Perpframes, Aligners and Hangers
Linear Systems, Pseudo-Inverse
Matrix Norm, Rank One
Matrix SubspacesIn the last section you battled to show that every matrix A could be written as A=(hanger)(stretcher)(aligner).
What's the pay off?
2x2 ExampleHere is an SVD of a 2 x 2 matrix :
Now watch what the matrix A does to an ellipse lined up on the perpframe .
The matrix stretches the ellipse and transfers it from the
perpframe to the
If you look carefully at the "during" plot you'll see that A
More generally A,
You can verify this by computing what each factor of A does.
Check what each factor of A does to :
(The aligner matrix brings to the y-axis.)
(The stretcher matrix stretches the y-axis by a factor of .)
(The hanger rotates the y-axis to the direction of .)
Theorem:If you build a 2x2 matrix A=(hanger)(stretcher)(aligner) using
These two equations are the fundamental equations
of matrices of the form (hanger)(stretcher)(aligner).
In words they say that A
Proof: Outlined above--follow the action of each factor of A on .
3 x 3 matrices.The following plot shows the perpframe in blue and the perpframe in red.
See what the matrix does to the surface lined up on .
Just as in the 2x2 case, the matrix A obeys the fundamental equations:
In fact, matrices of the form (hanger)(stretcher)(aligner) always obey these fundamental equations, regardless of their dimension.
This decomposition presents orthonormal bases for
Proofs:The Column space of A, Col[A], is the span of the columns of A.
(The column way to multiply A v.)
(Since the a's are a basis.)
This shows shows that Col[A] = span.
Since are perpendicular unit vectors they are an orthonormal basis for Col[A].
The nullspace of A, N[A], is the set of vector that A sends to the zero
(Linearity and Fundamental Equations.)
tells you that
and so .
This tells you that any vector in N[A] is a linear combination of and , i.e. .
On the other hand, ,
The row space of A is the span of the rows of A, which is the same as the column space of .
So finding a basis for the row space of A is the same as finding a basis
for the columns space of .
Here's a look at :
Now you have written as (hanger)(stretcher)(aligner),
i.e., you have an SVD of .
So from what you did above you know
CommentsThe proofs above easily generalize to show that
Definitions:The rank of A is usually defined as the dimension of the column space.
Since the columns of the hanger matrix corresponding to non-zero singular
values form a basis for the column space, you know that the rank of A is
equal to the number of non-zero singular values.
Similarly, since the nullity of A is the dimension of the null space, the number of zero singular values equals the nullity of A.
=(number of non-zero singular values)+(number of zero singular values)
=(total number of singular values)
=(number on non-zero singular values of A)
(number of non-zero singular values of )
(1) Since the stretcher matrix for
is the transpose of the stretcher matrix for A, the singular values
for A and
give orthonormal bases for
a. the column space of A
b. the row space of A
c. the nullspace of A
d. the nullspace of .
2. Let . Here is an SVD for A.
Based on the SVD, give an orthonormal basis for the subspace spanned by
3. Let . Here is an SVD for A.
Based on the SVD of A and the fundamental equations, compute
If you check your answer by using , you may not get agreement since the SVD for A has been rounded to two decimals.