More on Determinants
Linearly Independent Sets and Basis

Vector Space

Example 1.

We are given a vector \mathbf{u} =\left \langle a ,b ,c\right \rangle where a ,b and c are real numbers. We would like to write \mathbf{u} as a linear combination of vectors \left \langle 1 ,0 ,0\right \rangle ,\left \langle 0 ,1 ,0\right \rangle and \left \langle 0 ,0 ,1\right \rangle . This means we are looking to find three real numbers x ,y and z such that

(1)   \begin{equation*}\left \langle a ,b ,c\right \rangle =x\left \langle 1 ,0 ,0\right \rangle +y\left \langle 0 ,1 ,0\right \rangle +z\left \langle 0 ,0 ,1\right \rangle \end{equation*}

The right side of the above equality tells us

(2)   \begin{align*}x\left \langle 1 ,0 ,0\right \rangle +y\left \langle 0 ,1 ,0\right \rangle +z\left \langle 0 ,0 ,1\right \rangle =\left \langle x ,0 ,0\right \rangle +\left \langle 0 ,y ,0\right \rangle +\left \langle 0 ,0 ,z\right \rangle \\ =\left \langle x ,y ,z\right \rangle \end{align*}

This means

(3)   \begin{equation*}x =a ,y =b ,z =c \end{equation*}


(4)   \begin{equation*}\left \langle a ,b ,c\right \rangle =a\left \langle 1 ,0 ,0\right \rangle +b\left \langle 0 ,1 ,0\right \rangle +c\left \langle 0 ,0 ,1\right \rangle \end{equation*}

This seems like a very trivial example but it constitutes the main result in our future studies.


Example 2.


Suppose we are given the vector \left \langle -2 ,5\right \rangle and we would like to write this vector as a linear combination of two vectors \left \langle 1 ,1\right \rangle and \left \langle 1 , -1\right \rangle. We are looking for two values x and y such that

(5)   \begin{equation*}\left \langle -2 ,5\right \rangle =x\left \langle 1 ,1\right \rangle +y\left \langle 1 , -1\right \rangle \end{equation*}

The right side of the above equality tells us

(6)   \begin{align*}x\left \langle 1 ,1\right \rangle +y\left \langle 1 , -1\right \rangle =\left \langle x ,x\right \rangle +\left \langle y , -y\right \rangle \\ =\left \langle x +y ,x -y\right \rangle \end{align*}



(7)   \begin{equation*}\left \langle -2 ,5\right \rangle =\left \langle x +y ,x -y\right \rangle \end{equation*}

which means

(8)   \begin{align*}x +y = -2 \\ x -y =5\end{align*}

Of course we know from previous lessons how to solve this system of linear equations in many different ways to get

(9)   \begin{equation*}x =\frac{3}{2} ,y = -\frac{7}{2} \end{equation*}


(10)   \begin{equation*}\left \langle -2 ,5\right \rangle =\frac{3}{2}\left \langle 1 ,1\right \rangle -\frac{7}{2}\left \langle 1 , -1\right \rangle \end{equation*}


Example 3.


Suppose we are given the set V of all polynomials of the form p\left (x\right ) =ax^{2} +bx +c where a ,b and c are real numbers. Take the usual addition operation for polynomials and multiplication of a polynomial with a constant, together with the given set. We show the set V and the two operation constitute a vector space.


a. Suppose we have two polynomials p\left (x\right ) =a_{1}x^{2} +b_{1}x +c_{1} and q\left (x\right ) =a_{2}x^{2} +b_{2}x +c_{2}. Then

(11)   \begin{align*}p\left (x\right ) +q\left (x\right ) =a_{1}x^{2} +b_{1}x +c_{1} +a_{2}x^{2} +b_{2}x +c_{2} \\ =\left (a_{1} +a_{2}\right )x^{2} +\left (b_{1} +b_{2}\right )x +\left (c_{1} +c_{2}\right )\end{align*}

which is a polynomial in V .



(12)   \begin{align*}p\left (x\right ) +q\left (x\right ) =a_{1}x^{2} +b_{1}x +c_{1} +a_{2}x^{2} +b_{2}x +c_{2} \\ =a_{2}x^{2} +b_{2}x +c_{2} +a_{1}x^{2} +b_{1}x +c_{1} \\ =q\left (x\right ) +p\left (x\right )\end{align*}


c. Given another polynomial r\left (x\right ) =a_{3}x^{2} +b_{3}x +c_{3} we have

(13)   \begin{align*}p\left (x\right ) +\left (q\left (x\right ) +r\left (x\right )\right ) =a_{1}x^{2} +b_{1}x +c_{1} +(a_{2}x^{2} +b_{2}x +c_{2} +a_{3}x^{2} +b_{3}x +c_{3}) \\ =(a_{1}x^{2} +b_{1}x +c_{1} +a_{2}x^{2} +b_{2}x +c_{2}) +a_{3}x^{2} +b_{3}x +c_{3} \\ =\left (p\left (x\right ) +q\left (x\right )\right ) +r\left (x\right )\end{align*}


d. Take the zero polynomial 0 and observe that

(14)   \begin{equation*}p\left (x\right ) +0 =a_{1}x^{2} +b_{1}x +c_{1} +0x^{2} +0x +0 =a_{1}x^{2} +b_{1}x +c_{1} =p\left (x\right ) \end{equation*}

Note that the zero polynomial is equal to zero for all values of x.


