Page 362 - 35Linear Algebra

362 Sample Final Exam

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(h) When λ = µ and g/λ = R , we get the equation for a circle radius R in
the (z, w)-plane. When λ, µ and g are postive, we have the equation for
an ellipse. Vanishing g along with λ and µ of opposite signs gives a pair
of straight lines. When g is non-vanishing, but λ and µ have opposite
signs, the result is a pair of hyperbolæ. These shapes all come from
cutting a cone with a plane, and are therefore called conic sections.

13. We show that L is bijective if and only if M is invertible.
(a) We suppose that L is bijective.

i. Since L is injective, its kernel consists of the zero vector alone.
Hence
L = dim ker L = 0.
So by the Dimension Formula,

dim V = L + rank L = rank L.
Since L is surjective, L(V ) = W. Thus

rank L = dim L(V ) = dim W.

Thereby
dim V = rank L = dim W.
ii. Since dim V = dim W, the matrix M is square so we can talk
about its eigenvalues. Since L is injective, its kernel is the zero
vector alone. That is, the only solution to LX = 0 is X = 0 V .
But LX is the same as MX, so the only solution to MX = 0 is
X = 0 V . So M does not have zero as an eigenvalue.
iii. Since MX = 0 has no non-zero solutions, the matrix M is invert-
ible.
(b) Now we suppose that M is an invertible matrix.

i. Since M is invertible, the system MX = 0 has no non-zero solu-
tions. But LX is the same as MX, so the only solution to LX = 0
is X = 0 V . So L does not have zero as an eigenvalue.
ii. Since LX = 0 has no non-zero solutions, the kernel of L is the
zero vector alone. So L is injective.
iii. Since M is invertible, we must have that dim V = dim W. By the
Dimension Formula, we have

dim V = L + rank L

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