Question

Consider a 2×2 matrix

A=[ 2.00 0.00
0.00 2.00 ].
Find an invertible 2×2-matrix P and a diagonal 2×2-matrix D such that A=PDP⁻¹. In order to be accepted as correct, all entries of the matrix A−PDP⁻¹
must have absolute value smaller than 0.1 . Additional attempts available with new variants ?

Answer 1

To find an invertible 2x2-matrix P and a diagonal 2x2-matrix D such that A=PDP⁻¹, we need to find the eigenvalues and eigenvectors of matrix A.

Step-by-step explanation:

To find an invertible 2x2-matrix P and a diagonal 2x2-matrix D such that A=PDP⁻¹, we need to find the eigenvalues and eigenvectors of matrix A. The eigenvalues can be found by solving the characteristic equation det(A-λI) = 0, where I is the identity matrix and λ is the eigenvalue.

For matrix A=[2.00 0.00 0.00 2.00], we have det(A-λI) = (2-λ)(2-λ) - 0 = (2-λ)² = 0. Solving this equation, we get λ=2.

Now, to find the eigenvectors, we substitute the eigenvalue back into the equation (A-λI)v=0, where v is the eigenvector. Substituting λ=2 and solving the equation (A-2I)v=0, we get the eigenvector v=[1 0].

Since we have found the only eigenvalue and corresponding eigenvector, we can form matrix P by using the eigenvector as the first column. So, P=[1 0].

Finally, matrix D is a diagonal matrix with the eigenvalues on the diagonal, which in this case is D=[2 0; 0 2]. Therefore, we have PDP⁻¹ = [1 0] [2 0; 0 2] [1 0]⁻¹ = [2 0; 0 2].

To check if all entries of the matrix A−PDP⁻¹ have absolute value smaller than 0.1, we subtract the matrices and check each entry. In this case, all entries of A−PDP⁻¹ are 0, which have absolute value smaller than 0.1.