Let $V$ be a vector space and $W$ an inner product space with an inner product $\langle.,.\rangle$.
Let $T : V \rightarrow W$ be a linear image. Define $\langle u,v\rangle^{'} = \langle T(u),T(v)\rangle$ with $u,v \in V$
I already proved that $\langle .,.\rangle^{'}$ defines a inner product on $V$ if and only is T is one-to-one. But the second question about this confuses me.
b) Let $a \in \mathbb{R}$. Say that $V=W=\mathbb{R^2}$ and $T: \mathbb{R^2} \rightarrow \mathbb{R^2}: X \longmapsto AX$ with $$A = \begin{bmatrix}3&3a\\\ 0& a\\ \end{bmatrix}$$
Let $\langle .,.\rangle$ be the standard inner product on $\mathbb{R^2}$.
Determine all $a$ so that $\langle .,.\rangle^{'}$ defines an inner product on $\mathbb{R^2}$ and find for these values an orthonormal basis of $\mathbb{R^2}$ with the inner product $\langle .,.\rangle^{'}$.
I don't see how to handle this question, but I think for the second part I just need to use Gram Schmidt?
Notice that \begin{align} \textrm{$\langle \cdot,\cdot \rangle'$ is an inner product in $\mathbb R^2$} & \ \Leftrightarrow \ \textrm{$T$ is injective} \\ & \ \Leftrightarrow \ \textrm{the kernel of $T$ is trivial} \\ & \ \Leftrightarrow \ \textrm{if $T(X)=0$ then $X=0$} \\ & \ \Leftrightarrow \ \textrm{if $\begin{bmatrix}3&3a\\0&a\end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$ then $\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$} \\ & \ \Leftrightarrow \ \textrm{if $x_1 \begin{bmatrix}3\\0\end{bmatrix} + x_2 \begin{bmatrix}3a\\a\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$ then $x_1=0$ and $x_2=0$} \end{align} so, it suffices to find all the values of $a$ for which the vectors $$\textrm{$\begin{bmatrix}3\\0\end{bmatrix}$ and $\begin{bmatrix}3a\\a\end{bmatrix}$}$$ are linearly independent. You can use your favorite method for this! And for the second part, yes, you can apply the Gram-Schmidt process to the standard unit vectors $$\textrm{$\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$}$$ with the new inner product $\langle \cdot,\cdot \rangle'$.