Hi!!
1) Verify that W is not a subspace of the vector space (give a
specific example that violates the test for a vector subspace). W is
the set of all matrices in Mn,n such that A^2 = A.
The identity matrix I is in W (I^2 = I * I = I) but A = I + I = 2*I is not in W:
A^2 = A * A = (2*I) * (2*I) = (2*2)*(I * I) = 2*2*I = 2*(2*I) = 2*A
that is different to A.
Then W is not a subspace.
--------------------------
2) Given that {u1, u2, ... , un} is a linearly independent set of
vectors and that the set {u1, u2, ... , un, v} is linearly dependent,
prove that v is a linear combination of the ui's.
By definition we have that :
The set {u1, u2, ... , un, v} is linearly dependent iff there exist
scalars a1, a2, ..., an, av, not all zero, such that:
a1*u1 + a2*u2 + ... + an*un + av*v = 0
Now if av = 0 then we have that:
0 = a1*u1 + a2*u2 + ... + an*un + 0*v =
= a1*u1 + a2*u2 + ... + an*un
But the problem condition says that a1, a2, ...,an, are not all zero;
this cannot be possible because {u1, u2, ... , un} is a linearly
independent set of vectors.
Then is av non zero, so:
0 = a1*u1 + a2*u2 + ... + an*un + av*v
then:
-av*v = a1*u1 + a2*u2 + ... + an*un
if we call:
bi = ai/(-av) we have:
v = b1*u1 + b2*u2 + ... + bn*un and this is what we want to prove.
------------------------
3) Find the rank of the matrix:
_ _
| 1 2 3 ... n |
| n+1 n+2 n+3 ... 2n |
| 2n+1 2n+2 2n+3 ... 3n |
| . . . . |
| . . . . |
| . . . . |
|_n^2-n+1 n^2-n+2 n^2-n+3 n^2_|
for n= 2, 3, and 4. Can you find a pattern in these ranks?
To solve this problem we need to use elementary row operations to find
the row reduced form of each matrix.
See for references at:
"Elementary Row Operations" :
http://math.aubg.bg/materials/week01/rowops.htm
"Row Reduced Form":
http://cnx.rice.edu/content/m10295/latest/?format=pdf
Recall that the rank of a matrix A is the number of leading entries in
a row reduced form R for A. This also equals the number of nonrzero
rows in R.
n=2
_ _
| 1 2 |
|_ 3 4 _| --->
By replacing the second row R2 by 3*R1-R2 :
_ _
| 1 2 |
|_ 0 2 _|
The rank is 2.
n=3
_ _
| 1 2 3 |
| 4 5 6 | --->
|_ 7 8 9 _|
By replacing the second row R2 by 4*R1-R2 :
_ _
| 1 2 3 |
| 0 3 6 | --->
|_ 7 8 9 _|
By replacing the second row R3 by 7*R1-R3 :
_ _
| 1 2 3 |
| 0 3 6 | --->
|_ 0 6 12 _|
Since R3 = 2*R2 we can eliminate it:
_ _
| 1 2 3 |
| 0 3 6 | --->
|_ 0 0 0 _|
The rank is 2
n=4
_ _
| 1 2 3 4 |
| 5 6 7 8 |
| 9 10 11 12 | --->
|_13 14 15 16 _|
By replacing the second row R2 by 5*R1-R2 :
_ _
| 1 2 3 4 |
| 0 4 8 12 |
| 9 10 11 12 | --->
|_13 14 15 16 _|
By replacing the third row R3 by 9*R1-R3 :
_ _
| 1 2 3 4 |
| 0 4 8 12 |
| 0 8 16 24 | --->
|_13 14 15 16 _|
By replacing the third row R4 by 13*R1-R4 :
_ _
| 1 2 3 4 |
| 0 4 8 12 |
| 0 8 16 24 | --->
|_ 0 12 24 36 _|
The rank is 2
In general you can easyly verify that in the matrix
_ _
| 1 2 3 ... n |
| n+1 n+2 n+3 ... 2n |
| 2n+1 2n+2 2n+3 ... 3n |
| . . . . |
| . . . . |
| . . . . |
|_n^2-n+1 n^2-n+2 n^2-n+3 n^2_|
the rows have the following relationship:
R1 and R2 are independent;
R3 = 2*R2 - 1*R1
R4 = 3*R2 - 2*R1
Ri = (i-1)*R2 - (i-2)*R1
Rn = (n-1)*R2 - (n-2)*R1
Then for i>2, all rows Ri can be expresed as a linear combination of
R1 and R2, then the rank of all these matrices are 2.
