Diagonalization and Powers of A

June 9, 2019

A=$S\Lambda S^{-1}$

Diagonalize a matrix is

Suppose we have $n$ independent eigenvectors, and we put them into the columns matrix $S$. So let’s just call this matrix the eigenvector matrix. What happen we do $AS$? ($\lambda$ is the eigenvalue)

And we call the matrix filled with eigenvalues the eigenvalue matrix. What we have right now is

Since we have $n$ independent eigenvalues, we can invert the matrix $S$, writing it to the right hand side

Still, it’s possible that there’re some small number of matrices that do not have independent eigenvalues, as mentioned in the end of the last note.

The sufficient and necessary condition (if and only if) for matrix $A$ to be diagonalizable is it has $n$ linearly independent eigenvectors ($n$ is just A’s size).

Square eigenvalues

If $A\mathbf x=\lambda\mathbf x$, then

This is saying if the matrix is squared, the eigenvalues will get squared as well but the eigenvectors remain. We can also use the factorization learned above

This is true for any kth power.

Right now we can say $A$ is sure to be diagonalizable if it has all the $\lambda$’s different. But same $\lambda$ does not mean the matrix is not diagonalizable.

Difference Equation $u_{k+1}=Au_k$

Let’s have a difference equation $u_{k+1}=Au_k$, and rewrite it into

If we can, somehow, write $u_0$ into a combinations of $A$’s eigenvectors, then

And:

In general, there’re two types of questions. One is solve the Markov chain matrices to find the steady state (note 24), where we need to solve $\pi=P\pi$; in this setting, the transition matrix $P:=A$ is only to the 1st power (i.e. no exponential), we can just find the eigenvalues and then eigenvectors of $A$, and using the eigenvectors to find the coefficients $c_1,c_2,…$. The other type is we have $A^k$ difference equations. This is similar, we also just need to be find the eigenvalues of $A$ (not $A^k$) and then take it to the kth power.

Fibonacci Sequence

Fibonacci sequence is

If we manually add another equations:

We can rewrite this into matrix form:

By letting $\mathbf u_k=\begin{bmatrix} F_{k+1}\\ F_{k} \end{bmatrix}$, we can:

Thus $A=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix}$. The eigenvalue of the matrix is $\displaystyle \lambda_1=\frac{1}{2}(1+\sqrt5), \lambda_2=\frac{1}{2}(1-\sqrt5)$. And the eigenvectors are $\mathbf x_1=\begin{bmatrix} \lambda_1 \\ 1 \end{bmatrix},\mathbf x_2=\begin{bmatrix} \lambda_2 \\ 1 \end{bmatrix}$. And we know $\mathbf u_0=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. What’s left is to find out the right combination $c_1,\ c_2$. Note we only have $\lambda_1,\ \lambda_2$ two eigenvalues, so we only have two terms from (2).

The important idea here is that eigenvalues are dominating the growth. (Note that because $\mathbf x_1,\mathbf x_2$ are independent, they span the space, we can always find such combination $c_1,c_2$)