Look for the simplest possible motion: a solution that moves along a straight line through the origin, only stretching or shrinking as time goes on. Such a solution is a fixed vector times an exponential, and substituting it into the system gives a condition on the vector: the matrix must send it to a multiple of itself. That is exactly an eigenvector, and the multiple is its eigenvalue. Two independent eigenvectors give two straight-line solutions, and every solution is a combination of them.

Wind blowing along a straight valley. A leaf dropped on the valley floor is carried along the valley and never leaves it. The eigenvectors are the valleys of the matrix: directions in which it only pushes forwards or back.

$\mathbf{x}'=A\mathbf{x}$ with $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$. Along the eigenvector $(1,1)$ the solution $e^{3t}(1,1)$ moves straight out; along $(1,-1)$ the solution $e^{-t}(1,-1)$ moves straight in.

If $A\mathbf{v}=\lambda\mathbf{v}$ with $\mathbf{v}\ne\mathbf{0}$, then $\mathbf{x}(t)=e^{\lambda t}\mathbf{v}$ solves : a solution moving along the line through , outwards when and inwards when . The eigenvalues are the roots of . When they are real and different, the two eigenvectors are independent and every solution is the combination below, with its constants fixed by the starting vector.

• $A=\begin{pmatrix}1&2\\2&1\end{pmatrix}$: gives and , with eigenvectors and .

Differentiate the trial solution: the exponential brings down its rate, and the vector is constant. Apply the matrix instead: the exponential is a number and comes out, and the matrix acts on the vector. The two agree at every time exactly when the matrix sends the vector to the rate times itself — when the vector is an eigenvector and the rate its eigenvalue.

A fixed vector, stretched or shrunk exponentially.

The vector is constant, so only the exponential is differentiated.

The exponential is a number and comes out of the matrix product.

Dividing by the exponential, which is never zero, leaves the eigenvector equation.

If A sends v to λ times itself, then the exponential of λt times v solves the system: a point moving along the line through v.

Great work! You now know how to:

• find straight-line solutions from eigenvectors
• prove that an eigenvector gives a solution
• write the general solution for two real eigenvalues
• fit the constants to a starting vector