MARTINA BALAGOVIC: Hi.
Welcome to recitation.
Today's problem is about positive definite matrices.
And it's asking us, for which values of the parameter c,
which is sitting here in the matrix, is the matrix B, 2,
minus 1, minus 1, minus 1, 2, minus 1, minus 1, minus 1, 2
plus c, positive definite, and for which values of c is it
positive semidefinite?
I'm going to leave you alone with the problem.
You should pause the video and then come back and compare
your solution.
And we're back.
As you remember from the lectures, there are several
tests that you can do on matrices to find out if
they're positive definite and if they're positive
semidefinite.
And I'm going to demonstrate three to you.
First, I'm going to do the one that you should do in case you
have very little time and you're asked to do a problem
like this on the test, which is of course the determinant
test. The determinant test asks us to compute
determinants of the matrices in the upper left
corner of all sizes.
And it says that it's going to be positive definite if
they're all greater than 0 and positive semidefinite if some
0's sneak into into that sequence.
So let's calculate the determinants.
The first determinant is the determinant of this tiny
matrix here.
So it's just 2.
The next one is the determinant of this two by two
submatrix, 2, minus 1, minus 1, 2, which is equal to 4
minus 1, which is 3.
And finally, we have the determinant of B, which I'm
going to calculate for you.
I'm going to calculate it, I'm going to decompose it along
the first line, first row.
So it's 2 times the determinant of this submatrix,
2, minus 1, minus 1, 2 plus c, minus minus 1, this one here,
times this determinant, which is minus 1, minus 1,
minus 1, 2 plus c.
And then plus minus 1, this one here, times this
determinant, which is minus 1, 2, minus 1, minus 1.
And in total, that's 2 times 4 plus 2c minus 1 plus minus 2
minus c minus 1 and minus 1 plus 2.
And in total, this should give us 6 plus 4c minus 3 minus c
minus 3, which is 3c.
So let's look at the determinants.
2 is positive.
3 is positive.
3c is positive.
So the answer is the matrix is positive definite if c is
bigger than 0, and it's positive semidefinite if it's
either strictly bigger than 0 or equal to 0.
And that's all.
If you're on the test, this is everything that you should do.
Now let me show you two more tests to demonstrate that
first, they take longer, and second, to see these numbers
and their quotients show up in other tests and to try to
convince you that these tests really are equivalent.
Let me do the pivots test for you.
So we take our matrix B, 2, minus 1, minus 1, minus 1, 2,
minus 1, minus 1, minus 1, 2 plus c.
And let's pretend we're solving a system that has this
as a matrix.
So we multiply the first row by 1/2, and we add it to the
second and to the third row.
We get 2, minus 1, minus 1, 0, 3/2, minus 3/2, 0, minus 3/2,
3/2 plus c.
So the first column is good.
And then we just replace the third column with the third
column plus the second column.
And we get 2, minus 1, minus 1, 0, 3/2, minus
3/2, 0, 0, and c.
And so the pivots are 2, 3/2, and c.
And again, the answer is as before.
It's positive definite if c is strictly bigger than 0, and
it's positive semidefinite if c is greater or equal to 0.
But I want you to notice something else here.
So before, we had 2, 3, and 3c.
And now for determinants, as these pivots, we have 2, 3/2,
which is the second determinant over the first
determinant, and c, which can be thought of as 3c/3, so the
third determinant over the second determinant.
And something like this is always going to happen.
And finally, let me do the energy test, or
completing the square.
So one of the definitions of positive definite, one could
say the definition because of which we are really interested
in such matrices, is the following.
It's positive definite if when we consider this quadratic
form, so a form that maps x, y, and z to this expression
here that's going to be quadratic in x, y, and z, it's
positive semidefinite if this is always greater
or equal than 0.
And it's positive definite if, when we have an expression
like this and try to solve this equals 0, the only
solution is that x, y, and z all have to be 0.
So let's try calculating this form, completing the squares
on it, and seeing these numbers show up again.
So when I multiply this like this, put in a B, do the
multiplication, we get something like this.
2x squared plus 2y squared plus 2 plus c z squared minus
2xy minus 2xz minus 2yz.
And now let's try completing the squares using the formulas
that I prepared for you up here in this pop-up window in
the corner.
So let's try completing, we have a formula for the square
of a plus b plus c and the square of a plus b.
So first, we try to get something squared so that this
something has all the x's that appear here, so that we get
something squared plus some expression that only
takes y's and z's.
I'm not going to do the calculation in front of you,
further embarrass myself with it.
But let me tell you that what you get is 2 times x minus 1/2
y minus 1/2 z squared.
And this ate all the x's that showed up here.
The remainder only has y's and z's.
When you use the second of those formulas in a pop-up
window to complete the square of y's and z's, you get 3/2
times y minus z squared.
So this took up all the y's, and we're left just with a z
that comes as c times z squared.
And now, let's look at the following.
This is a square of some real number.
So that's positive.
This is a square of some real number, so that's positive.
And this is a square of some real
number, so that's positive.
They're all multiplied by positive numbers, 2, 3/2, and
c, which we've already seen here.
And so if c is bigger or equal than 0, this is certainly
always bigger or equal than 0.
Now to the question of if this can be equal to 0 without x,
y, and z all being 0, well, let's look at two cases.
If c is strictly bigger than 0, then let me write this
matrix here as 2, 3/2 and c times 1, minus 1/2, minus 1/2,
0, 1, minus 1, and 0, 0, 1.
Let's imagine c is strictly bigger than 0.
And let's see when can this expression be equal to 0.
Well, as we said, it's a sum of certain squares.
They're all greater or equal to 0, so they all have to be
equal to 0 for this expression to be 0.
In other words, z needs to be equal to 0.
y minus z needs to be equal to 0.
And x minus 1/2 y minus 1/2 z also needs to be equal to 0.
And since this matrix has all the pivots, this can only
happen if x, y, and z are all equal to 0.
On the other hand, if c is equal to 0, then let me write
this matrix here again.
Take 2 out here.
Take 3/2 out here.
And write again 1, minus 1/2, minus 1/2, 0, 1, minus
1, and 0, 0, 0.
And so imagining that c is 0, when can this whole expression
be equal to 0?
Well, the last bit is already 0.
We need to have y minus z equal to 0.
And we need to have 1 times x minus 1/2 times y minus 1/2
times z also equal to 0.
So we need to have x, y, z in the kernel of this matrix.
But this matrix only has two pivots.
It has one free variable.
So we can find a solution to this times x, y, z equals 0
that's not 0 itself.
Namely, if you remember how to solve systems like this, we
set z equal to 1.
From this, we calculate that y has to be 1.
And then from this, we calculate that x has to be 1.
And you can check that for in case c is 0, this thing here,
when you plug it in here, you really get 0.
In fact, this thing here is in the kernel of the matrix B. In
fact, in case c is 0, the columns of matrix B sum up to
0 because of this here.
And that's all I wanted to say for today.