# Prove that if $U_1 \subseteq U_2,$ then $U_1 = U_2.$ [duplicate]

I am trying to prove the following question:

Let $$U_1, U_2 \subseteq V$$ be subspaces of the same dimension. Prove that if $$U_1 \subseteq U_2,$$ then $$U_1 = U_2.$$

Here is an attempt:

Let $$U_1, U_2 \subseteq V$$ be subspaces of the same dimension. Assume that $$U_1 \subseteq U_2,$$ we want to show that $$U_1 = U_2.$$

Since $$U_1, U_2 \subseteq V$$ are subspaces, then they are vector spaces. Also, since they have the same dimension, then we can assume that $$\operatorname{dim} U_1 = \operatorname{dim} U_2 = n.$$ Now, since $$\operatorname{dim} U_1 = n,$$ then a basis for $$U_1$$ is a linearly independent subset of $$V$$ containing $$n$$ vectors. Similarly, since $$\operatorname{dim} U_2 = n,$$ then a basis for $$U_2$$ is a linearly independent subset of $$V$$ containing $$n$$ vectors.

Let $$\{x_1, x_2, \dots , x_n\}$$ be a basis for $$U_1.$$ Then, since $$U_1 \subseteq U_2,$$ then any linear combination of the vectors $$x_1, x_2, \dots , x_n$$ are in $$U_2.$$ i.e., $$x_1, x_2, \dots , x_n$$ are in $$U_2$$ and these vectors are linearly independent. Since $$\operatorname{dim} U_2 = n,$$ then these $$n$$ vectors form a basis for $$U_2$$ as well.

But then I do not know how to complete my proof, could anyone help me please?

• Isn't it already complete? Since the n vectors is a basis for $U_2$. Any element in $U_2$ is generated by the basis and hence is contained in $U_1$. Feb 5, 2022 at 4:20
• Abstract duplicate of Two vector spaces with same dimension and same basis, are identical?. The top voted answer there addresses the exact problem you have. Feb 5, 2022 at 4:26

Suppose $$U_1 \not = U_2$$. Then there is a vector $$v$$ in $$U_2$$ but not in $$U_1$$. The set of vectors $$\{x_1, ..., x_n, v\}$$ is linearly independent, and it is a subset of $$U_2$$. So we have found $$n+1$$ linearly independent vectors in a subspace of dimension $$n$$. This is a contradiction, since by definition, the dimension of a subspace is the largest number of linearly independent vectors you can find in it.
• @Brain Yes, this is one way to complete it. But in a way, your proof is already complete. You said yourself that $x_1, ..., x_n$ is a basis for both $U_1$ and $U_2$. So $U_1 = \text{span} \{x_1, ..., x_n \}$ and $U_2 = \text{span} \{x_1, ..., x_n \}$, so $U_1 = U_2$. Feb 5, 2022 at 13:38