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I was given the following problem on which I'm quite stuck at the moment.

Does the property $$(A^n)^T = (A^T)^n$$ hold for an arbitrary $m \times m$-matrix with $n \in \mathbb{Z}^+$?

I didn't have any intuition if this property actually holds so I started checking with an example $2 \times 2$-matrix.

$\begin{alignat*}{5}\left(\begin{bmatrix}1&1\\2&0\end{bmatrix}^2\right)^T &\rightarrow \left(\begin{bmatrix}1&1\\2&0\end{bmatrix} \times \begin{bmatrix}1&1\\2&0\end{bmatrix}\right)^T &&\rightarrow \left(\begin{bmatrix}3&1\\2&2\end{bmatrix}\right)^T &&&\rightarrow \begin{bmatrix}3&2\\1&2\end{bmatrix} \\ \left(\begin{bmatrix}1&1\\2&0\end{bmatrix}^T\right)^2 &\rightarrow \left(\begin{bmatrix}1&2\\1&0\end{bmatrix}\right)^2 &&\rightarrow \begin{bmatrix}1&2\\1&0\end{bmatrix} \times \begin{bmatrix}1&2\\1&0\end{bmatrix} &&&\rightarrow \begin{bmatrix}3&2\\1&2\end{bmatrix}\end{alignat*}$

These results led me to believe that this statement looks true on first sight. I started writing out the proof for it using both the definition of matrix multiplication and transpose.

$\begin{alignat*}{3}\left(\left(A^n\right)^T\right)_{i_{0}i_{n}} &= \left(\left(\prod_{i=1}^{n} A\right)^T\right)_{i_{0}i_{n}} = \left(\prod_{i=1}^{n} A\right)_{i_{n}i_{0}} &&= \sum_{i_{1}}^{m}\sum_{i_{2}}^{m}\cdots\sum_{i_{n-1}}^{m} A_{i_{0}i_{1}}A_{i_{1}i_{2}}A_{i_{2}i_{3}}\cdots A_{i_{n-2}i_{n-1}}A_{i_{n-1}i_{n}} \\ & &&=\sum_{i_{1},i_{2},\cdots ,i_{n-1}}^{m} A_{i_{0}i_{1}}A_{i_{1}i_{2}}A_{i_{2}i_{3}}\cdots A_{i_{n-2}i_{n-1}}A_{i_{n-1}i_{n}} \\ \left(\left(A^T\right)^n\right)_{i_{0}i_{n}} &= \left(\prod_{i=1}^{n} A^T\right)_{i_{0}i_{n}} &&= \sum_{i_{1}}^{m}\sum_{i_{2}}^{m}\cdots\sum_{i_{n-1}}^{m} A_{i_{0}i_{1}}^{T}A_{i_{1}i_{2}}^{T}A_{i_{2}i_{3}}^{T}\cdots A_{i_{n-2}i_{n-1}}^{T}A_{i_{n-1}i_{n}}^{T} \\ & &&= \sum_{i_{1}}^{m}\sum_{i_{2}}^{m}\cdots\sum_{i_{n-1}}^{m} A_{i_{1}i_{0}}A_{i_{2}i_{1}}A_{i_{3}i_{2}}\cdots A_{i_{n-1}i_{n-2}}A_{i_{n}i_{n-1}} \\ & &&=\sum_{i_{1},i_{2},\cdots ,i_{n-1}}^{m} A_{i_{1}i_{0}}A_{i_{2}i_{1}}A_{i_{3}i_{2}}\cdots A_{i_{n-1}i_{n-2}}A_{i_{n}i_{n-1}}\end{alignat*}$

However, this is where I'm stuck. Can one simply state that since all sums sum over the same range and the matrices don't differ, these two statements are equivalent?

Martin Sleziak
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Martijn
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2 Answers2

4

You are working too hard.

Use the property that $$ (AB)^T=B^TA^T $$ (in particular for $B=A$) and induction on $n$.

  • I need to start getting a good intuition on when I should use its definition or a simple property like this and employ induction... Thanks for your answer! – Martijn Dec 11 '16 at 16:25
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For $n=1$: $$ (A^1)^T = (A)^T = A^T = (A^T)^1 $$ Assuming the statement holds for $n \in \mathbb{N}$, we have for $n+1$: $$ (A^{n+1})^T = (A^n A)^T = A^T (A^n)^T = A^T (A^T)^n = (A^T)^{n+1} $$ so the statement holds for $n+1$ as well. With principle of induction it holds for all $n\in\mathbb{N}$.

mvw
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