Ok, so in my DiffEq class we've been doing problems which more or less amount to solving equations of the form:

$$\frac{dY}{dt} = AY$$

Where $A$ is just some $2\times2$ linear transformation and $Y$ is a parametric vector function defined more specifically as

$$Y(t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix}$$

The end result, assuming that there exists $\lambda_1, \lambda_2 \ne 0; \lambda_1 \ne \lambda_2$ which define the eigen values for A, is a definition for $Y(t)$ of the form,

$$Y(t) = k_1e^{\lambda_1t}\vec{V_1} + k_2e^{\lambda_2t}\vec{V_2}$$

Where $\vec{V_1}, \vec{V_2}$ are the corresponding eigen vectors to their respective eigen values and $k_1, k_2$ are just some constants.

For solutions which involve either $k_1 = 0$ or $k_2 = 0$, the end result is a straight-line solution. The rest are exponential curves within the vector space defined by the eigen vectors.

My understanding of eigen vectors, from a linear algebra class I took a year ago, so far is as follows (roughly):

  • geometrically speaking, an eigen vector is any vector whose direction after transformation by some matrix $A$ remains the same. It's only scaled and/or negated.

  • Every eigen vector for some matrix $A$ composes a subspace which in turn defines the eigen space for $A$'s vector basis.

  • therefore, the eigen vectors which $span(A)$ are linearly independent and define a coordinate space which also exists within $A$.

Regardless of whether or not the above is correct (if there's a mistake, any clarification/correction would be appreciated), what is it about eigen vectors specifically which allows for them to be used to solve these forms of differential equations?


Consider the vector ODE $$ \frac{d Y}{dt} = AY,\;\; Y(0)=Y_0, $$ where $Y$ is an $N$-vector function of $y$, where $Y_0$ is a constant $N$-vector, and where $A$ is a constant $N\times N$ matrix. This has a unique solution $$ Y(t) = e^{tA}Y_0\; \mbox{where}\; e^{tA}=\sum_{n=0}^{\infty}\frac{1}{n!}t^nA^n. $$ If $Y_0$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$ e^{tA}Y_0 = \sum_{n=0}^{\infty}\frac{1}{n!}t^nA^nY_0 = \left(\sum_{n=0}^{\infty}\frac{1}{n!}\lambda^n t^n\right)Y_0 = e^{\lambda t}Y_0. $$ If $A$ has a basis of eigenvectors $Y_1$, $Y_2$, then all solutions can be formed in this way because $Y_0 = \alpha Y_1+\alpha_2Y_2$ for unique $\alpha_1$, $\alpha_2$, which leads to the solution of the ODE in the form $$ Y(t)=\alpha_1 e^{\lambda_1 t}Y_1+\alpha_2 e^{\lambda_2 t}Y_2. $$ That's the basic story for this problem.

The History of eigenvector analysis traces back to Fourier's analysis of his Heat Equation, where he used separation of variables to solve $$ \frac{\partial u}{\partial t} = \frac{\partial^2u}{\partial x^2}. $$ His separation of variables technique involved finding all solutions of the form $u(t,x,y,z)=T(t)X(x)$, and trying to form a general solution out of sums of such solutions. That naturally led to eigenfunction/eigenvalue equations, long before the terminology even existed. Fourier looked for solutions of the form $u(t,x,y,z)=T(t)X(x)$ and he rearranged the terms to conclude that $$ \frac{T'(t)}{T(t)} = \frac{X''(x)}{X(x)} $$ In order for there to exist such a solution, he deduced that, for fixed $t$, the right side would have to be constant, which would then force the left side to be a constant as well. So he introduce a separation parameter $\lambda$ which became an eigenvalue of $\frac{d}{dt}$, and of $\frac{d^2}{dx^2}$: $$ \frac{T'(t)}{T(t)}=\lambda,\;\;\; \lambda=\frac{X''(x)}{X(x)}. $$ Fourier then went on to try to find the most general solution by superimposing linear combinations of the possible eigenvalue solutions: \begin{align} u(t,x) &= e^{\lambda_1 t}(A_1\cos(\sqrt{\lambda_1}x)+B_1\sin(\sqrt{\lambda_1}x)) \\ &+e^{\lambda_2 t}(A_2\cos(\sqrt{\lambda_2}x)+B_2\sin(\sqrt{\lambda_2}x))+\cdots. \end{align}

Fourier's superposition was the precursor of the first general definition of linearity and a linear space, and it was first done here in the context of infinite-dimensional linear spaces of functions.

