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, we need to prove three properties:

1. W must contain the zero vector.
2. W must be closed under addition.
3. W must be closed under scalar multiplication.

Let's go through each of these properties.

1. Zero Vector:
The zero vector in M22 is the 2x2 matrix with all elements being zero. Let's check if the zero vector is contained in W.
W = [ a , a+b ] and [a-b , b ]
To satisfy the zero vector condition, we need to have a = 0 and b = 0. Let's plug in these values:
W = [ 0 , 0+0 ] = [0, 0]
[ 0-0 , 0 ] [0, 0]
As we can see, the resulting matrix is indeed the zero vector. Therefore, the first condition is satisfied.

2. Closure under Addition:
To prove closure under addition, we need to show that if we take any two matrices from W and add them together, the resulting matrix will still be in W.
Let's take two arbitrary matrices from W:
A = [ a1 , a1+b1 ] and B = [a2-b2 , b2]
Now let's add these two matrices:
A + B = [ a1+a2-b2 , a1+b1+b2 ]
[a2-b2 , b2]
As we can see, the resulting matrix is still in the form specified for W. Therefore, the second condition is satisfied.

3. Closure under Scalar Multiplication:
To prove closure under scalar multiplication, we need to show that if we take any matrix from W and multiply it by any scalar, the resulting matrix will still be in W.
Let's take an arbitrary matrix from W:
C = [ a3 , a3+b3 ]
Now, let's multiply this matrix by a scalar k:
kC = [ k*a3 , k*(a3+b3) ]
[k*(a3-b3) , k*b3 ]
As we can see, the resulting matrix is still in the form specified for W. Therefore, the third condition is satisfied.

Since W satisfies all three properties of a subspace, we can conclude that W is indeed a subspace of M22.