Coordinates with Answers to practice problems
One of the most common techniques in Linear Algebra is to change coordinates to a more
convenient system. The most common coordinates are relative to the standard unit vectors. For
example, if you think of the vector
, that means
.
The idea you need to get is this: A vector is just a direction and length. We usually write them
down though as a column of numbers. Those are the coordinates of that vector in some basis.
Up to now, though, we’ve only used the one basis of standard unit vectors.
Now we try this for other bases. There’s nothing magical about the standard unit vectors other
than they are what everyone starts with.
So suppose you have a different basis,
of some 3 dimensional subspace V in
(We don’t know where they are or
Rn. In the new system,
their length.)
Notice the subscript to show these coordinates are in the new system.
Then any vector has coordinates in this system and coordinates in the standard system. Maybe V
is a subspace of R5 and so their coordinates in the bigger space have 5 components each. You
need to know how to go back and forth between the two systems. The idea behind how to do this
is one we’ve seen often in the course. Think of a vector
So that means there are
coordinates
where:
This means no matter what coordinates we use for our basis,
So if we put the vectors (in standard coordinates) as columns of a matrix S,
then
This is one important equation.
You can do this for any subspace of Rn
In the special case where V is all of Rn, note then that S is square and invertible (that would be a
good proof for the final) so to go the other way, multiply both sides of the above equation on the
left by
to get
This is also an important equation.
So in problems about this, you would be given vectors that form a new basis (or maybe asked to
come up with your own that have some special property). Then you’d be given a vector like
and be asked to find
.
For example, suppose V = R3 and we are given
and a vector
find
So first create the matrix
using the 3 vectors above as columns and find
So multiply this by
to get
(Continue below)
,
Transformations in new coordinate systems
One major reason to use a new coordinate system is that a new system may be more natural for
understanding and representing a transformation. Using the above matrices and
we can
do transformations in either the old standard system or switch vectors into the new basis, do the
transformation using a matrix in the new system and then switch the output vector back to the
old system.
To get the idea, any transformation T has a matrix
(how do we know that?). So we could use
the diagram to represent what we mean by a transformation:
In a new basis
, this transformation T must also have a matrix, call it
.
So, like with coordinate changes above, how do we get
if we know
or how do we get
if we know ? The answer to both these questions comes from considering the
following diagram:
So, If you were given
1. Build
and the new basis
, to find A:
using the basis vectors in standard coordinates as columns.
2. Calculate
3. Using the diagram
Of course, you can also go the other way,
by switching the vertical arrows and
matrices (or by multiplying both sides of the equation in 3 on the right by .
In general, note
Any two matrices
and
called similar matrices.
where there is an invertible matrix
such that
, are
P148, #38 is an example. It is in R2. We are given info that helps us write
in a basis we must
come up with by thinking. We begin with a line. (The one spanned by
.) We are told our
transformation T is reflection across this line. So for any point, we’d drop a perpendicular
to the line and go the length of that perpendicular on the other side to find T of the original
point. So if we build a basis using the given vector (call it
(like
and a perpendicular one
), then T leaves the first fixed and the second is multiplied by -1 in the new
coordinate system. So
Building
.
and calculating
, we can find
Notice
is a much nastier matrix than
is.
A few more example problems:
1. Find a basis
for R2 where
and
Solution: S is a matrix of 2 column vectors for our basis
equations:
and
and we know 2
(We are more used to the answer on the right.) So in
terms of the column vectors these become:
and
So we have 2 equations with 2 unknowns (the v’s)
So subtract 2 times each side of the first from the second. You get
that into the first equation above and solve for the second vector. You get:
two vectors form the basis.
Now substitute
So those
2. Is matrix
similar to
for all p and q?
Solution: Don’t know. Try to find
that works, that is:
so we’d have to have
Looking at the 4 resulting equations, we
see that we have to have:
and
has these two satisfied will do, so how about:
but then theses work. Any invertible matrix that
So they are similar.
3. Suppose we consider the plane of the solution set of x+2y+3z=0. A linear transformation can
be defined from all of R3 to this space by orthogonal projection. Just as in our earlier example,
that means for any vector, drop a perpendicular to the plane and map the vector to that point.
Find the matrix for this transformation in standard coordinates.
Solution: Suppose we can find a basis for R3 with 2 of the vectors in the plane of the solution set
of this equation and the third vector perpendicular to the plane. Then projection in coordinates
fro that basis just means dropping the third component. The only new (and a bit hard to do) part
is to find the 3rd vector. Once we do this, we’ll have S and B. We’ll calculate A as above.
To find 2 vectors that are a basis for the plane itself, just find 2 independent solutions, say
Now we need a vector
that is perpendicular to both these. So, using dot
products, we get 2 equations in 3 unknowns:
and solve it noting
is free, but lets just set
and
We could set up the matrix for this
and so
and (with a little work)
and we worked above to get
Finally,
and so
So our basis is: