Basis for a vector space $V$

Question

Let $B=\{v_1,v_2,v_3,v_4,v_5 \}$ be a basis for a vector space $V$. Let $w_1=v_4-v_2$ and $w_2=v_5-v_3$. Is $B'=\{v_1,v_2,v_3,w_1,w_2\}$ a basis for V?

Answer 1

Since $B=\{v_1,v_2,v_3,v_4,v_5\}$ and $B'=\{v_1,v_2,v_3,-v_2+v_4,-v_3+v_5\}$ and one can show that transformation matrix from $B$ to $B'$ is the following $$C=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$$ Since $C$ is block diagonal and it immediately follows that $\det C=1$. Since $C$ has a nonzero determinant then $B'$ is indeed a basis for vector space $V$.

Answer 2

Since

$v_4 = w_1 + v_2 \tag 1$

and

$v_5 = w_2 + v_3, \tag 2$

$V = \text{span}(\{v_1, v_2, v_3, v_4, v_5 \}) \subset \text{span}(\{v_1, v_2, v_3, w_1, w_2 \}) \subset V; \tag 3$

it also easy to see that $B'$ is a linearly independent set, since a linear dependence amongst the elements of $B'$ gives rise to a linear dependence of $B$ by simply substituting the given expressions for $w_1$, $w_2$ in terms of the $v_i$, $1 \le i \le 5$.

(3) shows that

$ V = \text{span}(B'), \tag 4$

and thus $B'$ is a basis for $V$.

Answer 3

According to the problem setting, one concludes that $\dim V = 5$.

Thus it suffices to show that $\mathcal{B}' = \{v_{1},v_{2},v_{3},w_{1},w_{2}\}$ is LI to conclude that it is a basis indeed.

In fact, one has \begin{align*} & a_{1}v_{1} + a_{2}v_{2} + a_{3}v_{3} + a_{4}w_{1} + a_{5}w_{4} =\\\\ & a_{1}v_{1} + (a_{2} - a_{4})v_{2} + (a_{3} - a_{5})v_{3} + a_{4}v_{4} + a_{5}v_{5} = 0 \end{align*}

Once $\mathcal{B} = \{v_{1},v_{2},v_{3},v_{4},v_{5}\}$ is LI, it results that $a_{1} = a_{2} = a_{3} = a_{4} = a_{5}$.

Thus $\mathcal{B}'$ is LI, which answers your question.

Hopefully this helps.