Necessary And Sufficient Condition For S S S To Be A Subspace Of $\mathbb R^n $ And Its Dimension

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Introduction

In linear algebra, a subspace is a subset of a vector space that is closed under addition and scalar multiplication. Given a real symmetric matrix A, we define the set S as the set of all vectors x in $\mathbb R^n$ such that $x^{T}Ax = 0$. In this article, we will discuss the necessary and sufficient condition for S to be a subspace of $\mathbb R^n$ and its dimension.

Preliminaries

Before we dive into the main topic, let's recall some basic concepts in linear algebra.

• A vector space is a set of vectors that is closed under addition and scalar multiplication.
• A subspace is a subset of a vector space that is closed under addition and scalar multiplication.
• A real symmetric matrix A is a square matrix that is equal to its transpose, i.e., $A = A^{T}$.
• The dot product of two vectors x and y is denoted by $x^{T}y$.

Necessary Condition for S to be a Subspace

To show that S is a subspace of $\mathbb R^n$, we need to verify that it satisfies the following properties:

1. Closure under addition: For any two vectors x and y in S, we need to show that $x + y$ is also in S.
2. Closure under scalar multiplication: For any vector x in S and any scalar c, we need to show that $cx$ is also in S.

Let's start with the first property.

Closure under addition

Suppose x and y are two vectors in S. Then, we have:

$$x^{T}Ax = 0 \quad \text{and} \quad y^{T}Ay = 0$$

We need to show that $(x + y)^{T}A(x + y) = 0$.

Using the properties of the dot product, we can expand $(x + y)^{T}A(x + y)$ as follows:

$$(x + y)^{T}A(x + y) = x^{T}Ax + x^{T}Ay + y^{T}Ax + y^{T}Ay$$

Since $x^{T}Ax = 0$ and $y^{T}Ay = 0$, we have:

$$(x + y)^{T}A(x + y) = x^{T}Ay + y^{T}Ax$$

Now, let's consider the second property.

Closure under scalar multiplication

Suppose x is a vector in S and c is a scalar. Then, we have:

$$x^{T}Ax = 0$$

We need to show that $(cx)^{T}A(cx) = 0$.

Using the properties of the dot product, we can expand $(cx)^{T}A(cx)$ as follows:

$$(cx)^{T}A(cx) = c^{2}x^{T}Ax$$

Since $x^{T}Ax = 0$, we have:

$$(cx)^{T}A(cx) = 0$$

Therefore, we have shown that S is closed under addition and scalar multiplication.

Sufficient Condition for S to be a Subspace

To show that S is a subspace of $\mathbb R^n$, we need to verify that it satisfies the following properties:

1. Closure under addition: For any two vectors x and y in S, we need to show that $x + y$ is also in S.
2. Closure under scalar multiplication: For any vector x in S and any scalar c, we need to show that $cx$ is also in S.

We have already shown that S is closed under addition and scalar multiplication in the previous section.

Dimension of S

To find the dimension of S, we need to find a basis for S.

A basis for S is a set of linearly independent vectors that span S.

Let's consider the following set of vectors:

$$\{v_{1}, v_{2}, \ldots, v_{k}\}$$

where $v_{i}$ is a vector in S for each $i = 1, 2, \ldots, k$.

We need to show that this set of vectors is a basis for S.

Linear independence

To show that the set of vectors is linearly independent, we need to show that the following equation holds:

$$c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k} = 0$$

implies that $c_{1} = c_{2} = \cdots = c_{k} = 0$.

Suppose we have:

$$c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k} = 0$$

Then, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

Using the properties of the dot product, we can expand $v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k})$ as follows:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{1}v_{1}^{T}Av_{1} + c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k}$$

Since $v_{1}^{T}Av_{1} = 0$, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k}$$

Similarly, we can show that:

$$v_{2}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{1}v_{2}^{T}Av_{1} + c_{3}v_{2}^{T}Av_{3} + \cdots + c_{k}v_{2}^{T}Av_{k}$$

and so on.

Since $v_{i}^{T}Av_{j} = 0$ for all $i \neq j$, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

$$v_{2}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

$$\vdots$$

$$v_{k}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

Since $v_{i}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$ for all $i = 1, 2, \ldots, k$, we have:

$$c_{1}v_{1}^{T}Av_{1} + c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k} = 0$$

$$c_{1}v_{2}^{T}Av_{1} + c_{2}v_{2}^{T}Av_{2} + \cdots + c_{k}v_{2}^{T}Av_{k} = 0$$

$$\vdots$$

$$c_{1}v_{k}^{T}Av_{1} + c_{2}v_{k}^{T}Av_{2} + \cdots + c_{k}v_{k}^{T}Av_{k} = 0$$

Since $v_{i}^{T}Av_{j} = 0$ for all $i \neq j$, we have:

$$c_{1}v_{1}^{T}Av_{1} = 0$$

$$c_{2}v_{2}^{T}Av_{2} = 0$$

$$\vdots$$

$$c_{k}v_{k}^{T}Av_{k} = 0$$

Since $v_{i}^{T}Av_{i} = 0$ for all $i = 1, 2, \ldots, k$, we have:

$$c_{1} = 0$$

$$c_{2} = 0$$

$$\vdots$$

$$c_{k} = 0$$

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Q: What is the necessary and sufficient condition for S to be a subspace of $\mathbb R^n$?

