Intersection of Subspaces and Linear independence

Question

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.

Answer 1

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 $U_1$.

Answer 2

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