# Intersection of Subspaces and Linear independence

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.

• 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 – peek-a-boo Jun 15 '19 at 5:42

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 $$U_1$$.
Note: $$U_1\cap U_2\subset U_1$$.
• 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? – Frank Swanton Jun 15 '19 at 6:19