Hey guys, in this video I'm going to do this
linear algebra proof. It says, let A and
B be two n by n matrices.
Suppose that B is skew-symmetric. Prove
that A times B times A transpose
is also skew-symmetric.
This is a short proof that
I had to do in my linear algebra midterm.
It's pretty straightforward. You just have to
know what a skew-symmetric matrix
is, the definition of that. And also,
some basic, like matrix
properties. So let me
first define
what skew-symmetric is in the proof
and we're going to use that definition later on to prove that
A B A transpose is skew-symmetric.
Let A and B be
two n by n
matrices. And suppose
B is skew-symmetric.
I'm just re-writing what we were given.
Then by assumption, we can say
that B transpose is equal to
negative B. And you can think of this as
a scalar. So you can think of that as negative 1 times
B. But it's basically going to be negative
the values of the matrix B. So that
is the definition of skew-symmetric.
We can assume that this is true by the assumptions that we're given
in the proof.
Then I said because square matrices of n by n size
can be multiplied to produce another n by n matrix
and the transpose of an n by n matrix is still
n by n, A B A transpose is
a valid matrix. So let me write all that down.
I was just very explicit because it was
a midterm problem.
This is kind of like, just writing out things that are
kind of obvious: that you can multiply an n by n matrix,
and just saying explicitly that A
B A transpose is a valid matrix, although we're kind of
given that it is. And so, what I'm going to do is I'm going to
take the transpose of what we're tying to prove
is symmetric and get that to be equal to
negative A B A transpose.
If A B A transpose transpose
is equal to negative A B A transpose,
then it falls in line
with the definition of being skew-symmetric
and therefore we are done with the proof.
I'm going to say, A B A
transpose transpose--And so for this step,
this is what I was saying about knowing certain
matrix properties because the order
when multiplying or taking transpose does matter.
A times B matrices is not
the same as B times A and similarly for transpose,
you kind of go in reverse.
A B A transpose transpose is equal
to A transpose transpose
times B transpose
times A transpose.
That is equal to A B
transpose, A transpose.
We are given that B is skew-symmetric.
We have already said that B transpose
is equal to negative B. So let's go ahead and plug that
in for B transpose.
We're given that. And then on the side, I said
we know that from the definition of skew-symmetric.
And then I said, let's think about this
as a scalar. So -1
times B A transpose.
If we think of it as a scalar, then we can move scalars
when we're multiplying with matrices.
That's going to be equal to either -1 or just
-A times B times
A transpose. Thus we have proven that
A B A transpose--the transpose of that
is equal to -A B A transpose.
To finish the proof we can say,
thus by definition, A B A
transpose is skew-symmetric
when B is skew-symmetric.
And we are done with this problem :0)
*chiptune music*