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Q62E

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Linear Algebra With Applications
Found in: Page 145
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Consider two subspaces V and W of n, where V is contained in W. Explain why dim(V)dim(W). (This statement seems intuitively rather obvious. Still, we cannot rely on our intuition when dealing with n.)

If two subspacesand Wof n, where V is contained in W , then V is a basis of W and therefore, dimVdimW .

See the step by step solution

Step by Step Solution

Step 1: Define the basis.

Independent vectors and spanning vectors in a subspace of

Consider a subspace V of n with dim(V)=m.

a. We can find at most m linearly independent vectors in V .

b. We need at least m vectors to span V.

c. If m vectors in are linearly independent, then they form a basis of V .

d. If m vectors in span , then they form a basis of V .

Number of vectors in a basis

All bases of a subspace V of n consist of the same number of vectors

Step 2: Find reason of dim(V)≤dim(W) .

Given that, two subspaces of n that are V and W, where V is contained in W.

So, from this, we have some vectors from basis of W that span V. Also, the number of vectors in a basis of a space is the same as the dimension of space.

Let us assume that BW={w1,w2,w3,...,wk} is a basis of W and it has m vectors.

Since V, is the subspace of W , it needs to be spanned by some vectors of W and those vectorsare linearly independent. So, V is a basis of W and it has n vectors.

Since nm, therefore, dimVdimW.

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