In this problem we will establish a few more relations between the idea of dimen

Question

In this problem we will establish a few more relations between the idea of dimension and the ideas of linear dependence and spanning sets. Some of the parts (like (e)) should make intuitive sense. Proving that such statements are actually true is a sign that our definition of dimension is a good one, and also means that we can use these properties freely in the future when we are thinking about subspaces.

Parts (a)–(d) can be solved by using the definition of dimension and the Key Lemma from the class of Thursday, November 17th (1st picture provided below) . Part (f) will likely require the proposition from Tuesday, November 15th (2nd picture provided below).

Suppose that W ? Rn is a subspace, that dim(W ) = d (i.e., W is of dimension d), and that ?v1,..., ?vm are vectors in W.

(a) Explain why you know that there is a set of d vectors in W which span W .

(b) If m > d prove that ?v1,. . . , ?vm must be linearly dependent.

(c) Explain why you know that there is a set of d vectors in W which are linearly independent.

(d) If m < d prove that ?v1,..., ?vm cannot span W.

Now suppose that V is another subspace of Rn, and that V ? W .

(e) Prove that dim(V ) =< dim(W ). (Suggestion: take a basis for V and use the Key Lemma.)

(f) If dim(V ) = d (i.e., the same dimension as W ) prove that V = W . (Suggestion: Try proving this by contradiction. Assume that V /= W then start with a basis for V and add to it a vector in W but outside of V. Then use part (b) of the Proposition from Tuesday, November 15th (2nd picture provided below), to get a contradiction.)

(g) Find the mistake in the following argument. Let
W = {(x, y, z) ? R3 | x + y + z = 0 }.

We have seen in class that (1,0,?1), (0,1,?1) span W. On the other hand, we know that ?e1, ?e2, and ?e3 are linearly independent. The Key Lemma says “(# in spanning set) >= (# in linearly independent set)”. Applying this to the spanning set (1, 0, ?1), (0, 1, ?1), and the linearly independent set ?e1, ?e2, ?e3 gives 2>= 3, which is clearly false. What went wrong?

Notes: (i?) If we take W = Rn (so that d = dim(W) = dim(Rn) = n), then 1(b) becomes the statement that “If we have more than n vectors in Rn, the vectors must be linearly dependent”, which was a lemma from the class of Friday, November 11th. That is, the lemma from that class is a special case of 1(b), which tells us what the general statement is for any subspace.

(ii?) As a result of (a) and (b), the number d is the largest size of any set of linearly independent vectors in W . As a result of (c) and (d) the number d is the smallest number of vectors in any set which spans W . Put together, this tells us that the following three numbers are equal :

? The number of elements in a basis for W.
? The smallest number of vectors in any spanning set for W.
? The largest size of any set of linearly independent vectors in W.

This common number is dimW.

Explanation / Answer

a) given that the subspace W has dimension d.

So it has a basis consists of d vectors.

And all the vectors in basis together will span W.

So there is one.

b) if possible ,that given m vectors are linearly dependent.then it can be extended to a basis for W.

Hence dimension of W will be greater. than d.

But it's impossible as dimension is unique.

C) as (a) there must be a basis and basis must be linearly independent

d) if possible that given m vectors with m<d can span W.

Then any set of d vectors cannot be linearly independent.

Hence W cannot have a basis of d vectors.contradiction.

e) as V is subspace of W. V can be spanned with same vectors that span W.hence V cannot have basis of more than d vectors.hence dim V <= dim W.

f) if possible ,assume V /= W and without loss of generality assume V is subspace of W.

Then there is atleast one vector in W that not in V.

Hece a vector in W cannot be spanned by spanning set of V.

so dimension of W must be greater than that of V.

It's a contradiction.assumption is wrong.

g) mistake is that e1,e2,e3 is not linearly independent in subspace W.but it's linearly independent in R3 only.

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