Question

How to prove a set is a basis of a matrix?
Let V=Span\left\\{ f_1,f_2,f_3 \right\\}, where ${{f}_{1}}=1$, ${{f}_{2}}={{e}^{x}}$, ${{f}_{3}}=x{{e}^{x}}$
$a)$ prove that S=\left\\{ {{f}_{1}},{{f}_{2}},{{f}_{3}} \right\\} is a basis of $V$.
$b)$ find the coordinates of $g=3+\left( 1+2x \right){{e}^{x}}$ with respect to $S$.
$c)$ is \left\\{ {{f}_{1}},{{f}_{2}},{{f}_{3}} \right\\}linearly independent?
Could somebody explain to me how to solve this problem please?

Answer 1

In this question we have given with a vector space represented as $V$which contains three elements in it. A vector space is defined as the set of all the possible linear combinations which can be made of its basis vectors, where the coefficients are taken from some other field represented as $\kappa$. We will look at all the points one by one and solve them using properties of vector space.

Complete step by step answer:
If our vector space $V$ has basis \left\\{ {{v}_{1}},{{v}_{2}}....,{{v}_{n}} \right\\} and the set of coefficients $\kappa =\mathbb{R}$, then the generic element $v\in V$ can be written in the format as:
$\Rightarrow v={{a}_{1}}{{v}_{1}}+{{a}_{2}}{{v}_{2}}+.........+{{a}_{n}}{{v}_{n}}$, where ${{a}_{1}}+.....+{{a}_{n}}\in \mathbb{R}$
And in the given question we have been given the vector space in the form of a span which means all the possible linear combinations of the three functions ${{f}_{1}}=1$, ${{f}_{2}}={{e}^{x}}$, ${{f}_{3}}=x{{e}^{x}}$.
This implies that every element in the space $V$ has the form:
$\Rightarrow {{a}_{1}}\cdot 1+{{a}_{2}}\cdot {{e}^{x}}+{{a}_{3}}\cdot x{{e}^{x}}$
Which can be simplified as:
$\Rightarrow {{a}_{1}}+{{a}_{2}}{{e}^{x}}+{{a}_{3}}x{{e}^{x}}$
Now consider $\left( a \right)$, prove that S=\left\\{ {{f}_{1}},{{f}_{2}},{{f}_{3}} \right\\} is a basis of $V$.
Since $V$ is defined as the span of the three functions, this means that every element in the span $V$ is defined as a linear combination of the three functions present therefore, we can say they are a basis of $V$.
Now consider $\left( b \right)$, find the coordinates of $g=3+\left( 1+2x \right){{e}^{x}}$ with respect to $S$.
We have the expression as:
$\Rightarrow g=3+\left( 1+2x \right){{e}^{x}}$
On simplifying the bracket, we get:
$\Rightarrow g=3+{{e}^{x}}+2x{{e}^{x}}$
On comparing this to the generic equation ${{a}_{1}}+{{a}_{2}}{{e}^{x}}+{{a}_{3}}x{{e}^{x}}$, we get:
${{a}_{1}}=3$, ${{a}_{2}}=1$ and ${{a}_{3}}=2$. Therefore, we can write that the coordinates of $g$ in $V$ are $\left( 3,1,2 \right)$.
Now consider $\left( c \right)$, is \left\\{ {{f}_{1}},{{f}_{2}},{{f}_{3}} \right\\} linearly independent?
Now to check linear independence, we can verify that none of the ${{f}_{i}}$ can be written as a linear combination of the other two, therefore, in this case:
We cannot find $2$ numbers ${{a}_{2}},{{a}_{3}}$ such that ${{a}_{2}}{{e}^{x}}+{{a}_{3}}x{{e}^{x}}=1$
We cannot find $2$ numbers ${{a}_{1}},{{a}_{3}}$ such that ${{a}_{1}}+{{a}_{3}}x{{e}^{x}}={{e}^{x}}$
We cannot find $2$ numbers ${{a}_{1}},{{a}_{2}}$ such that ${{a}_{1}}+{{a}_{2}}{{e}^{x}}=x{{e}^{x}}$
From the above statements, we can conclude that they are linearly independent.

Note: It is to be remembered that matrix and vectors are different. A matrix is a two-dimensional array of numbers which is represented in the form of rows and columns which are written as $m\times n$. A vector can be regarded as a special type of matrix. There exist two types of vectors, row vectors and column vectors. It has the size $m\times 1$.