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February 18th, 2018, 10:25 AM   #1
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EigenValues and differential equations.. Please check my math

Suppose P is the projection matrix onto the 45◦ line y = x in R2. Its eigenvalues are 1 and 0 with eigenvectors (1, 1) and (1,−1). If dy/dt = −Py (notice minus sign) can you find the limit of y(t) at t = ∞ starting from y(0) = (3, 1)?

My Solution:

dt/dy = -Py, the eigenvalues are now -1 and 0 and eigenvectors $\displaystyle u_1 = (1, 1)$ and $\displaystyle u_2 = (-1, 1)$

f(t) = c_1 e^{-t} u_1 + c_2 e^{0t} u_2 \\
f(0) = c_1 u_1 + c_2 u_2 = (3, 1), So. c_1 = 2 ~and~ c_2 = 1 \\
f(t) = 2 e^{-t} u_1 + u_2 \\
f(∞) = u_2 = (-1, 1)

Am I correct?
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February 18th, 2018, 11:14 AM   #2
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You seem to have a good method but you seem to be choosing different eigenvectors in different steps leading to a trivial mistake.

If you set $u_2 = (-1,1)$, then your $c_2$ value is wrong. It looks like you computed it with $u_2 = (1,-1)$ instead. It actually doesn't matter which vector you pick but it must be consistent. If you leave your $c_1,c_2$ alone, this corresponds to picking the eigenvector $u_2 = (1,-1)$ which you have initially at the top.

In this case, you should get the trajectory limits to $(1,-1)$. You can justify this geometrically as well since $-P$ generates a vector field with all vectors parallel to the vector $(1,1)$. Thus, every trajectory can be computed explicitly by using basic algebra to find the intersection of lines. In this example, the trajectory through $(3,1)$ is a line with slope 1 so it is parameterized by the line $y = x-2$. The trajectory approaches the line $y = -x$ asymptotically with the collision at $t = \infty$ given by the intersection of these lines which is easily seen to occur when $x = 1$ and thus $y = -1$.
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