Linear Algebra Exercises-n-Answers.pdf
Linear Algebra Exercises-n-Answers.pdf Linear Algebra Exercises-n-Answers.pdf
200 Linear Algebra, by Hefferon Because it is triangular, the fact that the characteristic polynomial is c(x) = (x − 1) 4 is clear. For the minimal polynomial, the candidates are m 1 (x) = (x − 1), ⎛ ⎞ 0 1 1 1 T − 1I = ⎜0 0 2 3 ⎟ ⎝0 0 0 3⎠ 0 0 0 0 m 2 (x) = (x − 1) 2 , m 3 (x) = (x − 1) 3 , ⎛ ⎞ 0 0 2 6 (T − 1I) 2 = ⎜0 0 0 6 ⎟ ⎝0 0 0 0⎠ 0 0 0 0 ⎛ ⎞ 0 0 0 6 (T − 1I) 3 = ⎜0 0 0 0 ⎟ ⎝0 0 0 0⎠ 0 0 0 0 and m 4 (x) = (x − 1) 4 . Because m 1 , m 2 , and m 3 are not right, m 4 must be right, as is easily verified. In the case of a general n, the representation is an upper triangular matrix with ones on the diagonal. Thus the characteristic polynomial is c(x) = (x−1) n+1 . One way to verify that the minimal polynomial equals the characteristic polynomial is argue something like this: say that an upper triangular matrix is 0-upper triangular if there are nonzero entries on the diagonal, that it is 1-upper triangular if the diagonal contains only zeroes and there are nonzero entries just above the diagonal, etc. As the above example illustrates, an induction argument will show that, where T has only nonnegative entries, T j is j-upper triangular. That argument is left to the reader. Five.IV.1.19 The map twice is the same as the map once: π ◦ π = π, that is, π 2 = π and so the minimal polynomial is of degree at most two since m(x) = x 2 − x will do. The fact that no linear polynomial will do follows from applying the maps on the left and right side of c 1 · π + c 0 · id = z (where z is the zero map) to these two vectors. ⎛ ⎞ ⎛ ⎞ Thus the minimal polynomial is m. ⎝ 0 0⎠ 1 ⎝ 1 0⎠ 0 Five.IV.1.20 This is one answer. ⎛ ⎝ 0 0 0 ⎞ 1 0 0⎠ 0 0 0 Five.IV.1.21 The x must be a scalar, not a matrix. Five.IV.1.22 The characteristic polynomial of ( ) a b T = c d is (a − x)(d − x) − bc = x 2 − (a + d)x + (ad − bc). Substitute ( ) 2 ( ) ( ) a b a b 1 0 − (a + d) + (ad − bc) c d c d 0 1 ( ) ( a = 2 + bc ab + bd a ac + cd bc + d 2 − 2 + ad ab + bd and just check each entry sum to see that the result is the zero matrix. ac + cd ad + d 2 ) + ( ad − bc 0 ) 0 ad − bc Five.IV.1.23 By the Cayley-Hamilton theorem the degree of the minimal polynomial is less than or equal to the degree of the characteristic polynomial, n. Example 1.6 shows that n can happen. Five.IV.1.24 Suppose that t’s only eigenvalue is zero. Then the characteristic polynomial of t is x n . Because t satisfies its characteristic polynomial, it is a nilpotent map. Five.IV.1.25 A minimal polynomial must have leading coefficient 1, and so if the minimal polynomial of a map or matrix were to be a degree zero polynomial then it would be m(x) = 1. But the identity map or matrix equals the zero map or matrix only on a trivial vector space.
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 ).
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200 <strong>Linear</strong> <strong>Algebra</strong>, by Hefferon<br />
Because it is triangular, the fact that the characteristic polynomial is c(x) = (x − 1) 4 is clear. For the<br />
minimal polynomial, the candidates are m 1 (x) = (x − 1),<br />
⎛ ⎞<br />
0 1 1 1<br />
T − 1I = ⎜0 0 2 3<br />
⎟<br />
⎝0 0 0 3⎠<br />
0 0 0 0<br />
m 2 (x) = (x − 1) 2 ,<br />
m 3 (x) = (x − 1) 3 ,<br />
⎛<br />
⎞<br />
0 0 2 6<br />
(T − 1I) 2 = ⎜0 0 0 6<br />
⎟<br />
⎝0 0 0 0⎠<br />
0 0 0 0<br />
⎛<br />
⎞<br />
0 0 0 6<br />
(T − 1I) 3 = ⎜0 0 0 0<br />
⎟<br />
⎝0 0 0 0⎠<br />
0 0 0 0<br />
and m 4 (x) = (x − 1) 4 . Because m 1 , m 2 , and m 3 are not right, m 4 must be right, as is easily verified.<br />
In the case of a general n, the representation is an upper triangular matrix with ones on the diagonal.<br />
Thus the characteristic polynomial is c(x) = (x−1) n+1 . One way to verify that the minimal polynomial<br />
equals the characteristic polynomial is argue something like this: say that an upper triangular matrix<br />
is 0-upper triangular if there are nonzero entries on the diagonal, that it is 1-upper triangular if the<br />
diagonal contains only zeroes and there are nonzero entries just above the diagonal, etc. As the above<br />
example illustrates, an induction argument will show that, where T has only nonnegative entries, T j<br />
is j-upper triangular. That argument is left to the reader.<br />
Five.IV.1.19 The map twice is the same as the map once: π ◦ π = π, that is, π 2 = π and so the<br />
minimal polynomial is of degree at most two since m(x) = x 2 − x will do. The fact that no linear<br />
polynomial will do follows from applying the maps on the left and right side of c 1 · π + c 0 · id = z<br />
(where z is the zero map) to these two vectors.<br />
⎛ ⎞ ⎛ ⎞<br />
Thus the minimal polynomial is m.<br />
⎝ 0 0⎠<br />
1<br />
⎝ 1 0⎠<br />
0<br />
Five.IV.1.20 This is one answer. ⎛<br />
⎝ 0 0 0<br />
⎞<br />
1 0 0⎠<br />
0 0 0<br />
Five.IV.1.21 The x must be a scalar, not a matrix.<br />
Five.IV.1.22<br />
The characteristic polynomial of<br />
( )<br />
a b<br />
T =<br />
c d<br />
is (a − x)(d − x) − bc = x 2 − (a + d)x + (ad − bc). Substitute<br />
( ) 2 ( )<br />
( )<br />
a b<br />
a b<br />
1 0<br />
− (a + d) + (ad − bc)<br />
c d<br />
c d<br />
0 1<br />
( ) (<br />
a<br />
=<br />
2 + bc ab + bd a<br />
ac + cd bc + d 2 −<br />
2 + ad ab + bd<br />
and just check each entry sum to see that the result is the zero matrix.<br />
ac + cd ad + d 2 )<br />
+<br />
(<br />
ad − bc 0<br />
)<br />
0 ad − bc<br />
Five.IV.1.23 By the Cayley-Hamilton theorem the degree of the minimal polynomial is less than or<br />
equal to the degree of the characteristic polynomial, n. Example 1.6 shows that n can happen.<br />
Five.IV.1.24 Suppose that t’s only eigenvalue is zero. Then the characteristic polynomial of t is x n .<br />
Because t satisfies its characteristic polynomial, it is a nilpotent map.<br />
Five.IV.1.25 A minimal polynomial must have leading coefficient 1, and so if the minimal polynomial<br />
of a map or matrix were to be a degree zero polynomial then it would be m(x) = 1. But the identity<br />
map or matrix equals the zero map or matrix only on a trivial vector space.
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