`q001. If a matrix A has a full set of eigenvectors, then the matrix T, whose columns are the eigenvectors, will transform a vector v expressed with respect to the standard basis, to the same vector expressed with respect to the eigenvector basis. The matrix T^-1 will transform a vector in the eigenvector basis back to the standard basis.
The matrix A transforms each eigenvector to a multiple of itself. If lambda is the eigenvalue, then the multiple is lambda times the eigenvector.
So if we 'shift' to the eigenvector basis, our matrix is just D, the diagonal matrix with the eigenvalues on the diagonal.
It follows that
D = T^-1 * A * T.
Show how this works for the matrix A = [0, 2; -2, 0].
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`q002. Now suppose you have the differential equation
y ' = A y + g.
If A has a full set of eigenvectors, then we can form the T matrix.
Let y = T * z. Then y ' = T * z '.
What equation do we get if we substitute T * z for y?
Solve that equation for z.
Why will this equation be easy to solve?
How will we then get our function y?
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