e. Take the polynomial -p\left (x\right ) = -a_{1}x^{2} -b_{1}x -c_{1} and observe that

(15)   \begin{align*}p\left (x\right ) +\left ( -p\left (x\right )\right ) =a_{1}x^{2} +b_{1}x +c_{1} +\left ( -a_{1}x^{2} -b_{1}x -c_{1}\right ) \\ =a_{1}x^{2} -a_{1}x^{2} +b_{1}x -b_{1}x +c_{1} -c_{1} \\ =0\end{align*}


f. Suppose k is a real number. Then

(16)   \begin{align*}kp\left (x\right ) =k\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) \\ =ka_{1}x^{2} +kb_{1}x +kc_{1}\end{align*}

which is a polynomial in V .


g. Observe that

(17)   \begin{align*}k\left (p\left (x\right ) +q\left (x\right )\right ) =k\left (a_{1}x^{2} +b_{1}x +c_{1} +a_{2}x^{2} +b_{2}x +c_{2}\right ) \\ =k\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) +k\left (a_{2}x^{2} +b_{2}x +c_{2}\right ) \\ =kp\left (x\right ) +kq\left (x\right )\end{align*}


h. Given a real number h ,

(18)   \begin{align*}\left (k +h\right )p\left (x\right ) =\left (k +h\right )\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) \\ =k\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) +h\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) \\ =kp\left (x\right ) +hq\left (x\right )\end{align*}


i. Observe that

(19)   \begin{align*}k\left (hp\left (x\right )\right ) =k\left (h\left (a_{1}x^{2} +b_{1}x +c_{1}\right )\right ) \\ =(kh)\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) \\ =\left (kh\right )p\left (x\right )\end{align*}


j. Last thing

(20)   \begin{align*}1p\left (x\right ) =1\left (a_{1}x^{2} +b_{1}x +c_{1}\right ) \\ =a_{1}x^{2} +b_{1}x +c_{1} \\ =p\left (x\right )\end{align*}

We checked all ten properties of a vector space.


Example 4.


The set of integers \mathbb{Z} and the two operations addition and multiplication do not construct a vector space since given the real number \frac{1}{2} and the element 3 of integers, we have

(21)   \begin{equation*}\frac{1}{2}\left (3\right ) =\frac{3}{2} \end{equation*}

which does not belong to the set of integers.


Example 5.


We want to determine if the set of all vectors \left (x ,y\right ) such that x +2y =0 is a subspace of \mathbb{R}^{2}. Note that in order to show a nonempty set U is a subspace of a vector space V, we need to show U is closed with respect to both operations inherited from V . First note that the set provided by this example is nonempty since \left (0 ,0\right ) belongs there. Now suppose \left (x ,y\right ) and \left (x_{1} ,y_{1}\right ) are two members of this set and k is a real number. Since x = -2y we can write

(22)   \begin{align*}\left (x ,y\right ) +\left (x_{1} ,y_{1}\right ) =\left ( -2y ,y\right ) +\left ( -2y_{1} ,y_{1}\right ) \\ =\left ( -2y -2y_{1} ,y -y_{1}\right ) \\ =\left ( -2\left (y -y_{1}\right ) ,y -y_{1}\right ) \\ =\left ( -2y_{0} ,y_{0}\right )\end{align*}

where y_{0} =y -y_{1}. Therefore the result belongs to the given set. Next,

(23)   \begin{align*}k\left (x ,y\right ) =k\left ( -2y ,y\right ) \\ =\left ( -2ky ,ky\right ) \\ =\left ( -2y_{2} ,y_{2}\right )\end{align*}

where y_{2} =ky . This shows the given line is a subspace of \mathbb{R}^{2} .


Example 6.


We would like to check if the set \{\left (\begin{array}{c}1 \\ 1\end{array}\right ) ,\left (\begin{array}{c} -1 \\ 1\end{array}\right )\} spans \mathbb{R}^{2} . Using the definition of an spanning set, we need to check if every vector in \mathbb{R}^{2} can be expressed as a linear combonation of the elements of the given set. Let’s pick an arbitrary vector \left (\begin{array}{c}a \\ b\end{array}\right )and write

(24)   \begin{equation*}x\left (\begin{array}{c}1 \\ 1\end{array}\right ) +y\left (\begin{array}{c} -1 \\ 1\end{array}\right ) =\left (\begin{array}{c}a \\ b\end{array}\right ) \end{equation*}


This means

(25)   \begin{equation*}\left (\begin{array}{c}x -y \\ x +y\end{array}\right ) =\left (\begin{array}{c}a \\ b\end{array}\right ) \end{equation*}

that is

(26)   \begin{align*}x -y =a \\ x +y =b\end{align*}


(27)   \begin{align*}x =a +y \\ x +y =a +y +y \\ =a +2y \\ =b \\ y =\frac{b -a}{2} \\ x =a +\frac{b -a}{2} =\frac{b +a}{2}\end{align*}


Therefore any vector \left (\begin{array}{c}a \\ b\end{array}\right ) in \mathbb{R}^{2} can be written as a linear combination

(28)   \begin{equation*}\frac{b +a}{2}\left (\begin{array}{c}1 \\ 1\end{array}\right ) +\frac{b -a}{2}\left (\begin{array}{c} -1 \\ 1\end{array}\right ) \end{equation*}

which means this set spans \mathbb{R}^{2} .

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