Hoffman, Kunze. Linear algebra (2ed, PH, 1971)(T)(415s).djvu: eKnigu

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Hoffman, Kunze. Linear algebra (2ed, PH, 1971)(T)(415s).djvu

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True or false? If the triangular matrix A is similar to a diagonal matrix, then
A is already diagonal...
Therefore, we may apply the
lemma before Theorem 5 to that operator and the invariant subspace W...
When we began this chapter, we
described our goal this way: To find an ordered basis in which the matrix
of T assumes an especially simple form...
Now suppose that we are given a linear operator T along with distinct
scalars c,- and non-zero operators Ei which satisfy (i), (ii) and (iii)...
This last comment,
together with F-16), F-17), and the fact that the a are distinct enables us
to apply Theorem 11 to conclude that T is diagonalizable...
What can you say about T?
9. Let V be the vector space of continuous real-valued functions on the interval
[—1, 1] of the real line...
If D denotes the differentia-
tion operator and p is the polynomial
p = xn + an-\xn~x + • • • + aix + Oo
then V is the null space of the operator p(D), because F-18) simply says
p(D)f = 0...
Returning to the differential equation F-18), which is
p(D)f = 0
p = (x - ci)ri • • • (a; - Ck)n
we see that the n functions Vet*1, 0 < m < r,- — 1, 1 < j < k, form a
basis for the space of solutions to F-18)...
Furthermore, if h is a polynomial of degree
less than k, we cannot have h{U) = 0, for then h(U)a = h{T)a = 0,
contradicting the definition of pa...
One of the consequences of Theo-
rem 3 will be the converse, so that admissibility characterizes those
invariant subspaces which have complementary invariant subspaces...
If
f/3 = ft + S gift, ft m Wi
f divides each polynomial gi and ft, = f70, where 70 ts in Wo-
284 The Rational and Jordan Forms Chap...
Let us begin with the very simple
case in which A is the companion matrix G.2) of a monic polynomial
P = Xn + Cn-\Xn~X + ¦ ¦ ¦ + C\X + Co...
As a matter of fact, in Chapter 5 we used the classical
adjoint to show that (i) and (ii) are equivalent...
For which matrices A is j\ ^ 1?
4. Construct a linear operator T with minimal polynomial x\x — IJ and charac-
teristic polynomial x3(x — IL...
An inner product on V is a function
which assigns to each ordered pair of vectors a, /3 in V a scalar (a|/3) in F in
such a way that for all a, /3, 7 in V and all scalars c
(a) (a + 0\y) = (a|7) + (|8|7);
(b) (ca|0) = c(a||8);
(c) (j3|a) = (a|J3), </ie bar denoting complex conjugation;
(d) (a|a) >0jJa^0...
When Q is the identity matrix, this inner product is essentially the same
as that in Example 1; we call it the standard inner product on FnX1...
, an}
is an ordered basis for V, and that we are given a particular inner product
on F; we shall show that the inner product is completely determined by
the values
(8-5) Gjk = (at|«y)
it assumes on pairs of vectors in (B...
By looking at this problem in R2 or in R3, one sees intuitively that a
best approximation to /3 by vectors in W ought to be a vector a in IF such
that j8 — a is perpendicular (orthogonal) to IF and that there ought to
284 Innw Product Spaces Chap...
Hence
It now suffices to prove that
IO8|
Oik)
This is straightforward computation in which one uses the fact that
(aj\ak) = Oforj 9*k...
Then {Tf\Tg) = [f\g], and so T preserves inner products; however, T is
not an isomorphism of V onto W, because the range of T is not all of W...
Since A* A = I if and only if A A* = /, we see that A is unitary
exactly when the rows of A comprise an orthonormal set of n-tuples in Cn
(with the standard inner product)...
The statement that a\ is a characteristic
vector for T simply means that W is invariant under T...