This lesson will prove Rn is a vector space.
In the previous lesson,
we defined the vector space as a set of vectors
with two operations defined,
addition and scalar multiplication,
which satisfy the axioms of addition
and scalar multiplication.
The five axioms of addition are shown below.
[keyboard clacking]
And here we have the five axioms of scalar multiplication.
[keyboard clacking]
In order to prove that Rn is a vector space,
we need to show that all 10 axioms hold true.
To begin we will let vector X, vector Y
and vector Z be three vectors in Rn.
We will begin by proving the axioms for vector addition.
So number one, to show the Rn is closed under addition,
we must show that for two vectors in Rn,
their sum is also in Rn.
Let's consider the sum of vector X and vector Y.
To find the sum we add the corresponding elements.
And notice the sum is a vector within entries showing
that it is in Rn, hence Rn is closed under vector addition.
We've now completed step one of 10 to prove Rn
is a vector space.
Number two, to show that addition is commutative,
we will again consider vector X plus vector Y.
To find the sum we add the corresponding elements.
Once we have the sums inside the column matrix,
we can use the commutative property of addition
for real numbers and change the order of the sums,
which we see here in the column matrix below.
And then we can write the column matrix
as a sum of two column matrices,
where now the first column matrix is vector Y
and the second column matrix is vector X.
We've now shown vector X plus vector Y
is equal to vector Y plus vector X.
Hence, addition of vectors in Rn is commutative.
We've not completed a step two of 10.
Number three, we will now show that addition of vector in Rn
is associative in a similar way.
We begin with parentheses around the sum of vector X
and vector Y and then we have plus vector Z.
We begin by determining the sum of the two vectors
are column matrices inside the parentheses.
And then we find the sum of these two column matrices,
which gives us this column matrix here.
And again, once we have everything inside one column matrix,
we can use the associative property of addition
for real numbers and change the grouping from being around
the sum of the X and Y components
to being around the sum of the Y and Z components.
Once we've used the associative property of addition
for real numbers,
we can break the column matrix
into a sum of two column matrices,
and then three column matrices,
where now we have grouping around
the sum of vector Y and vector Z.
And now we've shown the addition of vectors in Rn
is associative and we've completed step three of 10.
Number four, we need to show the existence
of the additive identity.
We begin by letting the zero vector be column matrix
within zero entries.
And then we consider a vector X plus the zero vector,
adding the corresponding elements
and simplifying we do get vector X showing
that vector X plus the zero vector is equal to vector X,
hence the zero vector is an additive identity.
We've not completed step four of 10.
Number five, we need to prove
the existence of an additive inverse.
To do this,
we let the opposite of vector X have components equal to
the opposite of the components of a vector X.
And then we consider the sum of vector X
and the opposite of vector X,
adding the corresponding element and simplifying.
Notice how we do get the zero vector,
hence the opposite of vector X is an additive inverse.
Now that we've proven the five axioms of addition,
we need to prove the five axioms related
to scalar multiplication.
To do this, we will let a and b be real numbers
and vector X and vector Y be vectors be vectors in Rn.
We first show that Rn is closed under scalar multiplication.
To do this, we will show that a times vector X
is also a vector with n entries.
So to find a times vector X,
we multiply a by each component of vector X
and notice how the result is a vector
within entries showing Rn is closed
under scalar multiplication.
Step six is complete.
Step seven, we need to show that a times
the sum of vector X and Y
is equal to a times vector X plus a times vector Y.
To begin, we find the sum of vector X and vector Y
in the parentheses by adding the corresponding elements
and then we multiply a by each of the sums.
Once we have everything inside of one column matrix,
we can use the distributive property
and distribute a to the sum of the X and Y values
inside the parentheses, which we see here in the next line.
And then from here, we can break the sum
into the sum of two column matrices,
where the first column matrix is a times vector X
and the second matrix is a times vector Y.
So now we've shown that a times the sum
of vector X and vector Y
is equal to a times vector X plus a times vector Y.
Step seven is complete.
Step eight, we wish to show that
the quantity a plus b times vector X
is equal to a times vector X plus b times vector X.
Because the a plus b is in parentheses,
we begin by performing scalar multiplication
and then once we have everything inside one column matrix,
we apply the distributive property.
And then once again, we can break this up into
the sum of two column matrices
where the first column matrix is a times vector X
and the second matrix is b times vector X,
which shows the quantity a plus b times vector X
is equal to a times vector X plus b times vector X.
Step eight is complete.
Step nine, we need to show that a times
the product of b in vector X is equal to
the product of a and b times vector X.
The first step is to perform scalar multiplication with b,
which we see here in the second line
and then we perform scalar multiplication with a.
Once we have everything inside one column matrix,
we can use the associative property multiplication
and change the grouping from being around the b's
and the X components to the a's and the b's.
And then we can factor out the product of a and b,
which gives us the product of a and b times vector X,
which is what we were trying to prove.
Step nine is complete.
And for our last step need to show
that one times vector X equals vector X.
Performing scalar multiplication with one,
here how we do it,
just get the components of vector X showing one
times vector X is equal to vector X.
And step 10 is complete.
So by the above proofs, it is clear that Rn satisfies
the vector space axioms,
hence Rn is a vector space under the usual operations
of the vector addition and scalar multiplication.
I hope you found this helpful.