Introductory Linear Algebra/Eigenvalues and eigenvectors

Motivation
Before discussing and , we provide some motivations to them.

We can see from this example that for some special matrices, their powers can be computed conveniently, by expressing in the form of $$PDP^{-1}$$ in which $$P$$ is invertible matrix and $$D$$ is diagonal matrix.

Naturally, given a matrix, we would be interested in knowing whether it can be expressed in the form of $$PDP^{-1}$$, and if it can, what are $$P$$ and $$D$$, so that we can compute its power conveniently. This is the main objective of this chapter.

Eigenvalues, eigenvectors and diagonalization
In view of the motivation section, we have the following definition.

The following are important and general concepts, which is related to diagonalizability in some sense.

The following theorem relates diagonalizable matrix with eigenvectors and eigenvalues.

Then, we will introduce a convenient way to find. Before this, we introduce a terminology which is related to this way of finding eigenvalues.

Then, we will introduce a concept which is related to.

After introducing these terminologies and concepts, we have the following algorithmic procedure for a diagonalization of an $$n\times n$$ matrix:
 * 1) compute all  of $$A$$ by solving $$\det(A-\lambda I)=0$$
 * 2) for each eigenvalue $$\lambda_1,\ldots,\lambda_k$$ of $$A$$, find a  $$\beta_1,\ldots,\beta_k$$ for the  $$E_{\lambda_1},\ldots,E_{\lambda_k}$$ respectively
 * 3) if $$\beta_1,\ldots,\beta_k$$ together contain $$n$$ vectors $$\mathbf v_1,\ldots,\mathbf v_n$$ (if not, $$A$$ is  diagonalizable), define $$P=\begin{pmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{pmatrix}$$
 * 4) we have $$A=PDP^{-1}$$ in which $$D$$ is a  matrix whose  entries are  corresponding to $$\mathbf v_1,\ldots,\mathbf v_n$$

In the following, we will discuss some mathematical applications of diagonalization, namely deducing the sequence formula, and solving system of (ODEs).

{{colored exercise| Solve the following system of ODEs with initial condition $$(x,y)=(2,3)$$ when $$t=0$$: $$\begin{cases}\frac{dx}{dt}=2x\\\frac{dy}{dt}=6y\end{cases}$$ + $$(x,y)=(2e^{2t},3e^{6t})$$. - $$(x,y)=(2e^{2t}+3e^{6t},3e^{2t}+2e^{6t})$$. - $$(x,y)=(2e^{6t}+3e^{2t},3e^{6t}+2e^{2t})$$. - $$(x,y)=(3e^{2t},2e^{6t})$$. - $$(x,y)=(3e^{6t},2e^{2t})$$.
 * type=""}
 * solving $$\frac{dx}{dt}=2x$$, we get $$x=C_1e^{2x}$$
 * solving $$\frac{dy}{dt}=6y$$, we get $$y=C_2e^{6y}$$
 * imposing initial condition, we have $$(C_1e^0,C_2e^0)=(2,3)\implies (C_1,C_2)=(2,3)$$
 * so the solution is $$(x,y)=(2e^{2t},3e^{6t})$$

{Find $$k$$ such that the following system of ODEs is inconsistent.$$\begin{cases}\frac{dx}{dt}=kx\\\frac{dy}{dt}=ky\end{cases}$$ + No such $$k$$ exists. - $$k=0$$ - $$k=1$$ - $$k$$ can be arbitrary real numbers.
 * type=""}
 * the solution is $$(x,y)=(C_1e^{kx},C_2e^{kx})$$ in which $$C_1,C_2$$ are arbitrary constants
 * so, for each $$k$$, the system is consistent

{Solve the following system with initial condition $$(x,y,z)=(2,3,8)$$ when $$t=0$$: $$\begin{cases}\frac{dx}{dt}+2\frac{dy}{dt}+5\frac{dz}{dt}=-6\\2\frac{dx}{dt}+7\frac{dy}{dt}+4\frac{dz}{dt}=3\\3\frac{dx}{dt}+8\frac{dy}{dt}+5\frac{dz}{dt}=0\end{cases}.$$ - It is inconsistent. - $$(x,y,z)=(2,3,8)$$ - $$(x,y,z)=(-6,1,2)$$ - $$(x,y,z)=(-6t,t,2t)$$ - $$(x,y,z)=(2t+2,3t+3,8t+8)$$ + $$(x,y,z)=(-6t+2,t+3,2t+8)$$ - $$(x,y,z)=(2t-6,3t+1,8t+2)$$ }}
 * type=""}
 * it is only true when $$t=0$$
 * $$\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)=(-6,1,2)$$, not $$(x,y,z)=(-6,1,2)$$
 * solving this system, we will get $$\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right)=(-6,1,2)\implies (x,y,z)=(-6t+C_1,t+C_2+2t+C_3)$$
 * imposing the initial condition, we have $$(C_1,C_2,C_3)=(2,3,8)$$