1
$\begingroup$

Let $\{e_1,\ldots,e_n\}$ be an arbitrary basis in a finite dimensional inner product space $V$. Prove there exists vectors $\{f_1,\ldots,f_n\}$ such that $(e_i,f_j)=\delta_{ij}$.

I tried using the the Gram-Schmidt process to obtain the $f_i$'s but the resulting $f_i$'s should apparently be uniquely determined and $(e_i,f_j)=\delta_{ij}$ won't hold.

What other methods could I try to obtain the $\{f_1,\ldots,f_n\}$? Perhaps we could use induction on $\dim(V)$ for a proof?

$\endgroup$
2
$\begingroup$

Let $A$ and $B$ be the matrices having as columns, respectively, the vectors $e_i$ and $f_i$. Your relation can be written in the form $A^TB=I$. Hence, $B=(A^T)^{-1}$. This shows that the vectors $f_i$ are the columns of the matrix $(A^T)^{-1}$.

$\endgroup$
2
  • $\begingroup$ Does this really show they exist or just that $(A^T)^{-1}$ would have the $f_i$ as columns if they did exist? $\endgroup$
    – mathjacks
    Feb 7 '16 at 16:27
  • $\begingroup$ Since $A$ is invertible (because you started with a basis), the matrix $(A^T)^{-1}$ is always well defined. In other words, there is always a solution. $\endgroup$
    – John B
    Feb 7 '16 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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