I'm self learning linear-algebra.
I'm having trouble understanding why if $A$ is a matrix where $F(x) = Ax$ is an orthogonal projection, it cannot be necessarily invertible.
Why is that the case, but when $A$ is a matrix such that $F(x) = Ax$ is an isomorphism, $A$ is necessarily invertible?
On
A projection (not even orthogonal) satisfies $A^2=A $. If $A $ is also invertible, you can multiply the equality by $A^{-1} $ to obtain $A=I $. In other words, the only invertible projection is the identity.
Isomorphism, on the other hand, implies that $A $ is bijective, which is the same as invertible.
Imagine you want to project all $\mathbb{R}^3$ onto a plane $\pi$: necessarily some points from $\mathbb{R}^3$ have to go to same points in $\pi$ (simply speaking there are too many points in $\mathbb{R}^3$, you have to squeeze $\mathbb{R}^3$ to become a plane), so this map cannot be invertible. Defining this map with a matrix $A$ means that the matrix cannot be invertible because you cannot map every point from $\pi$ from it's respective point, where they came from, in $\mathbb{R}^3$. The only thing you can say about a projection matrix is that if $A$ is a projection matrix, then it's idempotent ($A^2=A$). In fact if you take the same map as before and try to apply it to the plane $\pi$ it has to get you again to $\pi$. The only invertible projection is the identity, which can be easily proven with the fact that a projection is idempotent.
Conversely if $A$ is a matrix of an isomorphism, by definition an isomorphism is a map one-to-one, so it maps every point in the domain to one and one only point into the codomain. So if you want to come back from the codomain to the domain you don't have any choice over taking a point from the codomain and making it come back from the point where it came from the first place. So an isomorphism is clearly invertible and then you can compute the inverse matrix $A^{-1}$ that gives you the inverse map