Implementing intuition whether there exist bases such that: I encounter alot of questions asking me to determine whether there exist bases such that a condition takes place.
Example:
Let T(x,y) = \begin{bmatrix}2x + 3y\\3x + 4y\end{bmatrix} be a linear map. T: R^2 --> R^2
are there bases s.t. the matrix [T](from B to C) = \begin{bmatrix}1&2\\2&4\end{bmatrix}
The answer is no because the representing matrix is invertible\begin{bmatrix}2&3\\3&4\end{bmatrix}
and the matrix \begin{bmatrix}1&2\\2&4\end{bmatrix} is not.
First of all, I don't understand this statement. I would like if someone can explain that to me please.
and how do you know to identify wherther there are bases or not?
Thanks in advance! 
 A: A matrix's invertibility does not depend on what basis you choose. You can see this in a number of ways. First, if you know that the eigenvalues are invariant, remember that invertibility means zero is not an eigenvalue. If you know determinants are invariant, remember that invertibility is equivalent to the determinant being nonzero. Or if you want to start from the top, know that a change of basis is implemented as $$T\to A^{-1}TA $$ so if $T$ is invertable, then $$(A^{-1}TA)^{-1} = A^{-1}T^{-1}A$$ so the matrix due after changing basis is still invertible.
Thus since your matrix is invertible in the natural basis, it must be invertible in all bases and the non-invertible matrix is therefore not a candidate.
In general, two matrices are related by a change of basis if and only if they have the same Jordan normal form (up to rearrangements of blocks), (in other words the same structure of invariant subspaces). For diagonalizable matrices this is equivalent to the matrices having the same eigenvalues (in the same multiplicities).
Generally it's easier to prove they aren't related then they are. There are several invariants that must be the same under a change of basis. For instance if the matrices have a different trace (which is very easy to spot), then they aren't related by a change of basis. Or if they have a different determinant. Failing that, if they have a different eigenvalue.
