Question
Asked By StarryKnight48 at
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Spencer
Expert · 2.1k answers · 2k people helped
Step 1/3
(a) False As the vectors u and v are given to be linearly independent, they can never be parallel. Parallel vectors are linearly dependent.
(b) True
As W is a subset of V , and W is generated by span of two vectors, it implies it is an subspace of V
(c) False We only know that V is a 5 dimensional subspace but we do not know if V is `\mathbb{R}^5` , it can be a 5 dimensional subspace of `R^6` .Hence it is not correct to say that W is an subspace of `R^5` .(d) False
As explained in previous part, it is possible for u to be of size more than 5
Step 2/3
(e) False
As V is a 5-dimensional space then it has at least 5 coordinates, and hence u and v can not form a basis for `R^2` .(f) True
as W is spanned by u and v. So, {u,v} forms an ordered basis for W.
(g) True
As W is generated by 2 linearly independent vectors, so its dimension is 2 and it is also a subspace of V. This means W is a 2-dimensional subspace of V.
(h) False
W has more than 2 coordinates as u and v are elements of a 5-dimensional space. So u and v does not belong to `R^2` and hence W is not a subspace of `R^2` .Step 3/3
(i) True
As basis B has 2 element so `[w]_B` is an 2- dimensional vector and an element of `R^2`(j) True
In terms of ordered basis B. `u=[1;0] \text{ and } v=[0;1] . ` So, u+v=[1;1]Final Answer
a) False
b) True
c) False
d) False
e) False
f) True
g) True
h) False
i) True
j) True
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