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Suppose $U_1$ and $U_2$ are subspaces of a finite-dimensional vector space.

Let $u_1,...,u_m$ be a basis of $U_1\cap U_2$, thus dimension of the intersection is $m$.

$\textbf{The part I don't understand is:}$

Because $u_1,...,u_m$ is a basis of $U_1\cap U_2$, it is linearly independent in $U_1$.

Why is this true?

Reference:

Axler, Sheldon J. $\textit{Linear Algebra Done Right}$, New York: Springer, 2015.

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  • $\begingroup$ A basis for $U_1 \cap U_2$ is by definition as subset $\beta \subset U_1 \cap U_2 $ which is linearly independent and one which spans $U_1 \cap U_2$. Is there something else which is bothering you? Also, the concept of linear independence is one which depends on the underlying field of scalars, not on the space $V$ of consideration $\endgroup$ – peek-a-boo Jun 15 '19 at 5:42
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Since $u_1, ..., u_m$ is a basis for $U_1 \cap U_2$, it is linearly independent in $U_1 \cap U_2$ (by the definition of a basis).

Therefore, for scalars $c_1,..., c_m$ in the vector space, we have $c_1u_1 +... + c_mu_m = 0 \implies c_1=c_2=...=c_m = 0$ (by the definition of linear independence).

This implication holds true for the subspace $U_1$ as well, since each of $u_1,...,u_m$ are in $U_1$ (since $u_1,...,u_m$ are in $U_1$ and $U_2$). Therefore, $u_1,...,u_m$ is linearly independent in $U_1$.

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Note: $U_1\cap U_2\subset U_1$.

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  • $\begingroup$ Chris, thanks for the response. I was thinking about this in 2D and 3D. Start with two planes, $p1$ and $p2$. If we have non-intersecting and non-parallel lines in this section of two planes, it will be LI. This set of lines would be still LI in each of the plane, $p1$ and $p2$. Is this correct? $\endgroup$ – Frank Swanton Jun 15 '19 at 6:19
  • $\begingroup$ Well, for one thing, for a plane to be a subspace it must go through the origin. Then the two planes are either the same, or intersect in a line. You don't have to consider planes through the origin; but in your problem they must be. I guess my point is that you may need to go up a dimension (or so). $\endgroup$ – user403337 Jun 15 '19 at 6:30
  • $\begingroup$ How about the intuition? I am thinking of two big tables going through origin. Imagining two lines independent in the intersection of the two tables. If it is the case, it must be LI in each of the table. Is this intuition correct? Because table is a two-dim and lines are 1-d, so the analogy is apt, right? $\endgroup$ – Frank Swanton Jun 15 '19 at 10:35
  • $\begingroup$ Yeah but, again my only problem with that is that the planes (or tables, as you put it) must intersect in a single line. To get two (or more) lines in the intersection, you need higher dimensions. $\endgroup$ – user403337 Jun 15 '19 at 14:51

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