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 with mine.
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 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 over 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.
2 x squared plus 2 y squared
plus 2 plus c z squared minus
2x*y minus 2x*z minus 2y*z.
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
to 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.
