Certainly! To show that two matrices A and B have the same characteristic polynomial, we need to demonstrate that their eigenvalues are identical. The characteristic polynomial of a matrix A is given by det(A - λI), where λ is the eigenvalue and I is the identity matrix.
Given the characteristic polynomial P(A) = A^2 - 3A + 4, we'll first find its eigenvalues by solving the equation P(λ) = 0:
A^2 - 3A + 4 = 0
Now, let's solve for the eigenvalues λ:
λ^2 - 3λ + 4 = 0
Using the quadratic formula:
λ = (3 ± √((-3)^2 - 4*1*4)) / 2*1
λ = (3 ± √(9 - 16)) / 2
λ = (3 ± √(-7)) / 2
Since the characteristic polynomial results in complex roots, it indicates that the matrix A has complex eigenvalues.
Now, let's investigate the matrix B, where B is also given by the characteristic polynomial P(B) = B^2 - 3B + 4. We'll follow the same steps to find its eigenvalues λ:
B^2 - 3B + 4 = 0
λ^2 - 3λ + 4 = 0
Again, using the quadratic formula:
λ = (3 ± √((-3)^2 - 4*1*4)) / 2*1
λ = (3 ± √(9 - 16)) / 2
λ = (3 ± √(-7)) / 2
Notice that the eigenvalues for matrix A and matrix B are the same complex values. Therefore, matrix A and matrix B have the same characteristic polynomial, which is P(λ) = λ^2 - 3λ + 4.
