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Let V be a vector space over a field K, then the dimension of V(K) is equal to the number of the basis of the vector space V(K).

Since any subspace W of a vector space V(F) forms a vector space with respect to the same binary operation, then the dimension of the subspace W will also be equal to the number of the basis of the subspace W.

Also, $dim\left(W\right)猢�dim\left(V\right)$

Since V is a three-dimensional subspace of$R^5$ , then V has at least one basis.

Let $\mathfrak{I}=\left\{u,v,w\right\}$be the basis of V, then $\mathfrak{R}=\left\{c_1\stackrel{鈫�}{u},c_2\stackrel{鈫�}{v},c_3\stackrel{鈫�}{w}\right\}$will also be the basis of V, where $c_1,c_2,c_3$are non-zero.

And so the number of bases of the subspace V is infinite for different combinations of $c_1,c_2,c_3$.

If V is any three-dimensional subspace of$R^5$ , then V has infinitely many bases as we have $\mathfrak{R}=\left\{c_1\stackrel{鈫�}{u},c_2\stackrel{鈫�}{v},c_3\stackrel{鈫�}{w}\right\}$as the basis of V, where $c_1,c_2,c_3$are non-zero.