A: The necessary and sufficient condition for S to be a subspace of $\mathbb R^n$ is that S must be closed under addition and scalar multiplication. In other words, for any two vectors x and y in S, we must have $x + y$ in S, and for any vector x in S and any scalar c, we must have $cx$ in S.

Q: How do we show that S is closed under addition?

A: To show that S is closed under addition, we need to show that for any two vectors x and y in S, we have $(x + y)^{T}A(x + y) = 0$. Using the properties of the dot product, we can expand $(x + y)^{T}A(x + y)$ as follows:

$$(x + y)^{T}A(x + y) = x^{T}Ax + x^{T}Ay + y^{T}Ax + y^{T}Ay$$

Since $x^{T}Ax = 0$ and $y^{T}Ay = 0$, we have:

$$(x + y)^{T}A(x + y) = x^{T}Ay + y^{T}Ax$$

Now, let's consider the second property.

Q: How do we show that S is closed under scalar multiplication?

A: To show that S is closed under scalar multiplication, we need to show that for any vector x in S and any scalar c, we have $(cx)^{T}A(cx) = 0$. Using the properties of the dot product, we can expand $(cx)^{T}A(cx)$ as follows:

$$(cx)^{T}A(cx) = c^{2}x^{T}Ax$$

Since $x^{T}Ax = 0$, we have:

$$(cx)^{T}A(cx) = 0$$

Q: What is the dimension of S?

A: The dimension of S is equal to the number of linearly independent vectors in S. To find the dimension of S, we need to find a basis for S.

A basis for S is a set of linearly independent vectors that span S.

Let's consider the following set of vectors:

$$\{v_{1}, v_{2}, \ldots, v_{k}\}$$

where $v_{i}$ is a vector in S for each $i = 1, 2, \ldots, k$.

We need to show that this set of vectors is a basis for S.

Q: How do we show that the set of vectors is linearly independent?

A: To show that the set of vectors is linearly independent, we need to show that the following equation holds:

$$c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k} = 0$$

implies that $c_{1} = c_{2} = \cdots = c_{k} = 0$.

Suppose we have:

$$c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k} = 0$$

Then, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

Using the properties of the dot product, we can expand $v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k})$ as follows:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{1}v_{1}^{T}Av_{1} + c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k}$$

Since $v_{1}^{T}Av_{1} = 0$, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k}$$

Similarly, we can show that:

$$v_{2}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = c_{1}v_{2}^{T}Av_{1} + c_{3}v_{2}^{T}Av_{3} + \cdots + c_{k}v_{2}^{T}Av_{k}$$

and so on.

Since $v_{i}^{T}Av_{j} = 0$ for all $i \neq j$, we have:

$$v_{1}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

$$v_{2}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

$$\vdots$$

$$v_{k}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$$

Since $v_{i}^{T}A(c_{1}v_{1} + c_{2}v_{2} + \cdots + c_{k}v_{k}) = 0$ for all $i = 1, 2, \ldots, k$, we have:

$$c_{1}v_{1}^{T}Av_{1} + c_{2}v_{1}^{T}Av_{2} + \cdots + c_{k}v_{1}^{T}Av_{k} = 0$$

$$c_{1}v_{2}^{T}Av_{1} + c_{2}v_{2}^{T}Av_{2} + \cdots + c_{k}v_{2}^{T}Av_{k} = 0$$

$$\vdots$$

$$c_{1}v_{k}^{T}Av_{1} + c_{2}v_{k}^{T}Av_{2} + \cdots + c_{k}v_{k}^{T}Av_{k} = 0$$

Since $v_{i}^{T}Av_{j} = 0$ for all $i \neq j$, we have:

$$c_{1}v_{1}^{T}Av_{1} = 0$$

$$c_{2}v_{2}^{T}Av_{2} = 0$$

$$\vdots$$

$$c_{k}v_{k}^{T}Av_{k} = 0$$

Since $v_{i}^{T}Av_{i} = 0$ for all $i = 1, 2, \ldots, k$, we have:

$$c_{1} = 0$$

$$c_{2} = 0$$

$$\vdots$$

$$c_{k} = 0$$

Therefore, the set of vectors is linearly independent.

Q: What is the dimension of S?

A: The dimension of S is equal to the number of linearly independent vectors in S. In this case, the dimension of S is equal to the number of vectors in the set $\{v_{1}, v_{2}, \ldots, v_{k}\}$.

Therefore, the dimension of S is $k$.

Q: What is the significance of the dimension of S?

A: The dimension of S is an important concept in linear algebra. It represents the number of linearly independent vectors in S, which is a measure of the "size" of S.

The dimension of S is also used to determine the number of independent equations that can be formed from the set of vectors in S.

In summary, the dimension of S is an important concept in linear algebra that represents the number of linearly independent vectors in S.