For any square matrix $A$ and positive number $N$, let:
$A^N=AAA...A $ ($n$ factors)
How do I prove that:
$(A^N)^T = (A^T)^N$
Need help proving one property of matrix transpose
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$\begingroup$
linear-algebra
matrices
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0Do you know that $(BC)^T = C^T B^T$? If so, try letting $B=A$ and $C = A^{N-1} $ and see if you can catch the pattern. If not, first prove $(BC)^T = C^TB^T$ (perhaps by using the definition of matrix multiplication $(BC)_{ik} = \sum_J B_{ij}C{jk}$) – 2017-02-02
2 Answers
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Generally $(AB)^T = B^T A^T$. Can you work it out from there?
Edit:
It then logically follows that:
$$(A^2)^T = (AA)^T = A^T A^T = (A^T)^2$$
$$(A^3)^T = (AAA)^T = A^TA^T A^T = (A^T)^3$$
etc.
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0I demonstrated this part but I am stuck using this property to prove the next one. – 2017-02-02
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Are you asking how to prove $(A^{N})^{T}=(A^{T})^{N}$ from $(AB)^{T}=B^{T}A^{T}$ or how to prove $(AB)^{T}=B^{T}A^{T}$?
The first is clear. The second one, since your $A$ is square, you need to prove $(AA)^{T}=A^{T}A^{T}$. Take $[ij]$ entry in $AA$. It is $\sum_{k}a_{ik}a_{kj}$. This will be $[ji]$ entry in $(AA)^{T}$. What is $[ji]$ entry in $A^{T}A^{T}$? It is, denoting elements of $A^T$ by $a_{ij}'$, $\sum_{k}a'_{jk}a'_{ki}$. Since $a_{ij}=a'_{ji}$, the sum equals $\sum_{k}a_{kj}a_{ik}$.