xplain how one can check whether a subset of a space is a subspace. Consider two sets{(x, y) ∈ R2 | |x| = |y|} and {(x, y) ∈ R2 | |x| = −|y|}. One of them is a subspace of R2 and theother is not. Using any method you prefer, decide which set is a subspace of R2 and which is not.

To check whether a subset of a space is a subspace, we need to verify three properties:

1. The zero vector of the larger space is in the subset.
2. The subset is closed under vector addition. This means that for any two vectors in the subset, their sum is also in the subset.
3. The subset is closed under scalar multiplication. This means that for any vector in the subset and any scalar, the product of the scalar and the vector is also in the subset.

Now, let's consider the two sets:

Set A = {(x, y) ∈ R2 | |x| = |y|} Set B = {(x, y) ∈ R2 | |x| = −|y|}

1. For Set A, the zero vector (0,0) is in the set because |0| = |0|. For Set B, the zero vector is also in the set because |0| = -|0|.
2. For Set A, if we take two vectors (x1, y1) and (x2, y2) such that |x1| = |y1| and |x2| = |y2|, their sum (x1+x2, y1+y2) is not necessarily in the set because |x1+x2| is not necessarily equal to |y1+y2|. Therefore, Set A is not closed under vector addition. For Set B, it's not closed under vector addition either. If we take two vectors (x1, y1) and (x2, y2) such that |x1| = -|y1| and |x2| = -|y2|, their sum (x1+x2, y1+y2) is not necessarily in the set because |x1+x2| is not necessarily equal to -|y1+y2|.
3. For Set A, it's not closed under scalar multiplication. If we take a vector (x, y) such that |x| = |y| and a scalar c, the product (cx, cy) is not necessarily in the set because |cx| is not necessarily equal to |cy|. For Set B, it's not closed under scalar multiplication either. If we take a vector (x, y) such that |x| = -|y| and a scalar c, the product (cx, cy) is not necessarily in the set because |cx| is not necessarily equal to -|cy|.

Therefore, neither Set A nor Set B is a subspace of R2 because they don't satisfy all the three properties of a subspace.