (Ne_i)^TM^Te_j = e_i^T(N^TM^T)e_j $$ Thus, the matrices $MN$ and $N^TM^T$ have the same entries.
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Just look at the definition of matrix product and transpose.
If $A$ is $i$ by $j$ and $B$ is $j$ by $k$, then $(AB)_{u,v} =\sum_{w=1}^j A_{u,w}B{w,v} $.
Also, $A^T_{u,v} = A_{v,u} $.
Therefore, since $B^TA^T$ is $k$ by $i$ (because $B^T$ is $k$ by $j$ and $A^T$ is $j$ by $i$),
$\begin{array}\\ (B^TA^T)_{u,v} &=\sum_{w=1}^j B^T_{u,w}A^T_{w,v}\\ &=\sum_{w=1}^j B_{w,u}A_{v,w}\\ &=\sum_{w=1}^j A_{v,w}B_{w,u}\\ &\text{and}\\ (AB)^T_{u,v} &=(AB)_{v,u}\\ &=\sum_{w=1}^j A_{v,w}B{w,v}\\ &=(B^TA^T)_{u,v}\\ \end{array} $