Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 201
So in the nontrivial case the minimal polynomial must be of degree at least one. A zero map or
matrix has minimal polynomial m(x) = x, and an identity map or matrix has minimal polynomial
m(x) = x − 1.
Five.IV.1.26 The polynomial can be read geometrically to say “a 60 ◦ rotation minus two rotations of
30 ◦ equals the identity.”
Five.IV.1.27
For a diagonal matrix
⎛
⎞
t 1,1 0
0 t 2,2 T = ⎜
⎝
. ..
⎟
⎠
t n,n
the characteristic polynomial is (t 1,1 − x)(t 2,2 − x) · · · (t n,n − x). Of course, some of those factors may
be repeated, e.g., the matrix might have t 1,1 = t 2,2 . For instance, the characteristic polynomial of
⎛ ⎞
3 0 0
D = ⎝0 3 0⎠
0 0 1
is (3 − x) 2 (1 − x) = −1 · (x − 3) 2 (x − 1).
To form the minimal polynomial, take the terms x − t i,i , throw out repeats, and multiply them
together. For instance, the minimal polynomial of D is (x − 3)(x − 1). To check this, note first
that Theorem 1.8, the Cayley-Hamilton theorem, requires that each linear factor in the characteristic
polynomial appears at least once in the minimal polynomial. One way to check the other direction —
that in the case of a diagonal matrix, each linear factor need appear at most once — is to use a matrix
argument. A diagonal matrix, multiplying from the left, rescales rows by the entry on the diagonal.
But in a product (T − t 1,1 I) · · · , even without any repeat factors, every row is zero in at least one of
the factors.
For instance, in the product
⎛
(D − 3I)(D − 1I) = (D − 3I)(D − 1I)I = ⎝ 0 0 0
⎞ ⎛
0 0 0 ⎠ ⎝ 2 0 0
⎞ ⎛
0 2 0⎠
⎝ 1 0 0
⎞
0 1 0⎠
0 0 −2 0 0 0 0 0 1
because the first and second rows of the first matrix D − 3I are zero, the entire product will have a
first row and second row that are zero. And because the third row of the middle matrix D − 1I is zero,
the entire product has a third row of zero.
Five.IV.1.28 This subsection starts with the observation that the powers of a linear transformation
cannot climb forever without a “repeat”, that is, that for some power n there is a linear relationship
c n · t n + · · · + c 1 · t + c 0 · id = z where z is the zero transformation. The definition of projection is
that for such a map one linear relationship is quadratic, t 2 − t = z. To finish, we need only consider
whether this relationship might not be minimal, that is, are there projections for which the minimal
polynomial is constant or linear?
For the minimal polynomial to be constant, the map would have to satisfy that c 0 · id = z, where
c 0 = 1 since the leading coefficient of a minimal polynomial is 1. This is only satisfied by the zero
transformation on a trivial space. This is indeed a projection, but not a very interesting one.
For the minimal polynomial of a transformation to be linear would give c 1 · t + c 0 · id = z where
c 1 = 1. This equation gives t = −c 0 · id. Coupling it with the requirement that t 2 = t gives
t 2 = (−c 0 ) 2 · id = −c 0 · id, which gives that c 0 = 0 and t is the zero transformation or that c 0 = 1 and
t is the identity.
Thus, except in the cases where the projection is a zero map or an identity map, the minimal
polynomial is m(x) = x 2 − x.
Five.IV.1.29 (a) This is a property of functions in general, not just of linear functions. Suppose
that f and g are one-to-one functions such that f ◦ g is defined. Let f ◦ g(x 1 ) = f ◦ g(x 2 ), so that
f(g(x 1 )) = f(g(x 2 )). Because f is one-to-one this implies that g(x 1 ) = g(x 2 ). Because g is also
one-to-one, this in turn implies that x 1 = x 2 . Thus, in summary, f ◦ g(x 1 ) = f ◦ g(x 2 ) implies that
x 1 = x 2 and so f ◦ g is one-to-one.
(b) If the linear map h is not one-to-one then there are unequal vectors ⃗v 1 , ⃗v 2 that map to the same
value h(⃗v 1 ) = h(⃗v 2 ). Because h is linear, we have ⃗0 = h(⃗v 1 ) − h(⃗v 2 ) = h(⃗v 1 − ⃗v 2 ) and so ⃗v 1 − ⃗v 2 is
a nonzero vector from the domain that is mapped by h to the zero vector of the codomain (⃗v 1 − ⃗v 2
does not equal the zero vector of the domain because ⃗v 1 does not equal ⃗v 2 ).