Question

State if the following statements are true and which are false? Justify your answer with a
short proof or a counterexample.

If {v1, v2, .... , vn} is a basis for vector space V, then {v1+v2+ +vn, v2,.... ,vn} is also
a basis for V

Answer 1

Answer on Question #44918 – Math - Linear Algebra

Problem.

State if the following statements are true and which are false? Justify your answer with a short proof or a counterexample.

If {v1, v2, ..., vn} is a basis for vector space V, then {v1+v2+ +vn, v2,..., vn} is also a basis for V

Solution.

The statement is true.

We need to show that $[v_1 + v_2 + \dots + v_n, v_2, \ldots, v_n]$ is a set of linearly independent vectors and each element of $V$ can be represented as linear combination of vectors from set.

Suppose that vectors $v_{1} + v_{2} + \dots + v_{n}, v_{2}, \ldots, v_{n}$ are linearly dependent. Then there exist $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n} \in P$ (where $P$ is a field) such that

$$\lambda_ {1} \left(v _ {1} + v _ {2} + \dots + v _ {n}\right) + \lambda_ {2} v _ {2} + \dots + \lambda_ {n} v _ {n} = 0$$

or

$$\lambda_ {1} v _ {1} + (\lambda_ {1} + \lambda_ {2}) v _ {2} + \dots + (\lambda_ {1} + \lambda_ {n}) v _ {n} = 0,$$

but $\lambda_1v_1 + (\lambda_1 + \lambda_2)v_2 + \dots + (\lambda_1 + \lambda_n)v_n \neq 0$, as vectors $v_1, v_2, \ldots, v_n$ are linearly independent.

Hence $v_{1} + v_{2} + \dots + v_{n}, v_{2}, \ldots, v_{n}$ are linearly independent.

If $u \in V$, then there exist $\lambda_1, \lambda_2, \ldots, \lambda_n \in P$ ($[v_1, v_2, \ldots, v_n]$ is a basis), such that

$$u = \lambda_ {1} v _ {1} + \lambda_ {2} v _ {2} + \dots + \lambda_ {n} v _ {n}$$

or

$$u = \lambda_ {1} (v _ {1} + v _ {2} + \dots v _ {n}) + (\lambda_ {2} - \lambda_ {1}) v _ {2} + \dots + (\lambda_ {n} - \lambda_ {1}) v _ {n}.$$

Hence each element of $V$ can be represented as linear combination of vectors from set

$$\{v _ {1} + v _ {2} + \dots + v _ {n}, v _ {2}, \dots , v _ {n} \}.$$

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