If $e = (e\_1,e\_2,...,e\_n)$ is the basis of $\mathbb R^n$ and $P$ is a random invertible $n \times n$ matrix, proof that when $f\_j=Pe\_j (j=1,...,n)$ then $f = (f\_1,f\_2,...,f\_n)$ is another basis of $\mathbb R^n$

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What I have found:

Let's say 
$
P = 
\begin{pmatrix}
p\_{11} & p\_{12} & ... & p\_{1n}\\
p\_{21} & p\_{22} & ... & p\_{2n}\\
... & ... & ... & ...\\
p\_{n1} & p\_{n2} & ... & p\_{nn}\\
\end{pmatrix}
,
e\_j = 
\begin{pmatrix}
e\_{1j}\\
e\_{2j}\\
...\\
e\_{nj}\\
\end{pmatrix},
$ 
and
$
f\_j = 
\begin{pmatrix}
f\_{1j}\\
f\_{2j}\\
...\\
f\_{nj}\\
\end{pmatrix}
$
then 
$f\_j=Pe\_j= 
\begin{pmatrix}
p\_{11} & p\_{12} & ... & p\_{1n}\\
p\_{21} & p\_{22} & ... & p\_{2n}\\
... & ... & ... & ...\\
p\_{n1} & p\_{n2} & ... & p\_{nn}\\
\end{pmatrix}
e\_j=
\begin{pmatrix}
p\_{11} \cdot e\_{1j} + p\_{12} \cdot e\_{2j} + ... + p\_{1n} \cdot e\_{nj}\\
p\_{21} \cdot e\_{1j} + p\_{22} \cdot e\_{2j} + ... + p\_{2n} \cdot e\_{nj}\\
... \\
p\_{n1} \cdot e\_{1j} + p\_{n2} \cdot e\_{2j} + ... + p\_{nn} \cdot e\_{nj}\\
\end{pmatrix}
=
\begin{pmatrix}
f\_{1j}\\
f\_{2j}\\
...\\
f\_{nj}\\
\end{pmatrix}$

thus, we know that every element of $f\_j$ is linearly independent. 

But then, how do I know whether $f$ is linearly independent and span $\mathbb R^n$?