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 A is similar to B and we want to prove that A transpose is
also similar to B transpose. This is a linear algebra problem and I'm going to prove it using
the definition of matrix similarity and also using some basic matrix properties. Let's go ahead and
get started with the video. I'm going to write out the proof. The first thing I'm going to write is
basically just repeating what we're given, which is that we have two n-by-n matrices and that A and
B, the matrices, are similar to each other. And then I'm going to define what similarity means. In
symbolic notation when you say A is similar to B, you can write that as A tilde B and so I'm just
going to say again, let A and B be two n-by-n, so square, matrices. Suppose that A is similar to B,
then what similarity means is that by definition there exists an invertible n-by-n matrix P.
So there has to exist an invertible n-by-n matrix that we're calling P such that
A is equal to P times B times P inverse. We know that this is true and that this exists because
that is the definition of matrix similarity. This is also sometimes written equivalently as A times
P is equal to P times B but for the sake of this problem I'm just going to leave it like that.
This is what we are given and that we know is true and we just want to prove that it
will be true for the matrices A transpose and B transpose. We want to prove that if this is true,
then that implies that A transpose is similar to B transpose. To prove that this is true,
I'm basically just going to take the transpose of this equation and then simplify it to a way where
we can get it back into the similarity formula. And if we can do that, then we have proven that A
transpose is similar to B transpose as well. I'm going to say that we are going to have another
matrix that I'm going to call P tilde and that's going to be equal to P transpose. And we know
that P is an invertible matrix, so by definition if P is an invertible matrix then P transpose,
the transpose of an invertible matrix, is also invertible. We can say, by definition thus we
know that P tilde is invertible. Thus we can have the following. And then in the next part, I'm just
going to take the transpose on both sides. On the left hand side, the transpose is just going to be
A transpose and then I'm going to take the transpose of this entire right hand side. That's
going to be the transpose of P times B times P inverse. I'm going to take the transpose of that.
When you have multiple matrices multiplied to each other and then you take the transpose--so
let's say you have A times B times C and then you take the transpose of that--that is going
to be equal to C transpose times B transpose times A transpose. Basically the order of
the matrices matters and the order flips when you take the transpose. For the right hand side of our
equation, that's going to be P inverse transpose times B transpose times P transpose. We want to
plug in P tilde, which is P transpose. So P to the -1 times T is equal to P to the T times -1 times
B transpose times P transpose and we do that to be able to simplify this or plug in P tilde. So this
is equal to P tilde inverse times B transpose times P tilde and then in our final step, if
we compare the current definition that we have of A transpose being equal to this right hand side,
we want to get it into a form that looks like the original formula. A corresponds to A transpose, we
want P to correspond to P tilde inverse, B we have corresponding to B transpose and we want P inverse
to correspond to P tilde. The way we're going to do this is by rewriting this one more time.
We're going to think of this as being P inverse times B transpose and then this we want to have
being something equal to the -1. The way we're going to transform this is we're going to say P
tilde to the -1 times -1. And so by doing this, we now have P that corresponds to P tilde -1 and P
inverse corresponding to P tilde inverse times -1. By doing this, we now have it in the correct form
and we can conclude that A transpose is similar to B transpose. And we are done with this problem.
*chiptune music*