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Current time:0:00Total duration:14:10

Let's see if taking the
transpose of a matrix does anything to its determinant. So a good place to start is just
with the 2-by-2 scenario. So let's do the 2-by-2
scenario. So if I start with
some matrix here. Let me just take its
determinant. So a, b, c, d. And then let me take
its determinant. So this is going to be
equal to ad minus bc. Let me take the transpose of
this and then take its determinant. So that would be the determinant
of ac, the columns turn into the rows, and
then bd, the rows turn into the columns. What is this going
to be equal to? This is going to be equal
to ad minus bc again. The only thing that happened is
these two guys got swapped and they multiply times
each other anyway. So these two things
are equivalent. So at least for the 2-by-2 case,
the determinant of some matrix is equal to the
determinant of the transpose of that matrix. Now, that's just the
2-by-2 case. Now, I'm going to make an
inductive argument, or I could just say an argument by
induction, to show that this works for all n by
n, for all cases. And the way you construct an
argument by induction is you assume that it's true
for the n-by-n case. So let's assume that for
n by n, and let's say I have some matrix. Let me call it matrix B, and
let's say it's an n-by-n matrix, we assume that the
determinant of any matrix B that's n by n is equal to the
determinant of B's transpose. That's where we started off with
our inductive argument. And then we see if given this,
if given that, is it true of n plus 1 by n plus 1 matrix? Because if we can, if we can
say, look, given that it's true for the n-by-n case, it's
going to be true for the n plus 1 by n plus 1 case, then
we're done because we know it's true for the base case,
the 2-by-2 case, which you could say, well, that's
your first n by n. So if it's true for the 2 by 2
case, then it'll be true for the 3-by-3 case, because that's
just one increment. But then if it's true for the
3-by-3 case, then it'll be true for the 4-by-4 case. And if it's true for the 4-by-4
case, it'll be true for the 5-by-5 case, and you just
keep going up like that. So when you do a proof by
induction, you prove a base case, and then you prove that if
it's true for n, or in this case an n-by-n determinant, if
you can prove that given it's true for an n-by-n determinant,
it's going to be true for an n plus 1 by n plus
1 determinant or an n plus 1 by n plus 1 matrix, then you
have completed your proof. So let's see if this
is the case. So let me construct an n plus
1 by n plus 1 matrix. So let's say I have my matrix
A, my favorite letter to use for matrices, I think the
entirely linear algebra's favorite letter to
use for matrices. And let's say it's an n plus
1 by n plus 1 matrix. And just to simplify my notation
let's just say that m is equal to n plus 1. So we could call it
an m-by-m matrix. And what is it going
to look like? Let's draw its entries
right here. I'm going to draw more than the
normal amount of entries. a11, this is its first row,
a12, all the way to a1m. We have m columns, which is the
same thing as m plus 1. That's not m times 1 columns. This is m plus 1 as well. And then we do our second
row right here. a21, a22, a23, all
the way to a2m. Then you have your third row
right here: a31, a32, a33, all the way to a3m. And then you go all
the way down here. At the end you have your mth
row, which you could also say is your m plus 1 row. So it's your mth row, first
column, and then you have a sub m2, and then a sub m3,
all the way to a sub mm. Fair enough. Now, let me draw the
transpose of A. So a transpose is also going to
be an n plus 1 by n plus 1 matrix, which you could also
write as an m-by-m matrix. I'm just going to have to take
the transpose of this. So the transpose of that, this
row becomes a column, so it becomes a11, and this entry
right here is a12. It's this entry right there. Then you can go all the
way down to a1m. Then this pink row becomes a
pink column here, a21-- I wanted to do it in pink. You have a21, a22, you have an
a23, and it goes all the way down to a2m. You have your green row right
here, it's your third one, so it's a31, a32, a33, all
the way down to a3m. And then we can just skip a
bunch of rows in this case, but it's columns in this case. So you just draw some dots
and you have am1, am2. I'm just going down this guy,
but this row is now going to become-- which was
the last row. It's now going to become
the last column. am3, all the way down to amm. And I have my transpose. Now, let's take the
determinant of A. Let me do it in purple. So the determinant of my matrix
A, we could just go down this first row up here. It's going to be equal to a11
times the determinant of its submatrix, so it's the
determinant of this submatrix right here. We could call that the
submatrix a sub 11. We've seen this notation
before. So it's the determinant of a sub
11, and then it is minus a12 times the determinant of
its submatrix, so you cross out that row and that column. So that is going to be a sub 12,
and you're going to go all the way to-- and I don't what
the sign on that is, so we could call it negative 1 to the
1 plus m-- that'll give us the right sign for the
checkerboard pattern-- times the determinant of the submatrix
for this guy. So we call it a1m, where you
