I'm speechless 🩵

Hey there! So, you're diving into some linear algebra with Markov matrices and diagonalization, huh? Cool stuff! Let's break it down a bit.

First off, when we say a matrix $A$ is "positive Markov," we're talking about a special kind of matrix that's used to describe the transitions in a Markov process (like predicting weather changes or modeling how people move through different states of a game). The "positive" part means all the entries in the matrix are positive numbers, which makes sense because you can't really have negative probabilities, right?

Now, for a Markov matrix $A$ , the eigenvalue $\lambda_A=1$ is a given because the columns of $A$ add up to 1 (since they represent probabilities). This means that one of the eigenvalues has to be 1 to keep the total probability in the system constant over time.

The equation

$$\lim_{k\to\infty}{A^k}\vec{w}_0=c\vec{u}_A$$

is telling us that if you keep applying the Markov matrix $A$ over and over to some initial vector $\vec{w}_0$ , eventually the system will reach a steady state. That steady state is represented by $c\vec{u}_A$ , where $c$ is some constant and $\vec{u}_A$ is the eigenvector associated with the eigenvalue 1.

Now, onto the to-do about proving that a matrix $P$ is diagonalizable if $P^2 = P$ . This is a neat property of matrices called idempotency. When a matrix, when squared, gives you the same matrix back, it's called idempotent. For such matrices, it turns out they are diagonalizable. Here's a sketch of why that's true:

1. First, remember that a matrix is diagonalizable if you can find a basis of eigenvectors for the whole space it acts on.
2. Now, if $P^2 = P$ , then $P$ only has eigenvalues 0 and 1 because if you take any vector $\vec{v}$ and apply $P$ to it, the result after applying $P$ again doesn't change.
3. So, any vector that doesn't get killed by $P$ (sent to the zero vector) is an eigenvector with eigenvalue 1, and any vector that does get killed by $P$ is an eigenvector with eigenvalue 0.
4. Since $P$ is a linear transformation, the whole space can be split into these two kinds of vectors (those that get killed and those that don't), which are eigenspaces.
5. If you can find a basis for each of these eigenspaces, then you've got a full set of eigenvectors that span the whole space, which means $P$ is diagonalizable.

So, there you have it! A little bit of linear algebra magic to start your day. Isn't it neat how these abstract concepts can tell us so much about systems and transformations? Keep at it, and these ideas will become clearer and clearer.