Spring 97 Exam 2

(1) True or False

• False, consider a matrix with a 1 in the upper right and 0's elsewhere.
• False, consider projection onto the x-axis.
• False, trace = sum of eigenvalues, which are 1 or -1.
• True, since det is volume of parallelpiped.
• True, by Laplace expansion along first row.

(2) Add a copy of the last row of A_n to all the other rows of A_n. This does not change the determinant, and results is a matrix with 2's on or below the main diagonal (except for the last row of all 1's) and 0's above the main diagonal. The det of this lower diagonal matrix is the product of the diagonal entries, which is 2^(n-1).

(3) The eigenvalue .5 has algebraic and geometric multiplicity 1, and the eigenvalue 3 had algebraic and geometric multiplicity 2. For lambda=.5 an eigenvector is (1,1,5) and for lambda=3 eigenvectors are e_1 and e_2. These 3 eigenvectors form an eigen basis. The the first entry of A^t w is (.5)^t + 2 (3)^t, the second entry is (.5)^t + 3 (3)^t, and the third entry is 5 (.5)^t.

(4) If V is to be the kernel of the projection, then we are projecting onto V perpendicular. But we know the perpendicular space to the image of a matrix is just the kernel of the matrix's transpose. You can find this kernel to be spanned by (2,0,0,1) and (0,2,-1,0). The projection onto the lines through each vector, given by u u^T are, respectively,

4  0  0  2            0  0  0  0
0  0  0  0     and    0  4 -2  0
0  0  0  0            0 -2  1  0
2  0  0  1            0  0  0  0

4  0  0  2
0  4 -2  0
0 -2  1  0
2  0  0  1

(5)

• Since e_i is perpendicular to e_j for i not equal to j, the images of e_i and e_j are perpendicular. But these images are the columns of A.
• The dot product multipies out to e_i dot e_i - e_j dot e_j = ||e_i||^2 - ||e_j||^2 = 1 -1 = 0.
• Applying A to the dot product in the previous part, we get 0 = ||A e_i||^2 - ||A e_j||^2. So the columns of A must be of the same length.
• Let r be the length of any of the columns of A. Then dividing the columns of A by r results in a matrix with orthonormal columns, i.e. an orthogonal matrix. Clearly A is the product of this orthogonal matrix and r I.