Let A = eginbmatrix 1 -2 8 0 -1 0 0 0...

Let A = egin{bmatrix} 1 & -2 & 8 \ 0 & -1 & 0 \ 0 & 0 & -1 end{bmatrix} and P = egin{bmatrix} 1 & -4 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 end{bmatrix} Confirm that P diagonalizes A, and then compute each of the following powers of A. (a) A^{1000} (b) A^{-1000} (c) A^{2301} (d) A^{-2301}

Transcript

0:00 Hello.

00:01 Okay, so here we have that our given matrix is the matrix 110 and then 110 again and then 0 -0 -0.

00:12 The last row.

00:15 Okay, so you know the characteristic equation of, this is a matrix a, the characteristic equation then is lambda i, where i is the identity, minus a is equal to 0.

00:25 So if we do the determinant of lambda i minus a, we get lambda minus 1, negative 1 0, negative 1, lambda minus 1, 0, and then 0 lambda.

00:35 So then we're going to get, well, lambda times, while the quantity, lambda minus 1 squared.

00:44 So it's going to be lambda times, well, lambda squared minus 2 lambda, which gives us, here.

01:02 So, which is lambda squared times lambda minus two.

01:15 Now, the determinant of lambda i minus a equals zero implies that lambda squared times lambda minus two is equal to zero.

01:24 This implies that lambda is equal to either well zero twice, so zero, zero, or two.

01:33 Okay, so therefore then we have the eigen values of a, our lambda one and two, is equal to zero, and lambda three is equal to 2.

01:41 So then we get that by definition, x is equal to the column vector, x1, x2, x3 is an eigenvector of a corresponding to lambda, if and only if x is a non -trivial solution of lambda i minus a times x is equal to zero.

01:58 So we get, if lambda equals zero, we get just the matrix, negative 1, negative 1, negative 1 -0 -0 -0 -0 times the column vector, x1, x2, x2, is equal to the column matrix 0 -00.

02:17 So we then get that the eigenvectors of a corresponding to a equals 0 of the non -zero vectors of the form.

02:26 We're going to get x is equal to negative s -st, which is equal to negative s -t plus 0 -0 -t, which is equal to s times the column vector, negative 1, plus t times the column vector zero zero one so if we let u1 then be equal to the column vector negative one one zero and u2 be equal to the column vector zero one um we have that u1 and u2 are going to be linearly independent and these vectors then from a basis of the eigenspace corresponding to lambda equal zero so then if we have v1 is equal to u1 we consider uh q1 which is going to be equal to v1 over the magnitude of v1.

03:26 That's going to be equal to, so 1 over root 2 times the column vector negative 1 -110.

03:33 So that gives us negative 1 over root 2, negative 1 over root 2, negative 1 over root 2, negative 1 over root 2.

03:48 0.

03:54 And then v2 is then by graham schmidt...

Question

Let A = egin{bmatrix} 1 & -2 & 8 \ 0 & -1 & 0 \ 0 & 0 & -1 end{bmatrix} and P = egin{bmatrix} 1 & -4 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 end{bmatrix} Confirm that P diagonalizes A, and then compute each of the following powers of A. (a) A^{1000} (b) A^{-1000} (c) A^{2301} (d) A^{-2301}

Please give Ace some feedback

Let A = egin{bmatrix} 1 & -2 & 8 \ 0 & -1 & 0 \ 0 & 0 & -1 end{bmatrix} and P = egin{bmatrix} 1 & -4 & 1 \ 1 & 0 & 0 \ 0 & 1 & 0 end{bmatrix} Confirm that P diagonalizes A, and then compute each of the following powers of A. (a) A^{1000} (b) A^{-1000} (c) A^{2301} (d) A^{-2301}