Rational Canonical Form

[aznmaven]

Hi, can someone show me how one would go about finding the matrix Q in Q^(-1) A Q = RationalCanonicalForm(A). Please demonstrate using the example

{4, 1, 2, 0}
{-4, 0, 1, 5} = A
{0, 0, 1, -1}
{0, 0, 1, 3}

where the characteristic polynomial is (x-2)^4 and the minimal polynomial is (x-2)^3. To save you guys some time, I've computed

{2, 1, 2, 0}
{-4, -2, 1, 5} = (A-2I)
{0, 0, -1, -1}
{0, 0, 1, 1}

{0, 0, 3, 3}
{0, 0, -6, -6} = (A-2I)^2
{0, 0, 0, 0}
{0, 0, 0, 0}

Thanks in advance guys!

 

[lippyka]

The matrix Q can be found by computing the eigenvectors of A. Since the minimal polynomial is (x-2)^3, we know that the eigenvalue of A is 2 and the algebraic multiplicity is 3. Therefore, we need to find three linearly independent eigenvectors corresponding to the eigenvalue 2. We can use the fact that (A-2I)^3 = 0 to solve for eigenvectors. First, let's set up the system of equations:(A-2I)v1 = 0(A-2I)v2 = v1(A-2I)v3 = v2The first equation gives us (A-2I)v1 = 0, which can be written as:{2, 1, 2, 0}{-4, -2, 1, 5}{0, 0, -1, -1}{0, 0, 1, 1} * {x1} = {0} {x2} {x3} {x4}Solving this system of equations gives us: x1 = -2x3 + x4x2 = -3x3 + 2x4Therefore, the first eigenvector is v1 = {-2, -3, 1, 2}Now, let's solve for v2 using the second equation: (A-2I)v2 = v1which can be written as: {2, 1, 2, 0}{-4, -2, 1, 5}{0, 0, -1, -1}{0, 0, 1, 1} * {x5} = {-2} {x6} = {-3} {x7} {x8}Solving this system of equations gives us: x5 = -2