Math Problem Statement

Show formally, by going through the three requirements for subspaces, that N(A) = {x | Ax = 0} is a subspace. Do each part separately. If A is a m x n matrix, is N(A) a subspace of R^m or R^n?

Solution

Let's solve the problem step by step.

We are tasked with proving that N(A)={xAx=0}N(A) = \{x \mid Ax = 0\}, which is the null space of the matrix AA, is a subspace. We will use the three properties that define a subspace in vector spaces:

Requirements for Subspace:

To prove that N(A)N(A) is a subspace of Rn\mathbb{R}^n, we need to verify these three properties:

  1. The zero vector is in N(A)N(A).
  2. Closure under addition: If x1x_1 and x2x_2 are in N(A)N(A), then x1+x2x_1 + x_2 is also in N(A)N(A).
  3. Closure under scalar multiplication: If xx is in N(A)N(A) and cc is a scalar, then cxcx is also in N(A)N(A).

Step-by-step Verification:

1. The zero vector is in N(A)N(A):

We need to check if the zero vector 0\mathbf{0} is in N(A)N(A).

By definition, 0N(A)\mathbf{0} \in N(A) if: A0=0A\mathbf{0} = 0 Since multiplying any matrix AA by the zero vector gives the zero vector, we have: A0=0A\mathbf{0} = 0 Thus, the zero vector is indeed in N(A)N(A).

2. Closure under addition:

Let x1,x2N(A)x_1, x_2 \in N(A), which means: Ax1=0andAx2=0Ax_1 = 0 \quad \text{and} \quad Ax_2 = 0 We need to show that x1+x2N(A)x_1 + x_2 \in N(A), i.e., that A(x1+x2)=0A(x_1 + x_2) = 0.

Using the linearity of matrix multiplication, we have: A(x1+x2)=Ax1+Ax2=0+0=0A(x_1 + x_2) = Ax_1 + Ax_2 = 0 + 0 = 0 Thus, x1+x2N(A)x_1 + x_2 \in N(A), proving that N(A)N(A) is closed under addition.

3. Closure under scalar multiplication:

Let xN(A)x \in N(A) and cc be a scalar. We need to show that cxN(A)cx \in N(A), i.e., that A(cx)=0A(cx) = 0.

Again, using the linearity of matrix multiplication, we have: A(cx)=c(Ax)=c0=0A(cx) = c(Ax) = c \cdot 0 = 0 Thus, cxN(A)cx \in N(A), proving that N(A)N(A) is closed under scalar multiplication.


Subspace of Rm\mathbb{R}^m or Rn\mathbb{R}^n?

If AA is an m×nm \times n matrix, then xx (the elements of N(A)N(A)) are vectors from Rn\mathbb{R}^n, since AA acts on vectors from Rn\mathbb{R}^n. Thus, the null space N(A)N(A) is a subspace of Rn\mathbb{R}^n.


Conclusion:

We have verified that N(A)N(A) satisfies all three properties for being a subspace. Therefore, N(A)N(A) is a subspace of Rn\mathbb{R}^n if AA is an m×nm \times n matrix.


Would you like any more details on any of these steps or further clarifications?

Here are 5 follow-up questions to deepen understanding:

  1. How would you prove that the column space of AA is a subspace?
  2. What is the relationship between the null space and the rank of a matrix?
  3. How does the dimension of N(A)N(A) relate to the rank-nullity theorem?
  4. Can the null space of a matrix ever be all of Rn\mathbb{R}^n? If so, when?
  5. What happens to the null space if matrix AA is invertible?

Tip: Always remember that any subspace must include the zero vector, and checking this first is a good starting point in proving subspace properties.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Algebra
Subspaces
Null Space
Matrices

Formulas

Ax = 0

Theorems

Subspace Test (Closure under addition, scalar multiplication, and presence of zero vector)

Suitable Grade Level

Undergraduate (Linear Algebra)