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PROFESSOR: Hi, everyone.

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Today, we're going to talk
about linear transformations.

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So, we've seen linear
transformations incognito

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all the time until now.

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We've played around
with matrices.

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Matrices multiplying
vectors, say,

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in R^n and producing
vectors in R^m.

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So really, the language
of linear transformations

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only provides a
nicer framework when

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we want to analyze
linear operations on more

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abstract vector spaces, like
the one we have in this problem

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here.

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We're going to work with the
space of two by two matrices.

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And we're going to analyze the
operation, have the matrix A,

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and we produce its transpose.

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OK.

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So please take a few minutes
to try the problem on your own

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and come back.

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Hi, again.

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OK.

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So the first question
we need to ask ourselves

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is, indeed, why is
T a linear operator?

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So what are the
abstract properties

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that a linear
operator satisfies?

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Well, what happens when T acts
on the sum of two matrices, A

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and B?

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So it produces the matrix
the transpose of A plus B.

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But we know that this is
A transpose, B transpose.

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And so, this is
exactly T(A) plus T(B).

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So the transformation
that we're analyzing

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takes the sum of two
matrices into the sum

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of their transformations.

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OK.

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Similarly, it takes a
multiple of a transformation

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into the multiple of
the transformations.

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So it takes the matrix c times
A to c times A transpose, which

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is c T of A.

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OK.

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So it is a linear operator.

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Now, can we figure out
what its inverse is?

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Well, what does
the transpose do?

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It takes a column and
flips it into a row.

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So what happens if we apply
the operation once again?

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Well, it's going to
take the row and turn it

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back down to the column.

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So applying the
transformation twice,

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we come back to the
original situation.

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So therefore, T squared
is the identity.

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And from this, we
infer that the inverse

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is the transformation itself.

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Now, this was part one.

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Part two, we'll
compute the matrix

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of the linear transformation
in the following two bases.

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So the first basis
is, in fact-- it

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is the standard basis for the
space of two by two matrices.

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And the way we compute
the matrix, we first

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compute what T does to
each of the basis elements.

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So T of v_1.

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Let's go back.

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So here.

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So T takes the transpose
of this matrix.

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And we see that the transpose
of [1, 0; 0, 0] is [1, 0; 0, 0].

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So it's a symmetric matrix.

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So T of v_1 is v_1.

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What about T of v_2?

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Come back here.

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So this 1 comes here.

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0 comes here.

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And so we actually get v_3.

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So T of v_2 is v_3.

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Similarly, T of v_3 is v_2.

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And finally, T of v_4.

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Well, v_4 is a symmetric
matrix as well.

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So the transpose
doesn't change it.

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OK.

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Now, we encode this into a
matrix in the following way.

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Essentially, the first column
will tell us how T of v_1

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is expressed as a linear
combination of the basis

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elements.

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Well, in this case,
it's just v_1.

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So it's going to be 1 times
v_1 plus 0*v_2 plus 0*v_3 plus

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0*v_4.

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T of v_2 is v_3.

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So we have 0, 0, 1, 0.

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T of v_3 is 0*v_1,
1*v_2, 0*v_3, 0*v_4.

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And T of v4 is 0*v_1,
0*v_2, 0*v_3, plus 1*v_4.

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OK.

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So we've written down the matrix
of the linear transformation T

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in the standard basis.

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And you can check that this
is exactly what we want.

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The representation of some
matrix, say, [1, 2; 3, 4]

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in this standard basis is,
it's the vector [1, 2, 3, 4].

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T takes this to its
transpose, [1, 3; 2, 4].

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So this in the basis is
represented as [1, 3, 2, 4].

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Right?

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And it's not hard
to see that when

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M_T multiplies this vector,
we get exactly this vector.

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So we'll pause for a bit,
so that I erase the board.

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And we're going to return
with the representation of T

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in the basis w_1,
w_2, w_3, and w_4.

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OK.

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So let's now
compute the matrix T

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in the basis w_1,
w_2, w_3, and w_4.

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We played the same game.

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We look at how T acts on
each of the basis vectors.

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So T of w_1-- well, w_1
is a symmetric matrix.

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So T of w_1 is w_1.

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Similarly, with w_2 and w_3.

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They're all symmetric.

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What about w_4?

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Well, we see that the
1 comes down here,

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the negative one comes
up here, and in the end,

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we just get the negative of w_4.

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So, let me just write this out.

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We had T of w_1 equal to
w_1, T of w_2 equal to w_2,

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T of w_3 equal to w_3, and T
of w_4, was negative of w_4.

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So therefore, the matrix of
the linear transformation

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T, in this basis-- I'm going
to call the matrix M prime T--

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has a fairly simple expression.

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The only non-zero entries
are on a diagonal.

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And they're precisely
1, 1, 1, and negative 1.

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And finally, let's
tackle the eigenvalues

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slash eigenvectors issue.

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Well, you've seen what an
eigenvector for a matrix is.

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And the idea for an
eigenvalue, eigenvector

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for a linear transformation
is virtually the same.

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And we are looking for the
vectors v and scalars lambda

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such that T of v is lambda*v.
But if you guys look back

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to what we just did with
w_1, w_2, w_3, and w_4,

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you'll see precisely
that w_1, w_2,

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and w_3 are eigenvectors
for T with eigenvalue 1.

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And w_4 is an eigenvector for
T with eigenvalue negative 1.

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So yeah, we essentially have
solved the problem knowing

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a very, very nice
basis in which we

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computed the linear
transformation T.

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So I'll leave it at that.