I have labeled where the confusion of the solution is. I don't need an answer to the problem (also attached). I just need an explanation on how did we get this matrix that I've added a note in red?...

I have labeled where the confusion of the solution is. I don't need an answer to the problem (also attached). I just need an explanation on how did we get this matrix that I've added a note in red? from where did the [ 2/3  1/3 ... ] came?? what was done in order to get it? Thank you :)

Extracted text: -[- 2 -1 Let A 3 (a) Find the eigenvalues and corresponding eigenvectors of A. (b) Find a nonsingular matrix P such that D = P-' AP is diagonal.


Extracted text: [ 2 -1 1. Let A -2 Observe first that t – 2 1 Aa(t) = | - = (t – 2)(t – 3) – 2 = t2 – 5t + 6 – 2 = t? – 5t + 4 = (t – 4)(t – 1). %3D | t – 3 (a) Eigenvalues and corresponding eigenvectors of A. Since A4(t) = (t – 4)(t – 1), the eigenvalues areA= 1, 1 = 4. - - [ 1 with echelon -1 The eigenspace for A = 1 is the null space of the matrix 2 -2 -1 1 form A vector is in E if and only if -x + y = 0, so y = x. Hence 0 0 {[:] B, - {{:} E1 | x ER and a basis for E1 is B1 2 1 The eigenspace for A= 4 is the null space of the matrix with echelon form 1 A vector is in E4 if and only if 2x + y = 0, so y = -2x. Hence 0 0 {[ }} 1 E4 |x ER -2x and a basis for E4 is B4 -2 (b) Nonsingular matrir P such that D = P-'AP is diagonal. [ 1 1 Using the vectors in the bases for E1 and E4, we have P 1 -2 How 222 2/3 1/3 hence P-1 and 1/3 -1/3 2/3 1/3 2 -1 1 D = P-'AP 1/3 -1/3 ] 1 -2 -2 3 1 0 0 4 2/3 1/3 1 4 = D. 1/3 -1/3 1 -8 (c) f(A), where f(t) = t – 4t3 + 3t – 7. We have D = P-'AP, so A = PDP-l and f(A) = f(PDP-1) = Pf(D)P-1. (See diagonal factorization, text page 298.) Next, 4 1 0 0 4 1 1 1 0 0 4 1 0 +3 0 4 1 0 0 1 f(D) -7 3 + 0 -256 1 -4 0 256 + 0 12 -7 -7 0 0 5 1- 4+3 - 7 256 – 256 + 12 – 7 and so 1/3 [ -7 0 0 5 2/3 f(A) = Pf(D)P-1 1 -2 1/3 -1/3 -7 5 2/3 1/3 -3 -4 = f(A). -7 -10 1/3 -1/3 -8 1