Introduction

Section 2.1 Introduction

We let \(\R^n:=\{(x_1,x_2,\ldots,x_n):x_i\in \R, 1\leq i\leq n\}\text{.}\) Note that on \(\R^n\text{,}\) we can define addition and scalar multiplication defined as follows: for

\begin{equation*} x=(x_1,x_2,\ldots,x_n),y=(y_1,y_2,\ldots,y_n)\in \R^n, \text{ and } \alpha\in\R\text{.} \end{equation*}

\begin{align*} x+y:=\amp (x_1+y_1,x_2+y_2,\ldots,x_n+y_n)\\ \alpha \cdot x:=\amp (\alpha x_1,\alpha x_2,\ldots,\alpha x_n) \end{align*}

In the sequel, we will write \(\alpha\cdot x\) as \(\alpha x\text{.}\) An element \(x\in R^n\) is called a vector. It written as a column matrix and we also call it column vector.

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If \(A\) is \(m\times n\) matrix then columns of \(A\) can be thought of as vectors in \(\R^m\text{.}\) Similarly, each row can be thought of a vector in \(\R^n\) and is called a row vector.

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Proposition 2.1.1. Properties addition and multiplication in \(\R^n\).

We list the properties of vector addition and scalar multiplication in \(\R^n\) which can be easily proved. I encourage the readers to prove them.

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for all \(x,y\in \R^n\text{,}\) \(x+y=y+x\text{.}\) (Commutative property)
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for all \(x,y,z\in \R^n\text{,}\) \(x+(y+z)=(x+y)+z\text{.}\) (Associative property)
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The zero vector \(\overline{0}=(0,0,\ldots,0)\) has the property, for all \(x\in \R^n\text{,}\) \(\overline{0}=x+\overline{0}\text{.}\) This zero vector, we shall denote by 0 and is called the additive identity. One can show that zero vector is unique.
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for each \(x\in \R^n\text{,}\) there is a vector \(x'\in \R^n\text{,}\) such that \(x+x'=x'+x=\overline{0}\text{.}\) This \(x'\) is called the additive inverse of \(x\text{.}\) It is easy to see that \(x'=-x\text{.}\)
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for all \(\alpha\in \R\) and \(x,y \in \R^n\text{,}\) \(\alpha(x+y)=\alpha x+\alpha y\text{.}\)
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for all \(\alpha,\beta \in \R\) and \(x \in \R^n\text{,}\) \((\alpha+\beta) x=\alpha x+\beta y\text{.}\)
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for all \(\alpha,\beta \in \R\) and \(x \in \R^n\text{,}\) \((\alpha\beta) x=\alpha (\beta x)y=\beta(\alpha x)\text{.}\)
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for all \(x\in \R^n\text{,}\) \(1\cdot x=x\text{.}\)
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The set \(\R^n\) with addition and scalar multiplication along with the above eight properties is called a vector space over \(\R\text{.}\)

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Definition 2.1.2. Vector Subspaces in \(\R^n\).

A non empty subset \(V\subset \R^n\) is called a vector subspace of \(\R^n\) if \(V\) is closed under vector addition and scalar multiplication. That is,

for all \(x,y\in V, x+y\in V\) and
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for all \(x \in V, \alpha \in \R\text{,}\) we have \(\alpha x\in V\text{.}\)
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Note 2.1.3.

Note that sometime in stead of taking \(V\) to non-empty, one can insist that \(V\) conatins the zero vector. In particular, if \(V\) is non-empty and a vector subspace then \(0\in V\text{.}\)

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Example 2.1.4. Examples of vector subspaces in \(\R^n\).

\(\{0\}\) is a vectors subspace of \(\R^n\)
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Any line passing through origin in \(\R^2\) is a subspace of \(\R^2\text{.}\)
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Any line passing through origin in \(\R^3\) is a subspace of \(\R^3\text{.}\)
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Any plane passing through origin in \(\R^3\) is a subspace of \(\R^3\text{.}\)
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Example 2.1.5. Solution of homogeneous system as vector subpace.

Let \(A\) be \(m\times n\) real matrix. Then \(S_h=\{x\in \R^n:Ax=0\}\) is a vector subspace of \(\R^n\text{.}\) In fact any subspace of \(\R^n\) arises in this way. This subspace is also known as null space of \(A\text{.}\)

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Definition 2.1.6.

Let \(\alpha\) be a real number and \(S\subset\R^n\text{,}\) then we can define

\begin{equation*} \alpha S:=\{v\in \R^n: \exists x\in S, \text{ with }x =\alpha x\}. \end{equation*}

If \(W_1\) and \(W_2\) are two subsets of \(\R^n\text{,}\) then we can define

\begin{equation*} W_1+W_2:=\{x\in \R^n: \exists w_1\in W_1, w_2\in W_2 \text{ with }x=w_1+w_2\} \end{equation*}

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Example 2.1.7.

If \(W\) is a subsapce of \(\R^n\) then \(\alpha W\) is null space if \(\alpha =0\text{,}\) otherwise it is \(W\text{.}\)
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If \(W_1\) and \(W_2\) are \(x\)-axis and \(y\)-axis in the plane then \(W_1+W_2=\R^2\text{.}\)
By definition, \(W_1+W_2\subset \R^2\text{.}\) If \((x,y)\in \R^2\text{,}\) then \((x,y)=(x,0)+(0,y)\in W_1+W_2\text{.}\)
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Let \(W_1=\{\alpha (1,-1):\alpha \in \R\}\) and \(W_2=\{\beta (2,1):\beta\in \R\}\text{.}\) Then \(W_1+W_2=\R^2\text{.}\)
Again by definition \(W_1+W_2\subset \R^2\text{.}\) Suppose \((x,y)\in \R^2\text{.}\)Can we find \(\alpha,\beta\in \R\) such that \((x,y)=\alpha (1,-1)+\beta (2,1)\text{?}\) It is easy to see that it amount to solving a system of linear equations \(\alpha+2\beta =x\) and \(-\alpha+\beta=y\) for \(\alpha\) and \(\beta\) which does have a solution.
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Let \(W_1=\{(x,y,0):x,y\in \R\}\text{,}\)the \(xy-\)plane and \(W_2=\{(0,0,z):z\in\R\}\text{,}\) the \(z\)-axis. Then \(W_1+W_2=\R^3\text{.}\)
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Let \(W_1=\{\alpha (1,1,-1):\alpha\in \R\}\text{,}\)the line passining through \((1,1,-1)\) and the origin. Let \(W_2=\{\beta (2,-1,1):\beta\in \R\}\text{,}\)the line passining through \((2,-1,1)\) and the origin. Then \(W_1+W_2\) is the plane passing through \((1,1,-1)\) and \((2,-1,1)\) and the origin. Can you find its equation?
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Example 2.1.8.

Let \(W_1\) and \(W_2\) be two subspaces of \(\R^n\text{.}\) Can we generate more subspaces using \(W_1\) and \(W_2\text{?}\) Natural thing to look at are sunsets

(i) \(W_1\cap W_2\text{,}\)
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\(W_1\cup W_2\) and
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\(W_1+W_2\text{.}\)
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It is easy to see that (i) and (iii) are subspaces where as (ii) need not be a subspace. (justify). Under what condition (ii) is a subspace?

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