Need help seeing the utility of change of basis In my linear algebra class we are introduced to change of basis, but I failed to find it very insightful. It seems that we are simply multiplying a row vector by a matrix, which is a basic operation in linear algebra and does not deserve a name on its own
Can someone please provide me with some answer to the following? Much appreciated


*

*Let $z = Px$ where $x$ was the old basis, $z$ is the new basis and $P$ is the transform, what are some of the restrictions that needs to be placed on $P$?

*Can someone show me when this is useful? When is a new set of basis more beneficial in some way than the old ones?
 A: You'll have to elaborate on what you mean by 1; as written it isn't quite comprehensible.
Here's my answer for 2:
As is often the case when we impose an algebraic structure (such as, in this case, changes of basis), the interesting or "insightful" ideas don't come from looking at the space itself (i.e. what happens to a vector under this change of basis).  Rather, the most interesting ideas come from looking at the maps between spaces (i.e. what happens to a linear transformation under a change of basis).
In general, suppose we have a transformation $T:X \to Y$.  With respect to the standard basis, let's say that $T$ is represented by the matrix $A$.  Suppose we apply a change of basis to $X$ and $Y$ with corresponding matrices $Q$ and $P$ respectively.  Then, the new matrix corresponding to the same transformation, under this change, is given by
$$
P^{-1}AQ
$$
By choosing $P$ and $Q$ carefully, we can come up with useful ways of representing any transformation that we want to look at.
Examples:
Let's say $T:X \to X$, so that $A$ is a square matrix.  Apply change of basis $P$ to $X$, so that we end up with the new matrix
$$
J = P^{-1}AP
$$
It turns out that we can always choose a change of basis such that $J$ is in Jordan canonical form.  When $A$ is diagonalizable, $J$ is simply a diagonal matrix, giving us a much easier way to think of $T$.  In the diagonalizable case, this is an eigenvalue-eigenvector decomposition.
Suppose we again consider $T:X \to X$, but now make the restriction that our new basis has to be orthonormal.  That is, we state that $P$ must be a orthogonal (or unitary, in the complex case) change of basis.  In this case, we have $P^{-1} = P^T$.  So, our new matrix is simply
$$
U = P^TAP
$$
It turns out that we can always choose a change of basis such that $U$ is upper-triangular; this is called the Schur-decomposition of $A$ (or of $T$).  When $A$ is symmetric (i.e. $A = A^T$), it turns out that $U$ must be a diagonal matrix (this is called the "spectral theorem").
Finally, suppose we consider $T:X \to Y$, and make the restriction that both $P$ and $Q$ are orthogonal changes of basis.  We then end up with a matrix
$$
\Sigma = P^TAQ
$$
It turns out that, even when $A$ is not square, we can choose $P$ and $Q$ so that $\Sigma$ is diagonal with non-negative entries.  This gives us the so-called Singular-value decomposition, which is extremely useful in numerical linear algebra.
A: If you analyze your equation z = Px, as a matrix equation. P is the transformation matrix that transforms the x vector. If you want to solve the equation. You need to use the inverse of P, that is called P⁻¹.
P⁻¹ z= P⁻¹ P x = (P⁻¹ P) x = 1 x = x
P⁻¹ exists if and only if the determinant of P is not equal to zero. The determinant does not exist if P is not a square matrix.
When the inverse exists the transformation is one-to-one and onto. Such functions are called bijective. Bijections are functions that are both injective and surjective. This concept is fundamental for Abstract Algebra.
