Understanding basics of tensors I am trying to understand tensors to learn General relativity.
In the book that I am reading they claim that if the basis of a vector space undergoes a linear transformation $T$ then the components of the vector undergo the linear transformation $(T^{-1})^T$ (transpose of inverse of $T$).
This is how they prove it.

If the initial basis of a vector space is $\{e_i\}_1^n$ and the final basis $\{f_i\}_1^n$ is related by $$f_i=T_i{}^je_j$$ then for an arbitrary vector $v=x^ie_i=y^jf_j$ we have, $$y^k=x^j(T^{-1})_j{}^k=((T^{-1})^T)^k{}_jx^j$$

I am not able to understand the last equality at all.
I believe it is false.
Can somebody clarify if it is correct or wrong?
 A: I'll use a notation where I write vectors as $\vec{v}$ (because that's how \vec renders and I have no idea how to turn it into boldface in MathJax) and use primes to indicate the different basis: $v'^i$ is the components of $\vec{v}$ in the primed basis (some people use primes on the indices, which I find confusing).
So given some basis $\left\{\vec{e}_i\right\}$ we can write any vector $\vec{v}$ as
$$\vec{v} = v^i\vec{e}_i$$
Now consider a new basis $\left\{\vec{e'}_i\right\}$ which is related to the original basis by some nonsingular transformation matrix $T$:
$$\vec{e'}_j = T_j{}^i\vec{e}_i$$
(Note that this is just matrix multiplication using the ESC, and in particular $T$ is not a tensor.)
Well, we can express $\vec{v}$ in the new basis:
$$
\begin{align}
 \vec{v} &= v'^j\vec{e'}_j\\
         &= v'^jT_j{}^i\vec{e}_i\\
         &= v^i\vec{e}_i
\end{align}
$$
And comparing the second-last and last lines of this we get
$$v^i = v'^jT_j{}^i$$
OK, well we know $T$ is non-singular so there exists a $T^{-1}$, such that, in components 
$$T_j{}^i\left(T^{-1}\right)_i{}^k = \delta_j{}^k$$
So let's multiply the expression for $v^i$ above by this on the right:
$$
\begin{align}
 v^i\left(T^{-1}\right)_i{}^k
  &= v'^j T_j{}^i\left(T^{-1}\right)_i{}^k\\
  &= v'^j\delta_j{}^k\\
  &= v'^k
\end{align}
$$
or (big fanfare, and renaming indices gratuitously)
$$v'^i = v^j \left(T^{-1}\right)_j{}^i$$
And finally, we want to make this look like matrix multiplication on the left, so we need to diddle the indices of $T$:
$$v'^i = \left(\left(T^{-1}\right)^T\right)^i{}_j v^j$$
And we're done.
