Consider the row vectors $a = \begin{bmatrix}a_1 & a_2 & a_3 \\\end{bmatrix}$, $b = \begin{bmatrix}b_1 & b_2 & b_3 \\\end{bmatrix}$, $c = \begin{bmatrix}c_1 & c_2 & c_3 \\\end{bmatrix}$.

(a) Assume that $a,b,c \in \mathbb R^3$ are linearly independent and let $A = \begin{bmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \\ \end{bmatrix} $

Show that the matrix $AA'$ is symmetric (where $A'$ is the transpose of $A$).


closed as off-topic by Jack, Davide Giraudo, Namaste, Shailesh, user223391 Dec 10 '16 at 23:52

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    $\begingroup$ This follows from the fact that $(AB)^T = B^T A^T$. You don't need the linear independence assumption. $\endgroup$ – littleO Dec 7 '16 at 18:40
  • $\begingroup$ But can I prove it through linear independence assumption? $\endgroup$ – Furqan Dec 7 '16 at 18:41
  • $\begingroup$ What the symbol $*$ means? $\endgroup$ – Alex Silva Dec 7 '16 at 18:42
  • $\begingroup$ The linear independence assumption doesn't seem to help us at all, so I don't know why it was given. Have you tried computing $AA^T$ explicitly? Just multiply it out. $\endgroup$ – littleO Dec 7 '16 at 18:46

You should not be assuming that the rows (or columns) of $A$ are linearly independent because this may not be the case. The result holds for all matrices, not just ones with linearly independent rows. As littleO suggested, you just have to know that $(AB)^T=B^TA^T$ and $(A^T)^T=A$. Then you can check that $$ (AA^T)^T=(A^T)^TA^T=AA^T$$ which exactly says that $AA^T$ is symmetric.


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