Consider the matrix -6-1 1 1 1 A= 0 1 0 -2 Using Gershgorin Circle Theorem...

 

 

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Consider the matrix -6-1 1 1 1 A= 0 1 0 -2 Using Gershgorin Circle Theorem to determine the following. [a.] Does A have a dominant eigenvalue? Can we apply power methods to A? [b.] Does A have a dominant eigenvalue? Can we apply power methods to A-? [c.] Suppose we want to apply inverse iteration with shifts to find all the eigenvalues for the matrix A, give reasonable p for each case. Remark. Use the Gershgorin Circle Theorem, do not compute the eigenvalues. You will not get any credits if you compute the eigenvalues. 1. (Gershgorin Circle Theorem) In application, we do not have prior information of the eigenvalues for a matrix A. The following theorem is useful in estimating eigenvalues of a matrix A. Theorem (Gershgorin). Let the diagonal components of the matrix AER" be a11, a22,..., ann the absolute row sums of the off-diagonal part of A be ri=-1ji laj for 1i n, the absolute column sums of the off-diagonal part of A be c; = -1; |aj| for 1 j n. Then we have the following information on the eigenvalues of A using ri: i. The eigenvalues of A must be inside the union of the closed discs centered at ai with radius ri, for 1in. ii. Each ai is connected to one of the eigenvalues of A by a continuous curve that lies strictly inside the union of the closed discs in [i.], except for the endpoint (the eigenvalue) which may be on the boundary. Similarly we have the following information on the eigenvalues of A using cj: iii. The eigenvalues of A must be inside the union of the closed discs centered at ajj with radius cj, for 1 j n. iv. Each ajj is connected to one of the eigenvalues of A by a continuous curve that lies strictly inside the union of the closed discs in [iii.], except for the endpoint (the eigenvalue) which may be on the boundary. Gershgorin Circle Theorem is convenient to use. For example, consider the following ma- trix, 1 A 1 Here we have r = 1, r2 = 1. Thus by the theorem, we know our eigenvalues must be in the following two circles. Hence by [i.] and [ii.] we know that there must be one eigenvalue in each of the circle. Hence we have -2 0 and 22 4. Note that we cannot Consider the matrix -6-1 1 1 1 A= 0 1 0 -2 Using Gershgorin Circle Theorem to determine the following. [a.] Does A have a dominant eigenvalue? Can we apply power methods to A? [b.] Does A have a dominant eigenvalue? Can we apply power methods to A-? [c.] Suppose we want to apply inverse iteration with shifts to find all the eigenvalues for the matrix A, give reasonable p for each case. Remark. Use the Gershgorin Circle Theorem, do not compute the eigenvalues. You will not get any credits if you compute the eigenvalues. 1. (Gershgorin Circle Theorem) In application, we do not have prior information of the eigenvalues for a matrix A. The following theorem is useful in estimating eigenvalues of a matrix A. Theorem (Gershgorin). Let the diagonal components of the matrix AER" be a11, a22,..., ann the absolute row sums of the off-diagonal part of A be ri=-1ji laj for 1i n, the absolute column sums of the off-diagonal part of A be c; = -1; |aj| for 1 j n. Then we have the following information on the eigenvalues of A using ri: i. The eigenvalues of A must be inside the union of the closed discs centered at ai with radius ri, for 1in. ii. Each ai is connected to one of the eigenvalues of A by a continuous curve that lies strictly inside the union of the closed discs in [i.], except for the endpoint (the eigenvalue) which may be on the boundary. Similarly we have the following information on the eigenvalues of A using cj: iii. The eigenvalues of A must be inside the union of the closed discs centered at ajj with radius cj, for 1 j n. iv. Each ajj is connected to one of the eigenvalues of A by a continuous curve that lies strictly inside the union of the closed discs in [iii.], except for the endpoint (the eigenvalue) which may be on the boundary. Gershgorin Circle Theorem is convenient to use. For example, consider the following ma- trix, 1 A 1 Here we have r = 1, r2 = 1. Thus by the theorem, we know our eigenvalues must be in the following two circles. Hence by [i.] and [ii.] we know that there must be one eigenvalue in each of the circle. Hence we have -2 0 and 22 4. Note that we cannot