Fourier conjectured that solutions of such equations could be written in terms of such separated solutions, which is now translated into finding a basis of eigenvectors. The development of linearity, linear operators, selfadjoint operators, orthogonal eigenvector expansions, and eigenvector analysis evolved directly from the work of Fourier. And the infinite-dimensional problems of Fourier came before the finite-dimensional cases, making it even more confusing and obscure because its Historical context is lost when studying such analysis for the first time in finite-dimensional Linear Algebra.


Observe that $A = S \Lambda S^{-1}$, where $S = \begin{bmatrix}V_1 & V_2\end{bmatrix}$, $\Lambda = \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}$, and $S^{-1}=S^T$. Consequently, $$\frac{dY(t)}{dt}=AY(t) = S\Lambda S^{-1} Y(t). \tag{1}$$ Let $S^{-1}Y(t)=Z(t)$. Consequently, $(1)$ becomes $$\frac{dZ(t)}{dt}=\Lambda Z(t) \implies \begin{bmatrix}z_1'(t) \\ z_2'(t)\end{bmatrix}=\begin{bmatrix}\lambda_1 z_1(t)\\\lambda_2 z_2(t)\end{bmatrix} \implies Z(t)=\begin{bmatrix}k_1e^{\lambda_1t}\\k_2e^{\lambda_2t}\end{bmatrix}. \tag{2}$$ But $Y(t)=S Z(t)=\begin{bmatrix}V_1 & V_2\end{bmatrix}\begin{bmatrix}k_1e^{\lambda_1t}\\k_2e^{\lambda_2t}\end{bmatrix}=?$


$Y' = A Y\\ Y = e^{At}Y_0$

$e^{At} = \sum_\limits{n=0}^\infty \frac {A^nt^n}{n!}$

$A = PDP\\ A^n = PD^nP^{-1}$

$e^{At} = P\left(\sum_\limits{n=0}^\infty \frac {D^nt^n}{n!}\right)P^{-1}\\ Y = Pe^{Dt}P^{-1}Y_0$

$e^{Dt} = \begin{bmatrix} e^{\lambda_1 t}\\&e^{\lambda_2 t}\end{bmatrix}$

$PY_0 = \begin {bmatrix} k_1\\k_2\end{bmatrix}$

$Y = k_1V_1 e^{\lambda_1 t} + k_2V_2 e^{\lambda_2 t}$


The basic idea here is to find a new set of variables, $X(t)$ and $Y(t)$, related to the original variables $x(t)$ and $y(t)$ by a linear transformation, so that the differential equations for $X(t)$ and $Y(t)$ are decoupled: $$ \dot X = \lambda_1 X\, ,\qquad \dot Y=\lambda_2 Y\, . \tag{1} $$ In other words, assuming $$ \left(\begin{array}{c} x\\ y \end{array}\right)= M \left(\begin{array}{c} X\\ Y\end{array}\right) $$ where $M$ is invertible, we have \begin{align} \frac{d}{dt}\left(\begin{array}{c} x\\y \end{array}\right)&= A \left(\begin{array}{c} x \\ y\end{array}\right)\, ,\\ M \frac{d}{dt} \left(\begin{array}{c} X \\ Y\end{array}\right)&= A M \left(\begin{array}{c} X\\Y\end{array}\right)\, ,\\ \frac{d}{dt}\left(\begin{array}{c} X\\ Y\end{array}\right)&= M^{-1} A M \left(\begin{array}{c} X\\ Y\end{array}\right)\, .\tag{2} \end{align} Thus if $M^{-1} A M$ is diagonal with non-zero entries $\lambda_1,\lambda_2$, the system (2) is solved by (1), and $\lambda_1,\lambda_2$ are the eigenvalues of $A$ (by construction). The new coordinates $X$ and $Y$ are by construction eigenvectors of $A$. These new coordinates are “special” in that they have an especially simple evolution. Having solved in terms of the special coordinates $X,Y$, one can then go back to the original variable $x,y$ using $M$.


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