$\begingroup$If $A$ is a square matrix and $n$ a positive integer, is it true that $(A^n)^T = (A^T)^n$? Justify your answer.
Here, I referred to the list of properties for a transpose and there's one that says $(A^T)^{-1}$ is the same as $(A^{-1})^T$. Is that all I need to be able to prove this?
I have to write this as a proof.
$\endgroup$32 Answers
$\begingroup$The result holds for any positive integer $n$. Prove this by induction. This is true for $n=1$, as $(A^n)^T=A^T=(A^T)^n$. Suppose the result holds for some positive integer $n$. Then $(A^{n+1})^T=(A^n\cdot A)^T=A^T\cdot (A^T)^n=(A^T)^{n+1}$ and so the result holds for $n+1$. By induction conclude that the result is true for any positive integer $n$.
Note: $(A^T)^{-1}=(A^{-1})^T$ holds if $A$ is invertible.
$\endgroup$$\begingroup$A property of transposition is that for equal size square matrices, $A$ and $B$, $$ (A\cdot B)^T=B^T\cdot A^T. $$ This can be proven by looking at the matrix elements: $$ (AB)_{ij}=\sum_k a_{ik}b_{kj} .$$ Then $$ (AB)^T_{ij}=(AB)_{ji}=\sum_k a_{jk}b_{ki}=\sum_kb_{ki}a_{jk}=\sum_k B^T_{ik}A^T_{kj}=(B^TA^T)_{ij}. $$
Now then, from this comes that $$ (AA)^T=A^TA^T, $$ and from this, using the associativity of matrix product, your statement is proven.
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