Let $V$ and $V'$ be real (positive definite) inner product spaces of (same) finite dimensions with inner products $(-,-)_1$ and $(-,-)_2$.

Let $T:V\rightarrow V'$ be a $\mathbb{R}$-linear isomorphism of underlying vector spaces.

Then $T$ may not preserve the inner product, i.e. it may not be isometry.

Q. Is it always possible to replace the inner product $(-,-)_1$ by $r(-,-)$ for positive real $r$, and $(-,-)_2$ by $s(-,-)_2$ so that $T$ becomes isometry between $(V,r(-,-)_1)$ and $(V',s(-,-)_2)$?

  • $\begingroup$ Can you please clarify the question ? $\endgroup$ – Youem Apr 22 '18 at 5:06

No. For example, let $T$ be the identity map on $\mathbb{R}^2$. It is not an isometry between inner-product spaces with inner products $$ (x, y)_1 = x_1y_1+x_2y_2 $$ and $$ (x, y)_2 = 2x_1y_1 + x_2y_2 $$ and multiplying either of the inner products by a scalar cannot fix that.


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