Let Tn be the tridiagonal matrix whose diagonal entries are all equal to 2 and whose sub- and su

Let Tn be the tridiagonal matrix whose diagonal entries are...

Key Concepts

Error Estimation

Error estimation in the context of the Power Method involves determining how many iterations are required to achieve a desired accuracy (e.g., four decimal places). This estimation is based on the convergence rate and typically requires calculating or bounding the ratio of the subdominant to dominant eigenvalues. The analysis helps in predicting the number of iterations needed for the approximated eigenvalue to reach a specific precision.

Dominant Eigenvalue

The dominant eigenvalue of a matrix is the eigenvalue with the largest magnitude. For symmetric matrices, it is also the largest eigenvalue. In iterative methods like the Power Method, the accuracy of the dominant eigenvalue's approximation improves as the contribution of the other eigenvalues diminishes relative to this dominant term.

Convergence Rate

The convergence rate of the Power Method is determined by the relative size of the subdominant eigenvalue compared to the dominant eigenvalue. Specifically, if ?1 is the dominant eigenvalue and ?2 is the second largest in magnitude, then the error in the approximation generally contracts roughly by a factor of |?2/?1| with each iteration. A smaller ratio means faster convergence.

Power Method

The Power Method is an iterative algorithm used to approximate the dominant (largest in absolute value) eigenvalue and its corresponding eigenvector of a matrix. It involves repeatedly multiplying a nonzero initial vector by the matrix and normalizing the result until the vector converges in direction. The convergence rate depends on the ratio of the absolute value of the second largest eigenvalue to that of the dominant eigenvalue.

Tridiagonal Matrix

A tridiagonal matrix is a matrix that has nonzero entries only on the main diagonal and the first sub- and super-diagonals. Such matrices often arise in numerical solutions of differential equations and are particularly useful due to their sparsity, which allows for efficient computational methods, including specialized eigenvalue solvers that take advantage of the structure.

Transcript

00:01 We are given a matrix a, and we are asked to determine all item values in the corresponding item vectors.

00:10 The matrix a is 0 -4 -4 -9 -4 -0 -2 -4 -2 -4 -2 -2 -2 -4 -2 -2 -2.

00:19 Notice that negative -a is equal to 0 -9 -4 -4 -402.

00:39 And notice that this is the same.

00:43 As a transpose.

00:47 Therefore, by definition, matrix a is a skew -symmetric matrix.

01:04 The characteristic polynomial of a is the determinant of a minus lambda i, which is equal to the determinants of the matrix negative lambda, 4, negative 4, negative lambda, negative 2, 4, 2, negative to make this determinant easier to solve.

01:49 I'm going to use some properties of determinants.

01:52 So you know that if you add a multiple of one row to another or a multiple of a common column to another, that the determinant stays the same.

02:02 So this is equal to the determinant of the matrix.

02:07 And i'm going to add row two to row three.

02:12 So it becomes negative lambda four, negative four, negative four, negative 4, negative lambda, negative 2, and then negative 4 plus 0, negative rana plus 2 minus lambda, and negative lambda plus negative 2 is negative minus lambda.

03:03 This time i'm going to add the second column, the third column.

03:09 So this becomes negative lambda, negative 0, 4, negative lambda, 2 minus lambda, and the last column becomes 4 plus negative 4 to 0.

03:24 Negative lambda plus negative 2 is negative 2 minus lambda.

03:30 And negative 2 minus lambda plus 2 minus lambda is negative lambda minus lambda or negative 2 lambda.

03:56 And now i'll evaluate the term.

04:00 I'm going to expand across the bottom row.

04:04 So we have negative 2 minus lambda times the determine.

04:22 Negative lambda 0, negative 4, negative 2 minus lambda, minus 2 lambda times the determinant of negative lambda 4, negative 4, negative lambda.

