Suppose $\left \{ e_{j} \right \}$ is a basis for X; let $P = \left \{ p_{ij} \right \}$ be any nonsingular $n \times n$ matrix, and let $f_{j} = \sum p_{ij}e{i}$. Show that $\left \{ f_{j} \right \}$ is a basis for $X$.
To show that vector is a basis I know that I have to show two things:
- The vector spans the vector space, $X$
- The vector is linearly independent
To show (1), I've started from $Px$. The multiplying both sides by $P^{-1}$, I have $P^{-1}Px$, which is equal to $Ix$ assuming $P$ is nonsingular. Since $\left \{ e_{j} \right \}$ or $Ix = x$ is already a basis for $X$, $\left \{ f_{j} \right \}$ is also a basis for $X$.
For (2), I started from $Px=0$. Multiplying both sides by $P^{-1}$ we obtain the following $$\Rightarrow P^{-1}Px = P^{-1}0 \Rightarrow Ix = 0 \Rightarrow x = 0.$$
Does my solution look correct?
If for any vector $y$, we show it can be written as a linear combination of the $f_j$, then we will have shown $\{f_j\}$ spans $X$. The vectors $f_j$ are the columns of $P$, so we just need to find some $x$ such that $y=Px$. Taking $x=P^{-1} y$ works, since $P$ is non-singular.
Your solution looks correct, although you could be a little clearer about why those steps imply linear independence.