Question

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 ${ℝ}^n$.)

Short answer

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

Step-by-step solution

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 inare linearly independent, then they form a basis of$V$ .

d. If m vectors inspan, 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

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 $B_W=\{w_1,w_2,w_3,\mathrm{...},w_k\}$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, $dimV\le dimW$.