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Suppose we have a matrix A € Rmxn. Recall the Golub-Kahan bidiagonalisation pro- cedure and the Lawson-Hanson-Chan (LHC) bidiagonalisation procedure (Section 8. 2). Answer the following questions (5 marks each): (i) Workout the operation counts required by the Golub-Kahan bidiagonalisation. (ii) Workout the operation counts required by the LHC bidiagonalisation. (iii) Using the ratio , derive and explain under what circumstances the LHC is com- putationally more advantageous than the Golub-Kahan. (iv) Suppose we have a bidiagonal matrix B E Rnxn, show that both BTB and BBT are tridiagonal matrices. Hint: recall that the operation counts of the QR factorisation (using Householder reflec- tion) is about 2mn² - 3n³. You can relate those two bidiagonalisation procedures to the QR factorisation to work out their operation counts

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Answer:

(i) Operation counts required by the Golub-Kahan bidiagonalization:

The Golub-Kahan bidiagonalization procedure can be broken down into two steps:

1. Bidiagonalization of A using Householder reflections.

2. Reduction of the bidiagonal matrix to a diagonal matrix using QR iterations.

For the first step, the operation count is approximately 2mn² - 2n³. This is because the bidiagonalization process requires m Householder reflections for the rows and n Householder reflections for the columns, each involving approximately 2n operations.

For the second step, the operation count is approximately 8n³. This is because the reduction of the bidiagonal matrix to a diagonal matrix using QR iterations requires n-1 iterations, and each iteration involves approximately 8n operations.

Therefore, the total operation count for the Golub-Kahan bidiagonalization is approximately 2mn² + 6n³.

(ii) Operation counts required by the Lawson-Hanson-Chan (LHC) bidiagonalization:

The LHC bidiagonalization procedure can also be broken down into two steps:

1. Bidiagonalization of A using Householder reflections.

2. Reduction of the bidiagonal matrix to a diagonal matrix using singular value decomposition (SVD).

For the first step, the operation count is the same as in the Golub-Kahan bidiagonalization, which is approximately 2mn² - 2n³.

For the second step, the operation count is approximately 12n²(m - n) + 12n³. This is because the SVD involves finding the eigenvalues and eigenvectors of the bidiagonal matrix, which requires approximately 12n²(m - n) operations, and then constructing the singular value matrix, which requires approximately 12n³ operations.

Therefore, the total operation count for the LHC bidiagonalization is approximately 2mn² - 2n³ + 12n²(m - n) + 12n³.

(iii) The ratio between the operation counts of LHC and Golub-Kahan bidiagonalization is given by:

Ratio = (2mn² - 2n³ + 12n²(m - n) + 12n³) / (2mn² + 6n³)

Simplifying the expression, we get:

Ratio = (m + 6n) / (1 + 3n/m)

The LHC bidiagonalization is computationally more advantageous than the Golub-Kahan bidiagonalization when the ratio (m + 6n) / (1 + 3n/m) is smaller. This means that the LHC method is more efficient when the value of m (number of rows) is significantly larger than n (number of columns), or when the value of n/m is small.

(iv) To show that both BTB and BBT are tridiagonal matrices, we need to consider the structure of a bidiagonal matrix B.

A bidiagonal matrix B has nonzero entries only on the main diagonal and the first superdiagonal. Let's denote the nonzero elements on the main diagonal as di and the nonzero elements on the first superdiagonal as ei.

For BTB, the product of B with its transpose, the resulting matrix will have nonzero elements only on the main diagonal and the first two superdiagonals. The diagonal elements of BTB will be the squares of the diagonal elements of B (di^2), and the superdiagonal elements will be the products of adjacent diagonal and superdiagonal elements of B (di * ei). All other elements will be zero.

Similarly, for BBT, the resulting matrix will have nonzero elements only on the main diagonal and the first two subdiagonals. The diagonal

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