W= [ a , a+b ] (2x2 matrix)
a-b , b
a, b are element of R
Show that W is a Subspace of M22?
To show that W is a subspace of M22 (the set of 2x2 matrices), we need to verify three conditions: closure under addition, closure under scalar multiplication, and containing the zero vector.
1. Closure under Addition:
Let's assume we have two matrices A and B in W, where A = [a, a+b; a-b, b] and B = [c, c+d; c-d, d] (both matrices are elements of W).
Now, we will add these two matrices: A + B = [a+c, (a+b)+(c+d); (a-b)+(c-d), b+d].
Since a, b, c, and d are elements of the real numbers (R), the sum of a+c and a+b+c+d is also in R. Similarly, (a-b)+(c-d) and b+d are also in R.
Therefore, A + B is a valid 2x2 matrix, and it belongs to W.
2. Closure under Scalar Multiplication:
Let's take an arbitrary matrix A = [a, a+b; a-b, b] in W, where a and b are elements of R.
If we multiply this matrix by a scalar value k, we get kA = [ka, k(a+b); k(a-b), kb].
Again, since a, b, and k are in R, the entries ka, k(a+b), k(a-b), and kb are also in R.
Thus, kA is a valid 2x2 matrix, and it belongs to W.
3. Containing the Zero Vector:
The zero vector in M22 is the matrix [0, 0; 0, 0].
Let's substitute a = b = 0 in the matrix representation of W: [0, 0+0; 0-0, 0] = [0, 0; 0, 0].
As we can see, the zero vector is indeed in W.
Since W satisfies all three conditions, it is proven that W is a subspace of M22.