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I'm having a hard time early on in this linear algebra course, I'm a first year student in University. I'm reading my textbook right now and it gives the following differential equation as an example with a solution and I still can't understand how to solve it:

$x'_1 = 3x_1 + x_2 + x_3 $
$x'_2 = 2x_1 + 4x_2 + 2x_3$
$x'_3 = -x_1 - x_2 + x_3$

It goes on to say that $x: R \rightarrow R^3$ and $x(t)$ is a matrix with one column of the $x_1(t)$, $x_2(t)$, $x_3(t)$.

Then it gives me a matrix which is just to coefficients of the above system of equations.

Then it gives me a Q and D such that $Q^{-1}AQ=D$ and $A=QDQ^{-1}$.

I have no clue where these matrices are obtained, they are just given to me in my textbook which is really frustrating. I feel like the people that wrote the textbook may have overlooked some steps that they find trivial but new students are just learning. Can someone please explain how to solve a differential equation like this?

Thanks

James S. Cook
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Sam Creamer
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    This will help: http://mathworld.wolfram.com/MatrixDiagonalization.html – voldemort Jan 21 '14 at 06:29
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    Also take into account that you may not need to know about differential equations to pass a course in linear algebra. If anything the example of differential equations shows you how linear algebra permeates many areas of mathematics. – OR. Jan 21 '14 at 06:32
  • True, but I really want to grasp the concept of eigenvectors and eigenvalues, and I still don't. Yes when I did my final last semester I grinded and knew how to solve for them, but to this day I do not understand what they do and I think it's holding my learning back a bit – Sam Creamer Jan 21 '14 at 06:34
  • When they give you $Q$ and $D$ they are indeed skipping a lot of nontrivial steps. Learning to find $Q$ and $D$ is one of the major goals of a linear algebra course. Perhaps this question is meant to motivate why learning how to diagonalize $A$ will turn out to be useful. – littleO Jan 21 '14 at 07:56

1 Answers1

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Instead of just a bunch of unrelated equations, it's useful to consider your system of equations as an equation involving a matrix and a vector. First take your three $x$'s and think of them as the components of a vector: $$ {\bf x} = \pmatrix{x_1(t)\cr x_2(t)\cr x_3(t)\cr}$$ The left sides of your equations, the derivatives of the $x$'s, can be thought of as the derivative of the vector: $$ {\bf x}' = \pmatrix{x_1'(t)\cr x_2'(t)\cr x_3'(t)\cr}$$ The right sides can be obtained by multiplying a matrix $A$ of constants with your vector: $$ \pmatrix{ 3x_1 + x_2 + x_3\cr 2x_1 + 4x_2 + 2x_3\cr -x_1 - x_2 + x_3\cr} = \pmatrix{3 & 1 & 1\cr 2 & 4 & 2\cr -1 & -1 & 1\cr} {\bf x}$$ So now your differential equations can be written in a nice compact way as ${\bf x}' = A {\bf x}$.

But now how to solve this? Well, the key observation is that if instead of the complicated matrix $A$ you had a diagonal matrix of constants $$ D = \pmatrix{d_1 & 0 & 0\cr 0 & d_2 & 0\cr 0 & 0 & d_3\cr}$$ things would be very easy, because you could just separately solve the three equations: $$ \eqalign{x_1' &= d_1 x_1 \cr x_2' &= d_2 x_2\cr x_3' &= d_3 x_3\cr}$$ The next insight is that (at least sometimes) you can transform your system to a diagonal system: if $D = Q^{-1} A Q$ is a diagonal matrix for some invertible matrix $Q$, then ${\bf y} = Q^{-1} {\bf x}$ satisfies $$ {\bf y}' = (Q^{-1} {\bf x})' = Q^{-1} {\bf x}' = Q^{-1} A {\bf x} = Q^{-1} A Q Q^{-1} {\bf x} = D {\bf y}$$ So you could solve this for ${\bf y}$, and then get back to ${\bf x}$ by ${\bf x} = Q {\bf y}$.

And so the next question is, how can I find those $D$ and $Q$? But I'll leave that part up to your linear algebra textbook...

Robert Israel
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