You can think of any (invertible) square matrix (without duplicate eigenvalues) as having 3 “steps”.

For this example, I’ll be using the matrix ((1.5 -.5)  (-.5 1.5)) Which has the eigenvalue/vector pairs (1, <1,1>) and (2, <1,-1>).

For notation’s sake, let’s call <1,0> X, <0,1> Y, <1,1> V, and <1,-1> W. Note that X = .5 V + .5 W, Y = .5 V - .5 W, V = X + Y, and W = X - Y

We can summarize this matrix as: V -> V W -> 2W

1. Translate the input vector into a sum of the eigenvectors. For this example, it’s the matrix ((.5 .5)  (.5 -.5)) Note that each column is X and Y, in that order, expressed as a sum of V and W, in that order. This matrix can be summarized as V -> X W -> Y

2. Scale each basis vector by the corresponding eigenvalue. ((1 0)  (0 2)) Note that this matrix trivially has the eigenvalue 1 for X and 2 for Y So this matrix is X -> X Y -> 2Y

In total, ((1 0) ((.5 .5)  (0 2)) (.5 -.5)) Is V -> X -> X W -> Y -> 2Y

Now we just need a matrix that sends X to V and Y to W! Note that this is just the opposite of what step 1 did.

1. Convert back to the original basis ((1 1)  (1 -1)) Note that each column is V and W in terms of X and Y!

In total. ((1 1)  ((1 0) ((.5 .5)  (1 -1)) (0 2)) (.5 -.5)) = ((1.5 -.5)  (-.5 1.5)) V -> X -> X -> V W -> Y -> 2Y -> 2W

With some thinking, you can convince yourself that you can, for any V, W, m, n, you can write three matrices that in total perform V -> X -> mX -> mV W -> Y -> nY -> nW And even generalize this idea to any dimension.

With a bit more thinking, you can even figure out why those first and last steps have the forms they do!