04:48 This is equal to negative 2 minus lambda times negative lambda times negative 2 minus lambda is lambda times 2 plus lambda or 2 lambda plus lambda squared minus 0 times 0 0 0 0000 times negative lambda times negative lambda is lambda squared minus 4 times negative 4 is plus 16 factoring out the lambda from both of these i get lambda times negative line that over them.

05:43 I get negative lambda times two minus lambda times two plus lambda plus two times lambda squared plus 16.

06:04 So two lambda squared plus 32.

06:14 This is equal to negative lambda times two randa squared minus lambda squared is just lambda squared.

06:30 Zero lambda.

06:33 And then four plus 32 is 36.

06:39 And the complex numbers, this factors as negative lambda times lambda minus 6i times lambda plus 6i.

06:59 Therefore, the eigen values, they are lambda 1 equals 0, lambda 2 equals negative 6i, and lambda 3 equals 6i.

07:31 To find all eigenvectors associated with these eigenvalues, suppose that b1 is an eigenvector associated with eigenvalue of landa 1, and it must satisfy the equation a minus lambda 1 i times b1 equals the zero vector.

07:59 The matrix a minus lambda 1 i is the same as the matrix a, so 04 -94, negative 4, 0 ,0, 2.

08:17 4 to 0.

08:24 And with gaussian elimination, we can reduce this matrix to, we'll add the third row to the second row.

08:35 So this becomes 0 -1, negative 1, negative 4 is 0, 0 plus 2 is 2, and then 0 plus negative 2 is negative 2.

08:52 This is the rule of 4, 2, 0.

09:07 And this reduces to the matrix 1.

09:13 1 1 1 1 half 0 0 1 negative 1 0 0 0 and if we subtract half of row 2 from row 1 we get the rolling becomes 1 0 and 0 minus 1 1 1⁄2 or 0 plus 1 1 half or 1 half which gives us the system of equations v1 1 plus 1 1 1 1 1 1 1 3 equals 0 and b12 minus 3 3 equals 0 and therefore if we take v13 to the parameter r and you get that the general iindector v3 has the form negative 1 1 half r r is equal to r times negative 1 1 half 1, 1.

10:53 And if we were to call this parameter r, making a change of variables to s, that r is equals 2s, and this is the same as s times negative 1 to 2, 2.

11:15 So now there are vectors in whole numbers.

11:18 So this is igan vector v1 in its general form, and defines the iden vector associated with the lambda 2, it will satisfy the equation a minus the lambda 2 i v2 equals the 0 vector.

11:46 The matrix a minus lambda 2 i is going to be the matrix a plus 6i i.

11:56 So this is the matrix a, but now it's diagonals, are going to be 6i.

12:21 And with gaussian elimination, we can write this matrix as we divide each row by two you get three i four three i two negative two negative two three i negative one and two one three i if i add row two to row three i and i divide row one by three i i get one two over three i negative two over three i negative two over three i negative two over three i negative two three i negative 1, 0, 3i plus 1, and then 3i plus negative 1, 3i minus 1.

13:58 If i add 2 of row 1, 1, 1 2 over 1, then i get 1, 2 over 3i is the same as the multiply top and bottom by i, maybe 2 3rds i.

14:18 Likewise, negative 2 thirds i, negative 2 3i, negative 2 over 3i is positive.

14:22 Two thirds i and then negative two plus two is zero three i plus two times negative two -thirds i it's three i minus four -thirds i which is nine i minus four i over three which is five i over three and the negative one plus two times negative or plus two times two -thirds i is negative one plus four thirds i and the third row remains the same dividing the second row by five -thirds i i get zero one and negative one plus four -thirds i divided by five -thirds i is negative one over five -thirds i which is negative three over five i plus four i over three divided by five i over 3, which is same as 4i over 3 times 3 over 5i.

17:09 Cancel out the eyes, cancel out the 3s, this becomes 4 fifths, and 3 over 5i is the same as 3i over 5 squared or negative 3i over 5.

17:29 So this is the same as positive 3i over 5.

17:34 So this is 4 fifth plus 3i over 5...

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