Jordan Canonical Form Homework Solutions For each of the following, put the matrix in Jordan canonical form and find the matrix S such that S AS = J. [ ]. A = A λi = λ λ = ( λ) = λ λ = λ =, Since we have two distinct eigenvalues for a matrix in M, we know that the Jordan form is [ ] J = To find the similarity transformation matrix S, we need to find the associated eigenvectors. For λ =, we have A I = [ ] [ ] Solving this gives the basis for the eigenspace associated with λ = to be {[ ]} For λ = we have [ ] [ ] Solving this gives the basis for the eigenspace associated with λ = to be {[ ]}
So, we have S = [ ] And so,. B = [ ] [ ] J = S AS = [ ] [ ] [ ] B λi = λ λ = ( λ) = λ = Notice that the algebraic multiplicity is here. Now, we find the basis for the eigenspace. B I = [ ] [ ] Solving this, we see that x is a free variable and y =, so we have that the basis for the eigenspace associated with λ = is {[ ]} Let this vector be v. We only have a dimension of for this eigenspace, so we need to find another vector, namely the generalized eigenvector of grade. This gives rise to the system of equations Av = v + v [ ] [ ] [ ] [ ] x x = + y y { x + y = + x y = y
Solving, we see that y = and x can be any real number, so let x =. This gives [ ] v = Since we have one eigenvalue but the associated eigenvalues of grade and, we have and And so, J = S = [ ] [ ] 3 3 3. C = 3 3 [ ] = [ J = S AS () ] [ ] [ ] () C λi = λ 3 3 3 λ 3 λ = ( λ)[(3 λ)( λ) + 6] 3( λ + 3) + 3( + 3 λ) = ( λ)( λ + λ ) 3( λ) + 3( λ) = λ(λ )( λ) = λ =,, Since we have 3 distinct eigenvalues, we know that J will be a diagonal matrix, i.e. A is diagonalizable. We now need to find the bases for the eigenspaces associated with the three eigenvalues. For λ = 3 3 C I = 3 3
Solving this gives the basis for the eigenspace associated with λ = to be For λ = : 3 3 3 C I = 3 3 Solving this gives the basis for the eigenspace associated with λ = to be And for λ = we have 3 3 3 C I = 3 4 Solving this gives the basis for the eigenspace associated with λ = to be So, since we have 3 distinct eigenvalues, we have J = and 3 S =
so that 4. D = = 3 3 3 3 3 3 3 D λi = λ λ λ = ( λ)[ λ( λ)] + ( λ) = J = S AS 3 ( λ)(λ λ + ) = λ =, Since we have multiplicity for the eigenvalue λ =, we need to find the dimension of the basis of the associated eigenspace to see if we have linearly independent vectors or if we need to find the generalized eigenvector of grade. We know it will be of grade since the multiplicity is. D I = Solving this gives the basis for the eigenspace associated with λ = to be Now, for λ =, we have D I = Solving this gives the basis for the eigenspace associated with λ = to be
Since we don t have the dimension we want, A is not diagonalizable. So, we need to find the generalized eigenvector of grade. This gives rise to the system of equations Dv = v + v x x y = + y z z x + y = + x x = + y z = z Since z =, we have x + y = to contend with. As long as this equation holds, we can select any values for x and y. So, let x = and y =. Then, we have the vector Putting this together, we have J = and S = so that J = S AS =
5. E = E λi = λ λ λ ( λ) 3 = λ = Since we have an algebraic multiplicity of 3 for the only distinct eigenvalue, we need to see what the dimension is for the basis of the eigenspace. E I = Solving this gives the basis for the eigenspace associated with λ = to be Since we only have a dimension of, we need to find the generalized eigenvectors of grades and 3. So, let v be the eigenvector we found above. Then, This becomes the system of equations Ev = v + v x x y = + y z z x + z = x x + y + z = + y z = z When solving, we see that any real number works for y so we can set y =. And, equations and 3 five that z = as well. The second equation gives that x =, so we have
v = Now, we have This becomes the system of equations Ev 3 = v + v 3 x x y = + y z z x + z = + x x + y + z = y z = z Similar to before, we see that y can be any real number, so we set y =. The first equation gives that z = and using these with the second equation, we see that x =. this gives the vector Putting this all together, we have and where v 3 = J = S = J = S ES =