8/7/2019 EndsemLA_soln

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Part I Linear Algebra

1. Given: B is a diagonal matrix with distinct eigenvalues.

To prove: If  AB = BA then A is diagonlizable.

Proof:

(a) Step 1: Since B has distinct eigenvalues, B must be diagonalizable.

(b) Step 2: Let QBQ−1 = D where D is a diagonal matrix.

(c) Step 3: QAQ−1QBQ−1 = QBQ−1QAQ−1. Put Y  := QAQ−1.

(d) Step 4: Now, Y D = DY , D being diagonal.(e) Step 5: The above equation implies that (di − d j)yij = 0, where di and d j

are the diagonal entries of  D and Y  = [yij].

(f) Step 6: Eigenvalues of  B and QBQ−1 = D must be same, and since theyare distinct, di = d j. So, yij = 0 if i = j.

(g) Step 7: Since the off diagonal entries of  Y  are zero, Y  must be diagonal.Now since QAQ−1 = Y  and Y  is diagonal, A is diagonalizable,

Marking scheme: Up to step 2, 1 mark; and step 4 1 mark. Possible marks: 0,1,2, and 4.

2. Answer is FALSE.

(a) Step 1: Suppose there is a B which is symmetric such that A = B52. Notethat −1 is an eigenvalue of  A.

(b) Step 2: Let λ1, . . . , λ4 be the eigenvalues of B. Then λ52

1, λ52

2, λ52

3and λ4

52

are the eigenvalues of  A.

(c) Step 4: Thus, λ52

i = −1 for some i. This means that there is a real numberx such that x52 + 1 = 0 which is a contradiction.

Marking scheme: Step 1 carries 1 mark and step 2 carries 1 mark. Possiblemarks: 0,1,2 and 4.

3. Apply Gram-Schmidt process: One solution is:

{1/2(1, 1, 1, 1)T ,

√4√3

(−1/4,−1/4,−1/4, 3/4)T }.