PROBLEM 5

(a) T  We have  B = S' A S  where S is some invertible matrix and S' is
       the inverse of A.  Therefore B is the product of invertible
       matrices, and so is itself invertible.
	
(b) T  Pythagoras: 3*3 + 4*4 = 5*5  so the dot product of x with y is zero.

(c) T  we have  v = Au  for some vector u.  Therefore v.w = Au.w = u.Aw
       (because A is symmetric), and Aw=0 (w is in the kernel of A), so
       u.Aw = 0.

(d) F  The columns of an orthogonal matrix are unit vectors; therefore
       all the entries have absolute value at most 1.

(e) F  Similar matrices have the same determinant, but  det(3A) = 27 det(A),
       so det(A)=0 and A is not invertible.

(f) F  Any real eigenvalue of an orthogonal matrix is either 1 or -1
       because  ||Av|| = ||v||  for all vectors v.  (This is Example 4
       of 7.1 in the textbook, page 299.)

(g) T  The sum of the dimensions of the two eigenspaces is n
       (where A is an n-by-n matrix), so the geometric multiplicities
       add up to at least n, hence exactly n.

(h) F  If Av = 2v  then  0 = (AA - 4A + 3I) v = (2*2 - 4 + 3) v = 3v,
       so v=0.

(i) F  For example, A could be the non-diagonalizable matrix

         [ 1  1  0 ]
         [ 0  1  0 ]
         [ 0  0  2 ]

(j) F  Once we have n eigenvalues of an n-by-n matrix, we have them all,
       and the trace is just their sum.  Here n=6 and all the eigenvalues
       are positive, so the trace must be positive as well.