V As Subspace Of Rn: Proof & Explanation

by Admin 41 views
Proving V is a Subspace of Rn

Hey guys! Let's dive into a fundamental concept in linear algebra: proving that a particular set V is a subspace of Rn. Specifically, we're going to show that the set V = x ∈ Rn Ax = 0, where A is an m × n matrix, forms a subspace of Rn. This is a classic example and understanding it will give you a solid foundation for more advanced topics. So buckle up, and let's get started!

Understanding Subspaces

Before we jump into the proof, let's quickly recap what it means for a set to be a subspace. A subset V of a vector space Rn is a subspace if it satisfies three crucial conditions:

  1. The zero vector is in V: The zero vector, denoted as 0, must be an element of V. In other words, V must contain the origin.
  2. Closure under addition: If u and v are vectors in V, then their sum, u + v, must also be in V. This means that adding any two vectors within V will always result in another vector that's also inside V. Think of it like V being a closed-off world where addition doesn't let you escape.
  3. Closure under scalar multiplication: If u is a vector in V, and c is any scalar (real number), then the scalar multiple cu must also be in V. This means that scaling any vector within V by any scalar will still result in a vector that's inside V. Again, it’s like V has its own rules; multiplying vectors by numbers keeps them within its boundaries.

If V satisfies all three of these conditions, then it's officially a subspace of Rn. Now that we have a clear understanding of what it means to be a subspace, let's apply these conditions to our specific set V = x ∈ Rn Ax = 0.

Proof that V is a Subspace of Rn

Alright, let's get to the core of the matter! We want to demonstrate that V = x ∈ Rn Ax = 0 is indeed a subspace of Rn. To do this, we need to verify the three conditions we just discussed. Let's tackle them one by one.

1. The Zero Vector is in V

To show that the zero vector is in V, we need to prove that A0 = 0, where 0 represents the zero vector in Rn. Remember that matrix multiplication involves taking linear combinations of the columns of A using the entries of the vector we are multiplying by. When we multiply A by the zero vector, all the entries are zero. This means that the linear combination will have all coefficients equal to zero, which results in the zero vector in Rm. Mathematically:

A0 = 0

This is a fundamental property of matrix multiplication. Since A0 = 0, the zero vector satisfies the condition Ax = 0, and therefore, the zero vector is an element of V. So far, so good!

2. Closure Under Addition

Now, let's prove that V is closed under addition. This means that if we take two vectors u and v that are both in V, their sum (u + v) must also be in V. In other words, we need to show that if Au = 0 and Av = 0, then A( u + v ) = 0. Let's work through it:

Assume u ∈ V and v ∈ V. This means that:

Au = 0

Av = 0

Now, consider A( u + v ). Using the distributive property of matrix multiplication, we can write:

A( u + v ) = Au + Av

Since we know that Au = 0 and Av = 0, we can substitute these values:

Au + Av = 0 + 0 = 0

Therefore,

A( u + v ) = 0

This shows that the vector (u + v) satisfies the condition Ax = 0, which means that (u + v) is also an element of V. Thus, V is closed under addition. One more condition down!

3. Closure Under Scalar Multiplication

Finally, let's demonstrate that V is closed under scalar multiplication. This means that if we take a vector u that's in V and multiply it by any scalar c, the resulting vector (cu) must also be in V. In other words, we need to show that if Au = 0, then A( cu ) = 0. Here's how we can prove it:

Assume u ∈ V and let c be any scalar. This means that:

Au = 0

Now, consider A( cu ). Using the property of scalar multiplication with matrices, we can write:

A( cu ) = c(Au)

Since we know that Au = 0, we can substitute this value:

c(Au) = c(0) = 0

Therefore,

A( cu ) = 0

This shows that the vector (cu) satisfies the condition Ax = 0, which means that (cu) is also an element of V. Thus, V is closed under scalar multiplication. And that's all three conditions satisfied!

Conclusion

Boom! We've successfully shown that V = x ∈ Rn Ax = 0 is a subspace of Rn. We did this by verifying that V satisfies all three necessary conditions:

  • The zero vector is in V.
  • V is closed under addition.
  • V is closed under scalar multiplication.

Understanding this proof is crucial because it provides a concrete example of how to demonstrate that a set is a subspace. This concept pops up all the time in linear algebra, so having a solid grasp of it will definitely pay off. Plus, you can now confidently explain to your friends (or maybe even your professor) why this particular set is a subspace of Rn.

Keep practicing with different sets and matrices, and you'll become a subspace-proving master in no time. Keep up the great work, and happy studying!