To prove that the intersection of two subspaces, S and T, in a vector space V over a field F is also a subspace of V, you need to show that it satisfies the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and contains the zero vector. Let's go through each step:
1. **Closure under vector addition:**
Let's take two arbitrary vectors u and v that belong to S ∩ T. This means that u and v are in both S and T. Since S and T are subspaces, they are closed under vector addition. Therefore, u + v must also belong to both S and T. This implies that u + v is in their intersection, S ∩ T.
2. **Closure under scalar multiplication:**
Let's take an arbitrary vector u from S ∩ T and a scalar c from the field F. Since u belongs to both S and T, and both S and T are subspaces, c * u must belong to both S and T as well. This implies that c * u is in their intersection, S ∩ T.
3. **Contains the zero vector:**
Since S and T are subspaces, they both contain the zero vector, denoted as 0. Therefore, 0 is in both S and T, which means 0 is also in their intersection, S ∩ T.
Since S ∩ T satisfies all three properties of being a subspace, you have successfully shown that the intersection of two subspaces, S and T, is a subspace of the vector space V over the field F.
#####Show that S={a,b,c,d ∈ R^4 :2a-3b+5c-d=0} is a subspace of R4
To show that \(S = \{(a, b, c, d) \in \mathbb{R}^4 : 2a - 3b + 5c - d = 0\}\) is a subspace of \(\mathbb{R}^4\), we need to verify the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector.
1. **Closure under vector addition:**
Let \((a_1, b_1, c_1, d_1)\) and \((a_2, b_2, c_2, d_2)\) be two arbitrary vectors in \(S\). This means that they satisfy the equation \(2a_1 - 3b_1 + 5c_1 - d_1 = 0\) and \(2a_2 - 3b_2 + 5c_2 - d_2 = 0\).
Now let's consider the vector \((a_1 + a_2, b_1 + b_2, c_1 + c_2, d_1 + d_2)\). If we substitute the components of this vector into the given equation, we have:
\[2(a_1 + a_2) - 3(b_1 + b_2) + 5(c_1 + c_2) - (d_1 + d_2) =\]
\[2a_1 - 3b_1 + 5c_1 - d_1 + 2a_2 - 3b_2 + 5c_2 - d_2.\]
Since both \((a_1, b_1, c_1, d_1)\) and \((a_2, b_2, c_2, d_2)\) are in \(S\), their components satisfy the equation \(2a_1 - 3b_1 + 5c_1 - d_1 = 0\) and \(2a_2 - 3b_2 + 5c_2 - d_2 = 0\). Therefore, the above expression simplifies to \(0 + 0\), which means that the vector \((a_1 + a_2, b_1 + b_2, c_1 + c_2, d_1 + d_2)\) also satisfies the equation \(2a - 3b + 5c - d = 0\). Hence, the sum of two vectors from \(S\) is also in \(S\), and \(S\) is closed under vector addition.
2. **Closure under scalar multiplication:**
Let \((a, b, c, d)\) be a vector in \(S\), which means it satisfies the equation \(2a - 3b + 5c - d = 0\). Now, consider the vector \(k \cdot (a, b, c, d) = (ka, kb, kc, kd)\), where \(k\) is a scalar from \(\mathbb{R}\).
Substitute the components of \(k \cdot (a, b, c, d)\) into the equation \(2a - 3b + 5c - d = 0\):
\[2(ka) - 3(kb) + 5(kc) - (kd) = k(2a - 3b + 5c - d).\]
Since \((a, b, c, d)\) satisfies the equation, we know that \(2a - 3b + 5c - d = 0\). Therefore, the above expression simplifies to \(k \cdot 0\), which is \(0\). This means that the vector \(k \cdot (a, b, c, d)\) also satisfies the equation \(2a - 3b + 5c - d = 0\), and thus, \(S\) is closed under scalar multiplication.
3. **Contains the zero vector:**
The zero vector in \(\mathbb{R}^4\) is \((0, 0, 0, 0)\). Substituting the components of the zero vector into the equation \(2a - 3b + 5c - d = 0\) yields \(0\cdot 2 - 0\cdot 3 + 0\cdot 5 - 0 = 0\), which is true. Therefore, the zero vector is in \(S\).
Since \(S\) satisfies all three properties of being a subspace, namely closure under vector addition, closure under scalar multiplication, and containing the zero vector, \(S\) is indeed a subspace of \(\mathbb{R}^4\).
##### Show that the sets of vector
S={(1,1,1),(1,2,3),(2,-1,1)} are a subspace of R3
To determine if the set \(S = \{(1, 1, 1), (1, 2, 3), (2, -1, 1)\}\) is a subspace of \(\mathbb{R}^3\), we need to verify the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector.
1. **Closure under vector addition:**
Let \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) be two arbitrary vectors in \(S\). That is, they are both linear combinations of the given vectors \((1, 1, 1)\), \((1, 2, 3)\), and \((2, -1, 1)\). Now, let's consider the vector \((x_1, y_1, z_1) + (x_2, y_2, z_2)\):
\[(x_1 + x_2, y_1 + y_2, z_1 + z_2).\]
We need to check whether this vector is also in \(S\). We can express \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) as linear combinations of the given vectors in \(S\):
\((x_1, y_1, z_1) = a(1, 1, 1) + b(1, 2, 3) + c(2, -1, 1),\)
\((x_2, y_2, z_2) = d(1, 1, 1) + e(1, 2, 3) + f(2, -1, 1),\)
where \(a, b, c, d, e, f\) are scalars.
Now, the sum can be expressed as:
This shows that the sum is also a linear combination of the vectors in \(S\), and thus, it belongs to \(S\). Hence, \(S\) is closed under vector addition.
2. **Closure under scalar multiplication:**
Let \((x, y, z)\) be an arbitrary vector in \(S\). That is, it is a linear combination of the given vectors in \(S\):
\((x, y, z) = a(1, 1, 1) + b(1, 2, 3) + c(2, -1, 1),\)
where \(a, b, c\) are scalars. Now, let's consider the vector \(k \cdot (x, y, z) = (kx, ky, kz)\), where \(k\) is a scalar.
The scaled vector can be expressed as:
where , , and .
This shows that the scaled vector is also a linear combination of the vectors in \(S\), and thus, it belongs to \(S\). Therefore, \(S\) is closed under scalar multiplication.
3. **Contains the zero vector:**
The zero vector in \(\mathbb{R}^3\) is \((0, 0, 0)\). We can see that this vector can be expressed as a linear combination of the vectors in \(S\) by choosing all the scalars to be zero:
\((0, 0, 0) = 0 \cdot (1, 1, 1) + 0 \cdot (1, 2, 3) + 0 \cdot (2, -1, 1).\)
This shows that the zero vector is in \(S\).
Since \(S\) satisfies all three properties of being a subspace (closure under vector addition, closure under scalar multiplication, and containing the zero vector), \(S\) is indeed a subspace of \(\mathbb{R}^3\).
