If $A$ and $B$ are arbitrary $n \times n$ matrices, prove that $(A^TB^TBA)$ is symmetric

Question

My attempt:

$(A^TB^TBA)^T$=$(A^T)^T(B^T)^TB^TA^T$=$(AB)B^TA^T$ $\ne$ $(A^TB^TBA)$ therefore $(A^TB^TBA)$ is not symmetric.

Answer 1

As what @AlexR mentioned, we have $$(\underbrace{A^T}_{(1)}\, \underbrace{B^T}_{(2)}\, \underbrace{B}_{(3)}\, \underbrace{A}_{(4)})^T= \underbrace{A^T}_{(4)}\, \underbrace{B^T}_{(3)}\, \underbrace{B^{TT}}_{(2)}\, \underbrace{A^{TT}}_{(1)}=A^TB^TBA,$$ and therefore $A^TB^TBA$ is symmetric.

Answer 2

Note that $(AB)^T = B^TA^T$, unlike you assumed. With this in mind, redo what you did and you'll succeed.

Answer 3

Not the most elegant answers but I consider it a fun exercise even though it can be laborious for some. Remember that the multiplication of two matrices $B=(B_{ij})_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}}$ and $A=(A_{ij})_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}}$ is outer matrix given by $$ B\cdot A = (B_{ij})_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}} \cdot (A_{ij})_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}} = \left( \sum_{\alpha=1}^{n}B_{i\alpha}A_{\alpha j} \right)_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}} \quad ( \ast ) $$ And with the same notation the transpose of $A$ and $B$ are given by $$ (A_{ij})^{\;T}_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}}=(A_{ji})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \text{ and } (B_{ij})^{\;T}_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}}=(B_{ji})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \quad ( \ast \ast) $$ i.e. the permutation of the columns index $j$ with the index $i$ of the lines does not change the value of the entries of the matrix. And further note that, $$ A^T\cdot B^T = (A_{ji})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}}\cdot (B_{ji})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} = \left( \sum_{\beta=1}^{n}A_{j\beta}B_{\beta i} \right)_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \quad ( \ast\ast\ast ) $$ Now substituting $( \ast ), \; ( \ast\ast )$ and $(\ast\ast\ast)$ in $A^TB^TA\;B$ we have \begin{align} A^TB^TB\;A =& (A_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}}^T (B_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}}^T (B_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} (A_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \\ =& (A_{ji})_{\displaystyle_{1\leq j\leq n}^{1\leq i\leq n}} (B_{ji})_{\displaystyle_{1\leq j\leq n}^{1\leq i\leq n}} (B_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} (A_{ij})_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \\ =& \left( \color{red}{\sum_{\beta=1}^{n}A_{j\beta}B_{\beta i}} \right)_{\displaystyle_{1\leq i\leq n}^{1\leq j\leq n}} \left( \color{blue}{\sum_{\alpha=1}^{n}B_{i\alpha}A_{\alpha j}} \right)_{\displaystyle^{1\leq i\leq n}_{1\leq j\leq n}} \\ =& \left( \sum_{\gamma=1}^{n} \Big( \color{red}{\sum_{\beta=1}^{n}A_{\color{black}{j}\beta}B_{\beta \color{black}{\gamma}}} \Big) \Big( \color{blue}{\sum_{\alpha=1}^{n}B_{\color{black}{\gamma} \alpha}A_{\alpha \color{black}{j}}} \Big) \right)_{\displaystyle_{1\leq j\leq n}^{1\leq j\leq n}} \end{align} Now it's easy to see that $$ \left( \sum_{\gamma=1}^{n} \Big( \color{red}{\sum_{\beta=1}^{n}A_{\color{black}{j}\beta}B_{\beta \color{black}{\gamma}}} \Big) \Big( \color{blue}{\sum_{\alpha=1}^{n}B_{\color{black}{\gamma} \alpha}A_{\alpha \color{black}{j}}} \Big) \right)^{T}_{\displaystyle_{1\leq j\leq n}^{1\leq j\leq n}} = \left( \sum_{\gamma=1}^{n} \Big( \color{red}{\sum_{\beta=1}^{n}A_{\color{black}{j}\beta}B_{\beta \color{black}{\gamma}}} \Big) \Big( \color{blue}{\sum_{\alpha=1}^{n}B_{\color{black}{\gamma} \alpha}A_{\alpha \color{black}{j}}} \Big) \right)_{\displaystyle_{1\leq j\leq n}^{1\leq j\leq n}} $$