Calculus Diagonalization

Question

A = [-7 -0.25] Find an invertible matrix X and a diagonal matrix D such that X^-1AX=D

[48 1]

Answer 1

Hello Al G.,

To compute the diagonal matrix of A we must determine the Matrix P whose columns consist of the eigenvectors of A.

To get the eigenvectors, we must compute the eigenvalues first. To do this we compute the solutions of λ to the determinant of the following

|A-λ1|=0

where 1 is the unit matrix compatible with A. The determinant is

|A-λ1| = (-7-λ)(1-λ)-(48)(-.25) = -7+7λ-λ+λ²+12 = λ²+6λ+5 = (λ+5)(λ+1)

Thus the eigenvalues are λ=-1 and λ=-5

Recall that the eigenvectors are vectors where under the map of A, the result is a vector that is scaled by a quantity λ. To get the eigen vectors we go back to the matrix equation (A-λ1)X=0 and substitute λ for each of the eigenvalues we have obtained, then determine the row echelon form each time.

So for λ=-1 we get the augmented matrix form of the following matrix equation (A+1)X=0

[ -6 -.25 | 0 ]

[ 48 2 | 0 ]

Multiplying the top row by 8 and adding it to the bottom row gives

[ -6 -.25 | 0 ]

[ 0 0 | 0 ]

and making the top left unit gives

[ 1 1/24 | 0 ]

[ 0 0 | 0 ]

So for X^T=[a b] we have the resulting equation a+b/24=0, or b=-24a, so X^T=a [ 1 -24 ]

X^T= a [ 1 24 ] (written with transpose for simplicity)

We verify that this is indeed correct by substituting into the original matrix equation AX=-1X, which I will let you verify yourself rather than write it out. One last question might remain in your mind: "What is the little a" doing there? Well that just means that ANY multiple of X will work, this allows us to choose any size of

eigenvector we may want or need. Often in advanced courses we may need unit eigenvectors, so we will have to compute the magnitude of X and normalize X to get the unit eigenvector or normalized eigenvector. We may also need or want the eigenvectors to have special properties so that they might be right handed or left handed basis sets. Regardless of those technical details one last step remains.

We must now simply repeat this process for λ=-5, which gives the vector

X^T= s [ 1 -8 ]

for any value s.

GREAT, now we have out invertible matrix P

P = [ 1 1 ]

[ -24 -8 ]

and the diagonal matrix D relates via D=P^-1AP

I'll let you work out the matrix computation above, and just let you see the result

D=[ -1 0 ]

[ 0 -5 ]

So the diagonal matrix D which is composed of the eigenvalues of A along the diagonal. It is worth noting that if you included the 'unknown' or 'not determined' values multiplying each of the eigenvectors into the computation of D, the result matrix will be the same, but P will appear as follows.

P = [ a s ]

[ -24a -8s ]

Feel free to check that yourself if you'd like.