find the change of basis matrices from to and
from to .
Multiply the two.
Answer
For the matrix to change bases from to we need that
and that
.
Of course, the representation of a vector in with respect to
the standard basis is easy.
Concatenating those two together to make the columns of the change of
basis matrix gives this.
The change of basis matrix in the other direction can be gotten by
calculating and
(this job is routine)
or it can be found by taking the inverse of the above matrix.
Because of the formula for the inverse of a matrix,
this is easy.
This exercise is recommended for all readers.
Problem 2
Find the change of basis matrix for .
,
,
,
,
Answer
In each case, the columns
and
are concatenated to make the change of basis matrix
.
Problem 3
For the bases in Problem 2,
find the change of basis matrix in the other direction, from to .
Answer
One way to go is to find
and ,
and then concatenate them into the columns of the desired
change of basis matrix.
Another way is to find the inverse of the matrices that answer
Problem 2.
This exercise is recommended for all readers.
Problem 4
Find the change of basis matrix for each .
Answer
The columns vector representations
,
and ,
and
make the change of basis matrix
.
E.g., for the first column of the first matrix, .
This exercise is recommended for all readers.
Problem 5
Decide if each changes bases on .
To what basis is changed?
Answer
A matrix changes bases if and only if it is nonsingular.
This matrix is nonsingular and so changes bases.
Finding to what basis is changed
means finding such that
and by the definition of how a matrix represents a linear map,
we have this.
Where
we can either solve the system
or else just spot the answer
(thinking of the proof of Lemma 1.4).
Yes, this matrix is nonsingular and so changes bases.
To calculate , we proceed as above with
to solve
and get this.
No, this matrix does not change bases because it
is nonsingular.
Yes, this matrix changes bases because it is nonsingular.
The calculation of the changed-to basis is as above.
Problem 6
Find bases such that this matrix represents the identity map
with respect to those bases.
Answer
This question has many different solutions.
One way to proceed is to make up any basis for any space,
and then compute the appropriate (necessarily for the same space,
of course).
Another, easier, way to proceed is to fix the codomain as and
the codomain basis as .
This way (recall that the representation of any vector with respect
to the standard basis is just the vector itself),
we have this.
Problem 7
Conside the vector space of real-valued functions with basis
.
Show that
is also a basis for this space.
Find the change of basis matrix in each direction.
Answer
Checking that is a basis
is routine.
Call the natural basis .
To compute the change of basis matrix we must
find and , that is,
we need such that these equations hold.
Obviously this is the answer.
For the change of basis matrix in the other direction we could
look for and by solving these.
An easier method is to find the inverse of the matrix found above.
Problem 8
Where does this matrix
send the standard basis for ?
Any other bases?
Hint.
Consider the inverse.
Answer
We start by taking
the inverse of the matrix, that is, by deciding what is the inverse to
the map of interest.
This is more tractable than the representation the other way
because this matrix is the concatenation of these two column vectors
and representations with respect to are transparent.
This pictures the action of the map that transforms to
(it is, again, the inverse of the map that is the answer to
this question).
The line lies at an angle to the axis.
This map reflects vectors over that line. Since reflections are self-inverse, the answer to the question is: the original map reflects about the line through the origin with angle of elevation . (Of course, it does this to any basis.)
This exercise is recommended for all readers.
Problem 9
What is the change of basis matrix with respect to ?
Answer
The appropriately-sized identity matrix.
Problem 10
Prove that a matrix changes bases if and only if it is invertible.
Answer
Each is true if and only if the matrix is nonsingular.
What remains to be shown is that
left multiplication by a reduction matrix represents a
change from another basis to .
Application of a row-multiplication matrix translates a
representation with respect to the basis
to one with respect to , as here.
Applying a row-swap matrix translates a representation
with respect to
to one with respect to
.
Finally, applying a row-combination matrix changes a
representation with respect to
to one with respect to .
(As in the part of the proof in the body of this subsection, the various conditions on the row operations, e.g., that the scalar is nonzero, assure that these are all bases.)
This exercise is recommended for all readers.
Problem 12
Let be a nonsingular matrix.
What basis of does change to the standard basis?
Answer
Taking as a change of basis matrix
,
its columns are
and, because representations with respect to the standard basis
are transparent, we have this.
That is, the basis is the one composed of the columns of .
This exercise is recommended for all readers.
Problem 13
In with basis
we have this
represenatation.
Find a basis
giving this different representation for the same
polynomial.
State and prove that any nonzero vector
representation can be changed to any other.
Hint.
The proof of Lemma 1.4
is constructive— it not only says the bases change, it shows
how they change.
Answer
We can change the starting vector representation
to the ending one through a sequence of row operations.
The proof tells us what how the bases change.
We start by swapping the first and second rows
of the representation with respect to to get a representation
with respect to a new basis .
We next add times the third row of the vector
representation to the fourth row.
(The third element of is the third element of
minus times the fourth element of .)
Now we can finish by doubling the third row.
Here are three different approaches to stating such a result.
The first is the assertion: where is a vector space with
basis and is nonzero, for any nonzero column
vector
(whose number of components equals the dimension of )
there is a change of basis matrix such that
.
The second possible statement: for any (-dimensional)
vector space and any nonzero
vector , where
are nonzero, there are bases such that
and
.
The third is: for any nonzero member of
any vector space (of dimension ) and any nonzero column vector
(with components) there is a basis such that is
represented with respect to that basis by that column vector.
The first and second statements follow easily from the third.
The first follows because the third statement gives a basis
such that and then
is the desired .
The second follows from the third because it is just a
doubled application of it.
A way to prove the third is as in the answer to the first part
of this question.
Here is a sketch.
Represent with respect to any basis with a column
vector .
This column vector must have a nonzero component because
is a nonzero vector.
Use that component in a sequence of row operations to convert
to .
(This sketch could be filled out as an induction
argument on the dimension of .)
Problem 14
Let be vector spaces, and let be bases for
and be bases for .
Where is linear, find a formula relating
to .
Answer
This is the topic of the next subsection.
This exercise is recommended for all readers.
Problem 15
Show that the columns of an change of basis matrix
form a basis for .
Do all bases appear in that way: can
the vectors from any basis make the columns of a change of
basis matrix?
Answer
A change of basis matrix is nonsingular and thus has rank equal to the number of its columns. Therefore its set of columns is a linearly independent subset of size in and it is thus a basis. The answer to the second half is also "yes"; all implications in the prior sentence reverse (that is, all of the "if ... then ..." parts of the prior sentence convert to "if and only if" parts).
This exercise is recommended for all readers.
Problem 16
Find a matrix having this effect.
That is, find a that left-multiplies the
starting vector to yield the ending vector.
Is there a matrix having these two effects?
Give a necessary and sufficient condition for there to be a
matrix such that
and .
Answer
In response to the first half of the question,
there are infinitely many such matrices.
One of them represents with respect to
the transformation of with this action.
The problem of specifying two distinct input/output pairs is a bit
trickier.
The fact that matrices have a linear action precludes some possibilities.
Yes, there is such a matrix.
These conditions
can be solved
to give this matrix.
No, because
no linear action can produce this effect.
A sufficient condition is that
be linearly independent, but
that's not a necessary condition.
A necessary and sufficient condition is that any linear dependences
among the starting vectors appear also among the ending vectors.
That is,