Let us $H$ a Hilbert space and $v_1,\ldots,v_n$ vectors in $H$. Let put the matrix $A$ given by $a_{i,j}:=\langle v_i,v_j\rangle$ with $i,j=1,\ldots,n$. The proposition say: $\det A\neq 0$ if and only if $v_1,\ldots,v_n$ are linearly independent. I want to prove for $n=2$ the direction $\rightarrow)$
My attempt: Let us suppose that $\det A=0$ and i want to see that $v_1,v_2$ are L.D As $\det A=0$ implies that the columns are L.D then exist $\lambda\neq 0$ such that $$\langle v_1,v_1\rangle=\lambda \langle v_1,v_2\rangle$$ and $$\langle v_2,v_1\rangle=\lambda \langle v_2,v_2\rangle$$ and by hypothesis $\langle v_1,v_1\rangle\langle v_2,v_2\rangle-\langle v_1,v_2\rangle\langle v_2,v_1\rangle=0$. Equivalently i have $||v_1||^2=\overline{\lambda}\langle v_2,v_1\rangle=|\lambda|^2||v_2||^2=||\lambda v_2||^2$. Now my idea is to use parallelogram law because i did not used yet that i am working in a Hilbert space. Well i do not know how, but the other side i was thinking in $||v_1||^2-||\lambda v_2||^2=0$ and if $v_1$ and $v_2$ were L.I they satisfies the pitagoras law, is mean $||v_1-\lambda v_2||^2=||v_1||^2 +|\lambda|^2||v_2||^2$ and try to use the number $||v_1+\lambda v_2||^2$ and try to put together with the parallelogram law $$||u+v||^2+||u-v||^2= 2(||u||^2+||v||^2)$$. i am in the correct way? If somebody can help me , i will appreciate.
Thank you
Remark I thought that is wrong to think as following: If $\langle v_1-\lambda v_2,v_i\rangle=0$ with $i=1,2$ implies that $ v_1-\lambda v_2=0$ because i do not know that the basis of $H$ is $v_1,v_2$ right? I thought that my excercise is deifferent of this:The inner product matrix with zero determinant implies that all the vectors are linearly dependent. or not?
