Prove vectors create a basis

Question

Let $V$ be a vector space and $U,W,Z$ be it's subspaces where $V=Z \oplus U=Z\oplus W$. We know that $\beta_1,...,\beta_k$ is a basis of $U$ and $\beta_i=\gamma_i+\delta_i$ where $\gamma_i \in Z$ and $\delta_i \in W$ for $i \in \{1,2,...,k \}$. Prove that $\delta_1,..,\delta_k$ is basis of $W$.

We know from the direct sum that $\dim U=\dim W=k$ but I don't know how tho show $\delta_1,..,\delta_k$ is linear independent

Answer 1

We have $\delta_i=\beta_i-\gamma_i$. It remains to prove that the family $(\delta_i)$ is linearly independent. Let $a_i\in\Bbb R$ such that

$$\sum_i a_i\delta_i=0\iff \sum_ia_i\beta_i=\sum_i a_i\gamma_i$$ hence since $U\cap Z=\{0\}$ and $(\beta_i)$ is a basis for $U$ then $$\sum_i a_i\beta_i=0\implies a_i=0,\;\forall i$$

Answer 2

Suppose the $\delta_i$'s were not a basis. Then we would have

$$a_1\delta_1+\cdots+a_k\delta_k=0$$

with $a_j\neq0$ for at least one $j$. But then you would have

$$a_1\beta_1+\cdots+a_k\beta_k=-a_1\gamma_1-\cdots-a_k\gamma_k$$

The left-hand side is a vector in $U$. The right hand side is a vector in $Z$. These two vectors are equal and it isn't the zero vector, since at least one $a_j\neq 0$ and the $\beta_i$'s constitute a basis for $U$. But this contradicts the fact that $V=U\oplus Z$.