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Expert-verified Found in: Page 145 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Consider two subspaces ${\mathbf{V}}$ and ${\mathbf{W}}$ of ${{ℝ}}^{{n}}$, where ${\mathbf{V}}$ is contained in ${\mathbf{W}}$. Explain why ${\mathbf{dim}}\left(\mathbf{V}\right){\le }{\mathbf{dim}}\left(\mathbf{W}\right)$. (This statement seems intuitively rather obvious. Still, we cannot rely on our intuition when dealing with ${{ℝ}}^{{n}}$.)

If two subspacesand $W$of ${ℝ}^{n}$, where $\mathbf{V}$ is contained in $\mathbf{W}$ , then $V$ is a basis of $W$ and therefore, $dimV\le dimW$ .

See the step by step solution

## Step 1: Define the basis.

Independent vectors and spanning vectors in a subspace of

Consider a subspace $V$ of ${ℝ}^{n}$ with $dim\left(V\right)=m$.

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 ${ℝ}^{n}$ consist of the same number of vectors

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

Given that, two subspaces of ${ℝ}^{n}$ 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 ${B}_{W}=\left\{{w}_{1},{w}_{2},{w}_{3},...,{w}_{k}\right\}$ 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 $n\le m$, therefore, $dimV\le dimW$.

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