Exercise 3.7.1
Let be a field and let be the linear functional on defined by . For each of the following linear operators , let , and find .
(a) ;
(b) ;
(c) .
Solution:
(a) .
(b) .
(c) We have
Exercise 3.7.2
Let be the vector space of all polynomial functions over the field of real numbers. Let and be fixed real numbers and let be the linear functional on defined by
If is the differentiation operator on , what is ?
Solution: Let . Then
Exercise 3.7.3
Let be the space of all matrices over a field and let be a fixed matrix. If is the linear operator on defined by , and if is the trace function, what is ?
Solution: By exercise 3 in section 3.5, we know trace trace. Thus
Exercise 3.7.4
Let be a finite-dimensional vector space over the field and let be a linear operator on . Let be a scalar and suppose there is a non-zero vector in such that . Prove that there is a non-zero linear functional on such that .
Solution: Consider the operator . Then so . Therefore as an operator on . It follows that there’s a such that . Now , thus .
Exercise 3.7.5
Let be an matrix with real entries. Prove that if and only if trace.
Solution: Suppose is the matrix with entries and is the matrix with entries . Then the entry of is
Substituting for and for we get the entry of is (note that has dimension )
Thus the diagonal entries of are
for . Thus the trace is
If all then this sum is zero if and only if each because every one of them appears in this sum.
Exercise 3.7.6
Let be a positive integer and let be the space of all polynomial functions over the field of real numbers which have degree at most , i.e., functions of the form
Let be the differentiation operator on . Find a basis for the null space of the transpose operator .
Solution: The null space of consists of all linear functionals such that , or equivalently . Now for some consants . So
This sum equals zero for all vectors if and only if . While can be anything. Therefore the null space has dimension one and a basis is given by taking , which gives the function , the projection onto the coordinate.
Exercise 3.7.7
Let be a finite-dimensional vector space over the field . Show that is an isomorphism of onto .
Solution: Choose a basis for . This gives an isomorphism the space of matrices. And the dual basis gives an isomorphism . Now we know by Theorem 23 that the following diagram of functions commutes. This means if we start at and follow two functinos to the in the bottom left, it doesn’t matter which way around the diagram we go, we end up at the same place.
Both horizontal arrows are isomorphisms (by Theorem 12, page 88). Now clearly transpose on matrices is a one-to-one and onto function from the set of matrices to itself. Also and for any two matrices and . Thus transpose is also a linear transformation. Thus transpose is an isomorphism. Therefore three of the arrows in this diagram are isomorphisms and it follows that the fourth arrow must also be an isomorphism.
Exercise 3.7.8
Let be the vector space of matrices over the field .
(a) If is a fixed matrix, define a function on by . Show that is a linear functional on .
(b) Show that every linear functional on is of the above form, i.e., is for some .
(c) Show that is an isomorphism of onto .
Solution:
(a) This follows from the fact that the trace function is a linear functional and left multiplication by a matrix is a linear transformation from to . In other words
(b) Let be a linear functional. Let . Then
for some fixed constants .
Now let be any matrix. Then the element of is . Thus
Comparing () and () we see each apperas exactly once in each sum. So setting for all we get the appropriate matrix such that .
(c) Let be the function such that . Part (a) shows this function is into . Part (b) shows it is onto . We must show it is linear and one-to-one. Let and . Then We know the trace function itself is linear. Thus is linear. Now suppose trace . Fix . Let be the matrix with a one in the position and zeros elsewhere. Then by the proof of part (b) we know that trace. So if then . Thus must be the zero matrix. Thus is one-to-one. It follows that is an isomorphism.
From http://greggrant.org