cross out that guy's row, that guy's column, and you're
just left with all of this stuff over here. Fair enough. Now, let's look at the
determinant of a transpose. We learned earlier, you don't
have to go down the first row or you don't even
go down a row. You could go down a column. Let me be clear. So for our determinant of A, we
went down this row, and our submatrices, this was
my first submatrix. My second submatrix, you know
what it looks like. You would cross out the second
column and that row, and whatever's left over would be
the second submatrix and so on and so forth. But for our determinant of A
transpose, let's just go down this first column and get the
submatrices like that. So this is going to be equal
to-- let's get our first term right here. a sub 11 times the determinant
of its submatrix. So what's the determinant
of its submatrix? It's going to be-- its
submatrix, you cross out its row and its column, and you're
going to be left with this thing right here. Now, an interesting question
is how does this thing that I've just squared off, this
guy's submatrix, relate to this guy's submatrix? Well, if you look at it
carefully, this row from a22 to a2m has now become a column
from a22 to a2m. This row, which is the next one,
from a32 to a3m has now become a column from
a32 to a3m. If you keep going down this last
row, it has now become this column. So this guy's submatrix or the
thing we're going to have to take the determinant of right
here is equal to the transpose of this guy. So this is equal to a
sub 11 transpose. And if you go through it, we go
minus this guy, minus a12 times the determinant
of his submatrix. And if we cross out this guy's
row and that guy's column, what is that going
to look like? His submatrix is going
to look like this. It's going to have that there
and it's going to have that right there. How does that compare to a12? So a12 is if you crossed out
this and this, you're left with all of this right here. So once again, you see that this
row is the same thing as this column, that this row is
the same thing as this column, that that row is the same
thing as that column. So once again, the submatrix
we have to take the determinant of is equal
to the transpose of this thing over here. So it's equal to
a12 transpose. This thing-- I could draw it
shaded in-- is equal to the transpose of this thing, is
equal to the transpose of that thing right there. So, in general, each of these
submatrices when we go down this row is equal to the
transpose of each of these. So you keep going, so then
you're going to go all the way to plus minus 1. We're going to go all the way
down to the minus 1, 1 plus m times the determinant of--
it's going to be this guy transposed. You can even do it. If you go all the way to-- if
you cross that guy out and that guy out, you're left with
everything else on this matrix, and that's equal to the
transpose of if you cross this guy out and that guy out. This row turns into that column,
that row turns into that column. I think you see the point. I don't want to beat
a dead horse. So that's going to be equal
to a sub 1m transpose. Now, remember, going into this
inductive proof, or proof by induction, I assume that for--
remember, this is an n plus 1 by n plus 1 matrix. But going into it, I assumed
that for an n-by-n matrix, the determinant of B
is equal to the determinant of B transpose. Well, these guys right
here, these are n-by-n matrices, right? This guy right here is an
n plus 1 by n plus 1. Same thing for this
guy right here. But these guys right
here are n by n. So if we assume for the n-by-n
case that the determinant of a matrix is equal to the
determinant of a transpose-- this is the determinant of
the matrix, this is the determinant of its transpose--
these two things have to be equal. So we can then say that the
determinant of A transpose is equal to this term A sub 11
times this, but this is equal to this for the n-by-n case. Remember, we're doing the n
plus 1 by n plus 1 case. But these submatrices
are one dimension smaller in each direction. It has one less row and
one less column. So these two things are equal. So instead of writing this, I
can just write this, so times the determinant of a sub 11. Then you keep going. Minus a sub 12 times
the determinant. Instead of writing this, I
could write that because they're equal. Determinant a sub 12, all the
way to plus minus 1 to the 1 plus m times the determinant
of this. These two things are equal. That is equal to that. That was our assumption
in this inductive proof, a sub 1m. And then you see that, of
course, this line, this blue line here, is equivalent to
this blue line there. So we get that the determinant
of A, which is an n plus 1 by n plus 1, so this is the n
plus 1 by n plus 1 case. We get the determinant
of A is equal to the determinant of A transpose. And we got this assuming that it
is true-- let me write it-- assuming that it's true
for n-by-n case. And then we're done. We've now proved that this is
true in general, because we've proven the base case. We've proven it for the 2-by-2
case, and then we showed that if it's true for the n
case, it's true for the n plus 1 case. So if it's true for the 2 case,
it's going to be true for the 3-by-3 case. If it's true for the 3-by-3
case, it's true for the 4-by-4 case, so on and so forth. But the takeaway
is pretty neat. You can take the transpose, the determinant doesn't change.