0
$\begingroup$

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 $A$).

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jack, Davide Giraudo, Namaste, Shailesh, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.