Exercise 6.8.1

Answers

1.
No. A quadratic form is a function of one variable. But a bilinear form is a function of two variables.
2.
No. We have 4I = (2I)tI(2I). But 4I and 2I has different eigenvalues.
3.
Yes. This is Theorem 6.34.
4.
No. See Example 5 of this section. The matrix (01 1 0 ) is a counterexample when 𝔽 = ℤ2.
5.
Yes. Let H1 and H2 be two symmetric bilinear forms. We have
(H1 + H2)(x,y) = H1(x,y) + H2(x,y)
= H1(y,x) + H2(y,x) = (H1 + H2)(y,x).
6.
No. The bilinear forms
H1(x,y) = xt ( 10 0 1 )y,H2(x,y) = xt ( 11 0 1 )y

have matrix representations (10 0 1 ) and (11 0 1 ) respectively with the standard basis. But both of their characteristic polynomials are (1 − t)2.

7.
No. We must have H(0,0) = 0 since H(x,0) is a linear function of x.
8.
No. It’s n2≠2n for n≠2 by the Corollary 1 after Theorem 6.32.
9.
Yes. Pick a nonzero element u ∈ V arbitrarily. If H(x,u) = 0, then we have y = u. Otherwise pick another nonzero element v ∈ V such that {u,v} is independent. Thus we have
y = H(x,v)u − H(x,u)v≠0.

But we have

H(x,y) = H(x,v)H(x,u) − H(x,u)H(x,v) = 0.
10.
No. It needs one more condition that H is symmetric. For example, the matrix (01 0 0 ) has its congruent matrix
Qt ( 01 0 0 )Q = (ac b d ) (01 0 0 ) (ab c d ) = (acbc ad bd ).

If that congruent matrix is diagonal, we should have bc = ad = 0. If a = b = 0 or a = c = 0, then Q is not invertible. Similarly, it cannot be d = c = 0 or d = b = 0. So this bilinear form is not even diagonalizable.

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2011-06-27 00:00
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