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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?

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    $\begingroup$ 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 $U_1$. $\endgroup$
    – onRiv
    Feb 5, 2022 at 4:20
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    $\begingroup$ 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. $\endgroup$ Feb 5, 2022 at 4:26

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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.

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  • $\begingroup$ I think this is how I can complete my proof not the proof from the beginning. Am I correct? $\endgroup$
    – user965463
    Feb 5, 2022 at 9:14
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    $\begingroup$ @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 $U_1 = U_2$. $\endgroup$
    – user56202
    Feb 5, 2022 at 13:38