The following question is given in a section 2 lecture of linear algebra. The first section is about polynomial, so the lectures just started to talk about determinants and matrices.
Let $A$ be an $n\times n$ matrix over a number field $F$. Then there exists an invertible matrix $R$ such that $AR$ is symmetric.
I know that this question can be (elegantly) eliminated using Jordan canonical form.
But since the question is left to who just learn linear algebra, I don’t think Jordan form is necessarily required.
Then the question can be interpreted as the following:
Let $A$ be an $n\times n$ matrix over a number field $F$. Then $A$ can be changed to a symmetric matrix through elementary column operations.
The Jordan form method only establishes the existence of some invertible matrix satisfying this property, which (I think) makes it unclear how to relate it with row/column operations.
I think it may be dealt with by induction. Am I right? It is not very clear to me how to complete the inductive steps. Any help is sincerely appreciated.
Best Answer
It is not really clear from your question what type of proof is needed. To me it is also not clear how to proceed by some induction argument, so I suppose that some knowledge is required. As for the Jordan form, I think it does not solve this question in full generality since it requires the algebraic closedness of $F$.
The proof will be based on the well-known fact that arbitrary matrix $A$ can be represented in the form $$ A = SI_rQ,$$ where $S$ and $Q$ are invertible and $I_r$ can be represented in the block form as $$I_r = \left(\begin{matrix} I & 0 \\ 0 & 0 \end{matrix}\right) $$ with the $r \times r$ identity matrix as the left upper block. Using that we obtain $AQ^{-1}S^T = SI_rS^T$ is symmetric.
The fact that we used above is quite easy to prove by considering a pair of basises in $F^n$ in which the matrix has the form $I_r$. That is consider arbitrary basis $e_1, \dots, e_r$ in $\mathrm{im} A = \{Ax: x \in F^n\}$ (here $r = \mathrm{rank} A$). Let $f_1, \dots, f_r$ be arbitrary preimages of $e_1, \dots, e_r$. Let $f_{r+1}, \dots, f_n$ be arbitrary basis in $\ker A$. Also arbitrarily extend $e_1, \dots, e_r$ by $e_{r+1}, \dots, e_n$ to a basis in $F^n$. Then $A$ has the form $I_r$ in the pair of basises $e$ and $f$, that is $A = SI_rQ$, where columns of $Q^{-1}$ are the vectors $f_i$ and the columns of $S$ are $e_i$.