PROFESSOR: Hi, everyone.
Welcome back.
So today, I'd like to talk about positive definite matrices.
And specifically, we're going to analyze several properties
of positive definite matrices.
And specifically, we're going to look
at why each one of these following statements is true.
So first off, why every positive definite matrix is invertible.
Why the only positive definite projection matrix
is the identity matrix.
If D is a diagonal matrix with positive entries,
show that it must also be positive definite.
And then lastly, if S is a symmetric matrix where
the determinant S is bigger than 0,
show why this might not necessarily
imply that it's positive definite.
So I'll let you think about these for a moment.
And I'll come back in a second.
Hi, everyone.
Welcome back.
OK.
So let's take a look at part A. So part A
is asking why every positive definite matrix is invertible.
Well, let's just recall that if A is a matrix
and if A is invertible, then this necessarily
implies that the determinant of A is non-zero.
And I'm going to just write out det
A as the product of the eigenvalues of A. So lambda_1
to lambda_n are the eigenvalues of A.
OK.
In addition, if A is positive definite,
what does this say about the eigenvalues of A?
Well, it says that each eigenvalue of A, lambda_1,
lambda_2, dot, dot, dot, to lambda_n, each one of them
must be bigger than 0.
So this statement that each eigenvalue of A
is bigger than 0 is completely equivalent
to A being positive definite for symmetric matrices A.
So if I have a whole bunch of eigenvalues and each of them
are bigger than 0, what does this say about det A?
Well, I can take the product of all these eigenvalues.
And of course, the product of a whole bunch of positive numbers
must also be positive.
So if the quantity is positive, then it certainly cannot equal
0.
So this proves that det A is not equal to 0.
Hence, A must be invertible.
OK.
So for part B, we're asked to show
that the only positive definite projection
matrix is the identity matrix.
So again, how do we tackle this problem?
We're going to look at the eigenvalues.
So remember, if P is a projection,
what does it say about the eigenvalues of P?
Well, it says that the eigenvalues of P
are either 0 or 1.
So this is point one.
Point two, if P is a positive definite,
what does that say about the eigenvalues of P?
Well, as I've noted before, it means that the eigenvalues
are bigger than 0.
So if P is a projection and it's positive definite,
the only possible eigenvalues that
are both 0 and 1 and bigger than 0 are 1.
So the conclusion is that the eigenvalues of P
must all equal 1.
So which matrix has eigenvalues 1 and is also symmetric?
Well, the only matrix that satisfies this property
is the identity matrix.
Now you might ask, how do you actually show that?
Well, you could argue as follows.
If P is diagonalizable-- and every symmetric matrix
is diagonalizable, so I'm not making this up.
So if P is diagonalizable, then you
can always write P as some matrix, U,
times a diagonal matrix-- and we know in this case,
the diagonal matrix has eigenvalues 1,
so it's actually the identity matrix-- times the inverse
of the eigenvector matrix.
But of course, this is just U times U inverse,
which then gives me the identity at the end.
So U times the identity times U inverse.
This is just U times U inverse.
And of course, U and U inverse collapse back down
to the identity.
So this shows you that the only matrix that has eigenvalues
of 1 is the identity matrix.
So that's just to cross all the T's and dot all the I's.
OK.
For part C, we're given D as a diagonal matrix
with positive entries on the diagonal.
Now we have to show that it's positive definite.
OK.
So let me write D as follows.
I'll just write it like this.
I'm going to use a compact notation, which is sometimes
seen: diagonal d_1, d_2, d_n.
So D is a diagonal matrix whose elements along the diagonal
are d_1, d_2, dot, dot, dot, to d_n.
Now what does it mean for a matrix to be positive definite?
Well, it means that for any x, for any vector x,
I have to look at the product x transpose D*x.
And I have to show that it's bigger than 0.
And I should qualify this and say
that the vector we're looking at is x not equal to 0.
So we're now looking at the zero vector.
But for D to be positive definite,
we have to show that x transpose D*x is bigger than 0.
This is just one way to show that it's positive definite.
It's not the only way.
So if I write x out using components, x_1, x_2, dot, dot,
dot, to x_n-- I'll write it like this--
then you can work out the quantity x transpose D*x.
And we see that we get a sum of squares.
We get d_1 times x_1 squared plus d_2 x_2 squared plus dot,
dot, dot plus d_n x_n squared.
Now by definition, each coefficient is positive.
A sum of a square is positive.
so-- sorry.
A product of a positive number with a square is positive.
And then of course, a sum of positive numbers
is going to be positive.
So this means the whole thing is positive.
Now there's other more efficient ways
of getting at this using other tricks we know.
For example, if we're given a diagonal matrix,
we know its eigenvalues are already d_1,
d_2, dot, dot, dot, to d_n.
And we know that a matrix with positive eigenvalues
is already positive definite.
But this is kind of starting from the base
to show that it's positive definite.
And now lastly, let's look at part D.
So S is a symmetric matrix with det S bigger than 0.
Show that S might not necessarily
be positive definite.
So there's lots of counterexamples.
I only need to construct one.
So I'm looking at S, which is a symmetric matrix.
So I'm just going to throw in some numbers on some
off diagonals.
So I'll just pick one.
And now I need to pick some numbers along the diagonal.
And the easiest way to do it is just
to pick two negative numbers on the diagonal.
Because if there's a negative number on a diagonal,
then we know S can't be positive definite.
And I'll say more about that in a second.
So I can just pick negative 2 and negative 3.
So let's quickly check what det S is.
Well, it's negative 2 times negative 3 minus 1.
So this gives me 6 minus 1, which is 5.
So by construction, det S is positive, which is good.
And then, as I mentioned before, if there's
negative elements along the diagonal of the matrix,
that matrix can't be positive definite.
Well, why is that?
Well, suppose I wanted to take a look
at this upper left component, negative 3,
and it's negative, how do I show that that implies
S is not positive definite?
Well, what I can do is I can look at the product x transpose
S*x.
And what I do is I look at it.
I just take one value of x.
So we know that this has to be positive for every value of x.
So I can pick any value I want.
So I can take x, say, [1, 0] transpose.
And when I do this, we end up getting that x transpose S*x is
equal to negative 3.
So notice how by taking 1 in the first entry and 0
on the second entry, that picks out the upper left corner,
negative 3.
If I were to take 0 here and 1 here,
it would pick out the negative 2 entry in S.
So by picking 1 in entry i of a vector x,
and then computing this product x transpose S*x,
I pick out the i-th element along the diagonal.
And since that element is negative,
this shows me that along the direction [1, 0],
the product x transpose S*x is also negative.
And hence, S can't possibly be positive definite.
OK.
So just to summarize, we've taken a look
at a couple matrices and a couple different properties
of positive definite matrices.
And notably, we've used the eigenvalues
to get a handle of the positive definite matrices.
And I hope these provide some useful tricks.
And I'll see you next time.