Welcome to a lesson
on how to determine the fundamental subspaces of a matrix
given the singular value decomposition of the matrix.
Let A be an m by n matrix.
The fundamental subspaces of A are:
the null space of A, the null space of A transpose,
the column space of A, and the column space of A transpose,
which is also equal to the row space of A.
In the previous lesson,
we discussed how to find the fundamental subspaces
for any matrix.
In this lesson,
we'll focus on determining the fundamental subspaces
given the singular value decomposition of matrix A.
Where if A is equal to U times sigma times V transpose,
U times sigma times V transpose
is the singular value decomposition of matrix A.
And we can quickly and easily find the fundamental subspaces
if we have the singular value decomposition.
So in our previous lesson,
we found the singular value decomposition of matrix A,
which is shown below, where matrix U is the first matrix,
matrix sigma is the second matrix
and the matrix V transpose is the third matrix.
We will now use matrix U and the matrix V transpose
to find the four fundamental subspaces of matrix A.
So if A is an m by n matrix with rank A equal r,
then the fundamental subspaces of A are given below,
where the set containing the vectors U sub one to U sub r
is an orthonormal basis for the column space of A.
Remember vectors are orthonormal,
they're unit vectors
that are also orthogonal or perpendicular.
And we find the U vectors from the columns of matrix U.
In the set containing the vectors U sub r plus one
through U sub m
is an orthonormal basis for the null space of A transpose.
And the set containing the vectors V sub one
through V sub r
is an orthonormal basis for the column space of A transpose.
Remember the V vectors are from the columns of matrix B,
which would be the rows in the matrix V transpose.
And finally, the set containing the vectors V sub r plus one
through V sub n
is an orthonormal basis for the null space of A.
Let's begin by determining the column space of matrix A.
Well first notice how,
if we write matrix A in reduced row echelon form,
there are two lending entries,
which indicates the rank of A is equal to two,
and therefore r is equal to two.
Which means the set containing the vectors U sub one
and U sub two
is an orthonormal basis for the column space of A.
And again, the U vectors are from the columns of matrix U,
which is the first matrix.
Well notice how matrix U only has two columns,
and therefore these two columns
form an orthonormal basis for the column space of matrix A,
and therefore we can say the column space of matrix A
is equal to the span of the set containing
these two vectors.
So again these two vectors are orthonormal,
meaning they're unit vectors
and also orthogonal or perpendicular.
And now let's find the null space of A transpose,
remember r is equal to two
so the first vector would be the vector U sub three.
But notice matrix U only has two columns, not three columns,
this indicates the zero vector
is an orthonormal basis for the null space of A transpose,
or we can say the null space of A transpose
is equal to the span of the set containing the zero vector,
which of course would just give us a zero vector in r two.
So in the case where the column in matrix U
or the row in the matrix V transpose does not exist,
the result would be the zero vector.
And now it's find the column space of A transpose
as well as the null space of A on the next slide.
Notice the column space of A transpose
and the null space of A involve the V vectors.
We will find the V vectors in the rows of V transpose.
So the set containing the vectors V sub one through V sub r,
or in our case V sub one and V sub two
is an orthonormal basis for the column space of A transpose
or the row space of A.
We find these two vectors
using the first two rows of V transpose,
and therefore we can say
that the column space of A transpose or the row space of A
is equal to the span of the set containing the vectors
from the first two rows of the transpose.
And then finally, and then find the set
containing the vectors V sub r plus one through V sub n
is an orthonormal basis for the null space of A.
And again, if r is equal to three
and n is equal to three as well,
the set containing the vector V sub three
is an orthonormal basis for the null space of A.
We find the vector V sub three
in the third row of V transpose, this vector here,
and therefore the null space of A
is equal to the span of the set containing
this vector from the third row of V transpose.
So let's go ahead and summarize what we found.
We found the four fundamental subspaces of matrix A
using the singular value decomposition of A.
So not only do these sets above
form a basis for the indicated subspace,
they are also orthonormal,
which again means the vectors are unit vectors
and also orthogonal or perpendicular.
I hope you found this helpful.