pp. 247-248
- For each of the following matrices, determine a basis for each of the subspaces R(AT), N(A), R(A), and N(AT)
(a)
Change A to row-echelon form
Basis of R(AT) is the basis of row space of A which is .
N(A) is the null space of A which is the solution of
.
Hence, the basis of N(A) is
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
Basis of R(A) is the basis of row space of AT which is .
N(AT) is the null space of A which is the solution of
.
Hence, the basis of N(AT) is .
(b)
To find the basis of R(AT) we have A to its reduced row-echelon form.
Bases of R(AT) is the basis of row space of A which are .and
N(A) is the null space of A which is the solution of
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
Bases of R(A) is the basis of row space of AT which are .and
Since, the rank of A is 2 which is equal to number of column of 2, the dimension of N(AT) is equal to number of column minus number of rank = 2-2 =0. This means that N(AT) consists of only. Hence, N(AT) does not have basis.
(c)
To find the basis of R(AT) we have A to its reduced row-echelon form.
Bases of R(AT) is the basis of row space of A which are .and
Since, the rank of AT is 2 which is equal to number of column of 2, the dimension of N(A) is equal to number of column minus number of rank = 2-2 =0. This means that N(A) consists of only. Hence, N(A) does not have basis.
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
Bases of R(A) is the basis of row space of AT which are .and
N(AT) is the null space of AT which is the solution of
.
Here, we have
,
which lead to
.
Hence,
(d)
To find the basis of R(AT) we have A to its reduced row-echelon form.
Bases of R(AT) is the basis of row space of A which are , .and
N(A) is the null space of A which is the solution of
which can be written in the form of augmented matrix as
Here, we have
,
and
which lead to
Hence,
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
Bases of R(A) is the basis of row space of AT which are , .and
N(AT) is the null space of AT which is the solution of
.
Here, we have
,
and
which lead to
which lead to
Hence,
2. Let S be the subspace spanned by
Let A= be a matrix of size 1 by 3. We observe that R(AT), row space of A, is spanned by x. By Theorem 5.2.1, the .
Null space of A is determined by solving
From the above equation, we have
or
Hence, the basis of is and .
4. Let S be the subspace in spanned by , . Find a basis of
Let A= be a matrix of size 2 by 4. We observe that R(AT), row space of A, is spanned by x. By Theorem 5.2.1, the .
Null space of A is determined by solving
,
which can be written in the augmented form as
From the above equation we can conclude that
and
This leads to
Hence, the basis of is and .
pp. 258-259
1. Find the least square solution to each of the following systems.
(a)
and
We have that
and
.
The least square solution is determined by solving
,
which can be written in the augmented matrix form as
.
Hence, the solution is and
(b)
and
We have that
and
.
The least square solution is determined by solving
,
which can be written in the augmented matrix form as
.
Hence, the solution is and
(c)
We have that
and
.
The least square solution is determined by solving
,
which can be written in the augmented matrix form as
.
Hence, the solution is , and
2.
For (1.a), we have
and
.
Check that by determining. If then .
For (1.b), we have
and
.
Check that by determining . If then
For (1.c), we have
and
.
Check that by determining . If then
5.
(a) Find the best least square fit by a linear function to the data
Liner function implies that where a and b are to be determined by the data. Hence, we write
,
or
.
Obviously, the above equation is inconsistent. We use the least square approach to solve the problem.
and
The least square solution is determined by solving
,
which can be written in the augmented matrix form as
Hence, the solution is and
(b)
7. Given a collection of points (x1,y1)…(xn,yn) and Let
and
and let be the linear function that gives the best least square fit to the points.
We can write in the form of equations as
or in the form of matrix as
Obviously, the above equation is inconsistent. We use the least square approach to solve the problem.
and
Since we have
The least square solution is determined by solving
,
which can be written in the augmented matrix form as
Hence, the solution is and .
pp. 268
7.
(a)
(b)
(c)
8.
(a)
Hence,
(b) p is the projection of 1 onto x
Hence, we have
(c)
So, we have
.
9
Hence, are orthogonal
Since are orthogonal, the Pythagorean law holds. As a result, we have
Hence distance between two vectors is .
pp. 286-289
2
(a)
, ,
(b)
./
Hence, we have
and
9.
(a)
(b.i)
(b.ii)
(b.iii)
(b.iv)
26.
(a)
Hence, 1 and 2x-1 are orthogonal .
(b)
Hence , we have
Hence , we have .
(c)
We can approximate as