Question 3698: 1. Find bases for the following subspaces of F5:
W1 = {(a1, a2, a3, a4, a5) Î F5: a1 – a3 – a4 = 0} and
W2 = {(a1, a2, a3, a4, a5) Î F5: a2 = a3 = a4 and a1 + a5 = 0}.
What are the dimensions of W1 and W2?
2. The set of all upper triangular n x n matrices is a subspace W of
M n x n (F). Find a basis for W and determine its dimension.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! 1. Sol:
About W1, since a1 – a3 – a4 = 0, or a1 = a3 + a4.
We see that {(s+t,r,s,t,u)|r,s,t,u in F} is a general
form of vectors in W1.
Choose a1=1, a3=-a4 = 1 and then a4=0, a1=a3=1
we see that (1,a2,1,-1,a5), (1,a2,1,0,a5) are linear independent
of W1 for all scalars a2,a5 in F.
Thus, {(1,1,1,-1,0), (1,0,1,-1,1),(1,1,1,0,0),(1,0,1,0,1)}
is a set of 4 independent vectors.
Also, note W1 is a proper subspace of F5 and hence dimW1 <=4.
We claim that {(1,1,1,-1,0), (1,0,1,-1,1),(1,1,1,0,0),(1,0,1,0,1)}
a basis of W4 and hence Dim W1 = 4.
For W2, since a2 = a3, a3 = a4 and a1 + a5 = 0.
We see that {(s,t,t,t,-s)|s,t in F} is a general
form of vectors in W2.
Clearly, dim W2 = 2 from this general form.
By choosing a2=a3 = 1, a3=a4=1, and a1=a5 = 0.
And, choose a2=a3=a4 =0, and a1= -a5=1.
We obtain two independent vectors (1,1,1,0,0) and
(1,0,0,0,-1) in W2.
Thus, {(1,1,1,0,0),(1,0,0,0,-1)} is a basis of W2 and
dim W2 = 2.
2. Note the dimension of M n x n (F) isn nxn = n^2.
There are (1+2+3+..+n) = n(n+1)/2 free entries in
an upper upper triangular of M n x n (F).
Hence, dim W = n(n+1)/2.
[Discussion of question 1:
First of all , try to find the dimensions of W1 and W2.
Since the restraint of W1 is the linear equation
a1 – a3 – a4 = 0, so dim W1 = 5 -1 = 4.
For W2, there are three linear restraints, namely,
a2 = a3, a3 = a4 and a1 + a5 = 0
if they are independent,then dim W2 = 5 -3 = 2.
Since, clearly,two of the 3 equations are independent,
we see that dim W2 = 2 or 3. ]
Kenny
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