----------------------
3) Let A and B be n x n matrices with A does not equal to 0 and B does
not equal to 0.
Prove that if A is symmetric and B is skew-symmetric (B^T = -B), then
{A,B} is a linearly independent set.
A square matrix A is symmetric if A = A^T, that is a(i,j)=a(j,i).
A square matrix B is skew-symmetric if B = -B^T, that is b(i,j)=-b(j,i).
This means that the diagonal elements or B are all zero:
In effect b(i,i)=-b(i,i) implies that b(i,i)= 0 .
Suppose that {A,B} is a linearly dependent set, then exist a nonzero
scalar k that satisfies:
A = k*B
Then for all (i,j) is:
a(i,j) = k*b(i,j)
and
a(j,i) = k*b(j,i)
Since B is skew-symmetric we have that
a(i,j) = k*[-b(j,i)] = -k*b(j,i)
Then, for all (i,j):
-k*b(j,i) = k*b(i,j) = a(i,j) = a(j,i) = k*b(j,i)
Then is:
k*b(j,i) = 0 for all (i,j)
And since not all b(i,j) are zero results that k=0, this contradict
the initial asumtion that k is different to zero and the supposition
of the linear dependence of A and B results in this absurd, then the
set {A,B} is linearly independence.
------------------------
4) Use the Gram-Schmidt orthonormalization process to transform the
given basis for R^n into an orthonormal basis. Use the Euclidean inner
product for R^n and use the vectors in the order in which they are
given.
B = {(3,4,0,0), (-1,1,0,0), (2,1,0,-1), (0,1,1,0)}
For references on the Gram-Schmidt method see:
"The Gram-Schmidt Algorithm":
http://www.math.hmc.edu/calculus/tutorials/gramschmidt/
"physics - Gram-Schmidt process":
http://www.physicsdaily.com/physics/Gram-Schmidt_process
"Gram-Schmidt Orthogonalization":
This one is a nice way to understand what are you doing via the
Gram-Schmidt process.
http://www.mste.uiuc.edu/exner/ncsa/orthogonal/
We want to find a base of orthonormal vectors e1, e2, e3, e4.
We have the vectors:
u1 = (3,4,0,0)
u2 = (-1,1,0,0)
u3 = (2,1,0,-1)
u4 = (0,1,1,0)
Set v1 = u1
Then:
e1 = v1 / ||v1||
= u1/sqrt(3^2+4^2+0^2+0^2) =
= u1/sqrt(25) =
= (3,4,0,0)/5 =
= (3/5,4/5,0,0)
v2 = u2 - <u2,e1>*e1 =
= (-1,1,0,0) - (-3/5+4/5+0+0)*(3/5,4/5,0,0) =
= (-1,1,0,0) - 1/5*(3/5,4/5,0,0) =
= (-1,1,0,0) - (3/25,4/25,0,0) =
= (-28/25,21/25,0,0)
Then
e2 = v2 / ||v2|| =
= (-28/25,21/25,0,0)/(7/5) =
= (-4/5,3/5,0,0)
v3 = u3 - <u3,e1>*e1 - <u3,e2>*e2 =
= (2,1,0,-1) - (6/5+4/5)*(3/5,4/5,0,0) - (-8/5+3/5)*(-4/5,3/5,0,0) =
= (2,1,0,-1) - (6/5,8/5,0,0) - (4/5,-3/5,0,0) =
= (0,0,0,-1)
Then
e3 = v3 / ||v3|| =
= (0,0,0,-1) / 1 =
= (0,0,0,-1)
v4 = u4 - <u4,e1>*e1 - <u4,e2>*e2 - <u4,e3>*e3 =
= u4 - 4/5*(3/5,4/5,0,0) - 3/5*(-4/5,3/5,0,0) - 0*e3 =
= (0,1,1,0) - (12/25,16/25,0,0) - (-12/25,9/25,0,0) =
= (0,0,1,0)
Then
e4 = v4 / ||v4|| =
= (0,0,1,0) / 1 =
= (0,0,1,0)
B' = {(3/5,4/5,0,0), (-4/5,3/5,0,0), (0,0,0,-1), (0,0,1,0)}
---------------------------------------------------------
I hope that this helps you. If you find something unclear, incomplete
or a mistake (why not, it is 4am here) please use the clarification
feature in order to request for further assistance and/or a
correction. I will gladly respond to your requests.
Best regards.
livioflores-ga |
,
0 Comments
)