Understanding whether a set of vectors forms a basis for a vector space is a fundamental concept in linear algebra. A set of vectors acts as a basis for a vector space if it meets two key conditions:

• Linear Independence: The vectors in the set must be linearly independent, meaning no vector in the set can be written as a linear combination of the others.
• Spanning: The vectors must span the entire vector space, meaning any vector in the space can be expressed as a linear combination of the vectors in the set.

For a vector space like ℝ², which has a dimension of 2, a set of two vectors forms a basis if and only if they are linearly independent. We can determine linear independence by checking the determinant of the matrix formed by these vectors.

Basis Analysis for Subset S₁

Let's consider the first subset, $\text{S}_1 = \{[1, -2], [3, 5]\}$. To check if these two vectors are linearly independent, we can form a matrix using these vectors as columns (or rows) and calculate its determinant. If the determinant is non-zero, the vectors are linearly independent, implying they form a basis for ℝ².

The matrix formed by these vectors is:

$$\mathbf{A} = \begin{pmatrix} 1 & 3 \\ -2 & 5 \end{pmatrix}$$

Now, let's calculate the determinant of matrix $\mathbf{A}$:

$$\text{det}(\mathbf{A}) = (1 \times 5) - (3 \times -2)$$ $$\text{det}(\mathbf{A}) = 5 - (-6)$$ $$\text{det}(\mathbf{A}) = 5 + 6$$ $$\text{det}(\mathbf{A}) = 11$$

Since the determinant of $\mathbf{A}$ is $11$, which is not equal to zero ($11 \neq 0$), the vectors $[1, -2]$ and $[3, 5]$ are linearly independent. As ℝ² has dimension 2, and S$_1$ contains 2 linearly independent vectors, S$_1$ forms a basis for ℝ².

Basis Analysis for Subset S₂

Next, let's examine the second subset, $\text{S}_2 = \{[1, 1], [0, 0]\}$. Again, we can form a matrix with these vectors and calculate its determinant to check for linear independence.

The matrix formed by these vectors is:

$$\mathbf{B} = \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}$$

Now, let's calculate the determinant of matrix $\mathbf{B}$:

$$\text{det}(\mathbf{B}) = (1 \times 0) - (0 \times 1)$$ $$\text{det}(\mathbf{B}) = 0 - 0$$ $$\text{det}(\mathbf{B}) = 0$$

Since the determinant of $\mathbf{B}$ is $0$, the vectors $[1, 1]$ and $[0, 0]$ are linearly dependent. A set containing the zero vector is always linearly dependent because the zero vector can be expressed as any scalar multiple of any other vector, for example, $0 \times [1,1] = [0,0]$. Because S$_2$ contains linearly dependent vectors, it cannot form a basis for ℝ².

Conclusion on Vector Subsets as Bases

Based on our analysis of the subsets S$_1$ and S$_2$ for ℝ²:

• Subset S$_1$ consists of two linearly independent vectors. Therefore, S$_1$ is a basis for ℝ².
• Subset S$_2$ contains linearly dependent vectors (specifically, it includes the zero vector). Therefore, S$_2$ is not a basis for ℝ².

Thus, S$_1$ is a basis for ℝ² but S$_2$ is not a basis for ℝ².