I need help with this problem:
Determine if for the real parameter t there are values for which the function $(x1, x2),(y1, y2)) → (x1, x2)$ $ \begin{pmatrix} 6 & e^{2t} \\ 2e^t-1 & 2 \\ \end{pmatrix}$ $\begin{pmatrix} y1 \\ y2 \end{pmatrix}$ is a scalar product in in $R^2$ :
The answer is $t=0$. Here is why.
This matrix has to be 1) symmetric 2) positive definite, which amounts to say that
Point 1) $$2e^t-1=e^{2t} \ \ \Leftrightarrow \ \ (e^t-1)^2=0 \ \ \Leftrightarrow \ \ e^t=1 \ \ \Leftrightarrow \ \ t=0$$
meaning that the matrix is necessarily $\pmatrix{6&1\\1&2}$. But is it positive definite ?
Point 2) Let us expand the matrix product :
$$\pmatrix{x&y}\pmatrix{6&1\\1&2}\pmatrix{x \\ y}=6x^2+2xy+2y^2=5x^2+y^2+(x+y)^2$$
which is $>0$ unless $x=y=0.$
As the quadratic form associated with the given bilinear form is always $>0$ but for the null case, it is associated with a scalar (=dot) product.