Can vectors be independent linearly and also base 2?

Vectors can indeed be linearly independent and also form a basis for a vector space, such as $\mathbb{R}^2$. Let's break down these concepts:

1. Linear Independence: A set of vectors is said to be linearly independent if no vector in the set can be written as a linear combination of the others. In simpler terms, none of the vectors in the set is redundant. For example, in $\mathbb{R}^2$, the vectors $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ are linearly independent because neither can be expressed as a scalar multiple or combination of the other.

2. Basis: A basis of a vector space is a set of linearly independent vectors that span the entire space. For $\mathbb{R}^2$, any two linearly independent vectors can serve as a basis. The standard basis for $\mathbb{R}^2$ is $\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$.

3. Base 2: In the context of vector spaces, "base 2" typically refers to the binary number system, which is unrelated to the concept of a basis in linear algebra. However, if you're referring to a basis in $\mathbb{R}^2$, then two linearly independent vectors can indeed form a basis for this space.

Final Answer

Yes, vectors can be linearly independent and also form a basis for a vector space, such as $\mathbb{R}^2$. The standard basis for $\mathbb{R}^2$ is $\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}$, which is both linearly independent and spans the space.