Characteristic Equations

I noticed a pattern recently where you can solve problems by "guessing" the form of the solution then solving a polynomial

Differential equations

Lets say we have a differential equation of the form

y+py+qy=0y'' + py' + qy = 0

Guess y=erty = e^{rt}

r2ert+prert+qert=0r^2e^{rt} + p \cdot re^{rt} + q \cdot e^{rt} = 0 \\

Factor out erte^{rt}

ert(r2+pr+q)=0e^{rt}(r^2 + pr + q) = 0

Quadratic equation! solve for rr then use intial conditions to find pp and qq

This also shows there is always a solution erte^{rt} for a diffeq of the form

cny(n)++c1y+c0y=0c_ny^{(n)} + \dots + c_1y' + c_0y = 0

Since you can let y=erty = e^{rt} then factor out and solve the polynomial. a solution always exists if rCr \in \mathbf C

Recurrance relations

pn+1=cpn+dpn1p_{n+1} = cp_{n} + dp_{n-1}

Guess pi=xip_i = x^i

xn+1=cxn+dxn1x^{n+1} = cx^{n} + dx^{n-1} \\

Divide by xn1x^{n-1}

x2=cx+dx^2 = cx + d

Quadratic equation! solve for xx then use initial conditions to find cc and dd.

Sadly we must verify our solution for xx works, since unlike the differential equations example we don't know if pi=xip_i = x^i is true or not.

Linear algebra

A=XΛX1A = X\Lambda X^{-1}

By definition

eA=limnk=0n1k!Ak=limnX(k=0n1k!Λk)X1e^A = \lim_{n \to \infty} \sum_{k=0}^n \frac{1}{k!} A^k = \lim_{n \to \infty} X\left(\sum_{k=0}^n \frac{1}{k!} \Lambda^k\right)X^{-1}

Exponentiating Λ\Lambda just raises each λik\lambda_i ^k then divides and sums them up, so we get

eA=XeΛX1e^A = Xe^{\Lambda}X^{-1}


eΛ=[eλ1eλn]e^{\Lambda} = \begin{bmatrix} e^{\lambda_1} & & \\ & \ddots & \\ & & e^{\lambda_n} \end{bmatrix}


Eigenstuff, connection with markov matrices. is diagonalizing a markov matrix the same as solving the recurrance?