To prove that if $S = \{v_1, v_2, \ldots, v_n\}$ is a basis for a vector space $V$ and $c$ is a nonzero scalar, then the set $S_1 = \{cv_1, cv_2, \ldots, cv_n\}$ is also a basis for $V$, we need to show two things:
1. $S_1$ spans $V$: Every vector in $V$ can be expressed as a linear combination of vectors in $S_1$.
2. $S_1$ is linearly independent: The only way to obtain the zero vector by forming a linear combination of vectors in $S_1$ is by setting all the scalars to zero.
Proof:
1. $S_1$ spans $V$:
Let $v$ be an arbitrary vector in $V$. Since $S$ is a basis for $V$, $v$ can be expressed as a linear combination of vectors in $S$. Therefore, there exist scalars $a_1, a_2, \ldots, a_n$ such that:
Now, consider the vector $cv$:
Thus, $cv$ can be expressed as a linear combination of vectors in $S_1$. This shows that $S_1$ spans $V$.
2. $S_1$ is linearly independent:
To show that $S_1$ is linearly independent, we need to prove that the only solution to the equation:
is $x_1 = x_2 = \ldots = x_n = 0$. Suppose there exists a nontrivial solution to the equation, where at least one $x_i$ is nonzero. Without loss of generality, assume $x_1 \\eq 0$. Then we have:
Since $c$ is a nonzero scalar, and $x_1 \\eq 0$, we have $cx_1v_1 \\eq 0$. However, the equation states that the left side is equal to $0$, which contradicts the assumption. Therefore, the only solution to the equation is $x_1 = x_2 = \ldots = x_n = 0$, confirming that $S_1$ is linearly independent.
Based on the above proofs, we can conclude that if $S = \{v_1, v_2, \ldots, v_n\}$ is a basis for a vector space $V$ and $c$ is a nonzero scalar, then the set $S_1 = \{cv_1, cv_2, \ldots, cv_n\}$ is also a basis for $V$.