PROFESSOR: Hi, everyone.
Today, we're going to talk
about linear transformations.
So, we've seen linear
transformations incognito
all the time until now.
We've played around
with matrices.
Matrices multiplying
vectors, say,
in R^n and producing
vectors in R^m.
So really, the language
of linear transformations
only provides a
nicer framework when
we want to analyze
linear operations on more
abstract vector spaces, like
the one we have in this problem
here.
We're going to work with the
space of two by two matrices.
And we're going to analyze the
operation, have the matrix A,
and we produce its transpose.
OK.
So please take a few minutes
to try the problem on your own
and come back.
Hi, again.
OK.
So the first question
we need to ask ourselves
is, indeed, why is
T a linear operator?
So what are the
abstract properties
that a linear
operator satisfies?
Well, what happens when T acts
on the sum of two matrices, A
and B?
So it produces the matrix
the transpose of A plus B.
But we know that this is
A transpose, B transpose.
And so, this is
exactly T(A) plus T(B).
So the transformation
that we're analyzing
takes the sum of two
matrices into the sum
of their transformations.
OK.
Similarly, it takes a
multiple of a transformation
into the multiple of
the transformations.
So it takes the matrix c times
A to c times A transpose, which
is c T of A.
OK.
So it is a linear operator.
Now, can we figure out
what its inverse is?
Well, what does
the transpose do?
It takes a column and
flips it into a row.
So what happens if we apply
the operation once again?
Well, it's going to
take the row and turn it
back down to the column.
So applying the
transformation twice,
we come back to the
original situation.
So therefore, T squared
is the identity.
And from this, we
infer that the inverse
is the transformation itself.
Now, this was part one.
Part two, we'll
compute the matrix
of the linear transformation
in the following two bases.
So the first basis
is, in fact-- it
is the standard basis for the
space of two by two matrices.
And the way we compute
the matrix, we first
compute what T does to
each of the basis elements.
So T of v_1.
Let's go back.
So here.
So T takes the transpose
of this matrix.
And we see that the transpose
of [1, 0; 0, 0] is [1, 0; 0, 0].
So it's a symmetric matrix.
So T of v_1 is v_1.
What about T of v_2?
Come back here.
So this 1 comes here.
0 comes here.
And so we actually get v_3.
So T of v_2 is v_3.
Similarly, T of v_3 is v_2.
And finally, T of v_4.
Well, v_4 is a symmetric
matrix as well.
So the transpose
doesn't change it.
OK.
Now, we encode this into a
matrix in the following way.
Essentially, the first column
will tell us how T of v_1
is expressed as a linear
combination of the basis
elements.
Well, in this case,
it's just v_1.
So it's going to be 1 times
v_1 plus 0*v_2 plus 0*v_3 plus
0*v_4.
T of v_2 is v_3.
So we have 0, 0, 1, 0.
T of v_3 is 0*v_1,
1*v_2, 0*v_3, 0*v_4.
And T of v4 is 0*v_1,
0*v_2, 0*v_3, plus 1*v_4.
OK.
So we've written down the matrix
of the linear transformation T
in the standard basis.
And you can check that this
is exactly what we want.
The representation of some
matrix, say, [1, 2; 3, 4]
in this standard basis is,
it's the vector [1, 2, 3, 4].
T takes this to its
transpose, [1, 3; 2, 4].
So this in the basis is
represented as [1, 3, 2, 4].
Right?
And it's not hard
to see that when
M_T multiplies this vector,
we get exactly this vector.
So we'll pause for a bit,
so that I erase the board.
And we're going to return
with the representation of T
in the basis w_1,
w_2, w_3, and w_4.
OK.
So let's now
compute the matrix T
in the basis w_1,
w_2, w_3, and w_4.
We played the same game.
We look at how T acts on
each of the basis vectors.
So T of w_1-- well, w_1
is a symmetric matrix.
So T of w_1 is w_1.
Similarly, with w_2 and w_3.
They're all symmetric.
What about w_4?
Well, we see that the
1 comes down here,
the negative one comes
up here, and in the end,
we just get the negative of w_4.
So, let me just write this out.
We had T of w_1 equal to
w_1, T of w_2 equal to w_2,
T of w_3 equal to w_3, and T
of w_4, was negative of w_4.
So therefore, the matrix of
the linear transformation
T, in this basis-- I'm going
to call the matrix M prime T--
has a fairly simple expression.
The only non-zero entries
are on a diagonal.
And they're precisely
1, 1, 1, and negative 1.
And finally, let's
tackle the eigenvalues
slash eigenvectors issue.
Well, you've seen what an
eigenvector for a matrix is.
And the idea for an
eigenvalue, eigenvector
for a linear transformation
is virtually the same.
And we are looking for the
vectors v and scalars lambda
such that T of v is lambda*v.
But if you guys look back
to what we just did with
w_1, w_2, w_3, and w_4,
you'll see precisely
that w_1, w_2,
and w_3 are eigenvectors
for T with eigenvalue 1.
And w_4 is an eigenvector for
T with eigenvalue negative 1.
So yeah, we essentially have
solved the problem knowing
a very, very nice
basis in which we
computed the linear
transformation T.
So I'll leave it at that.
