Consider the set of Vectors

\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\)

Our objective is to determine Whether the set S is linear independent subset of Vector Space \(\displaystyle{R}^{{3}}\)

No, S is not linearly independent subset of Vector Space \(\displaystyle{R}^{{3}}\)

We know,

Dimension of Finite Dimensional Vector Space be n.

And Basis the vector space be B

So, Using Converse of Replacement Theorem

\(\displaystyle{\left|{B}\right|}={n}\)

Let linear independent subset of Vector Space be L, it contains m elements

And Generating subset of Vector Space be G

So, Using Replacement Theorem

\(\displaystyle{\left|{L}\right|}\le{\left|{G}\right|}\)

So, \(\displaystyle{\left|{L}\right|}\le{n}={\left|{B}\right|}\)

Hence, \(\displaystyle{\left|{S}\right|}={4}{>}\dim{\left({R}^{{3}}\right)}={3}\)

So,

\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\) is not linearly independent.

\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\)

Our objective is to determine Whether the set S is linear independent subset of Vector Space \(\displaystyle{R}^{{3}}\)

No, S is not linearly independent subset of Vector Space \(\displaystyle{R}^{{3}}\)

We know,

Dimension of Finite Dimensional Vector Space be n.

And Basis the vector space be B

So, Using Converse of Replacement Theorem

\(\displaystyle{\left|{B}\right|}={n}\)

Let linear independent subset of Vector Space be L, it contains m elements

And Generating subset of Vector Space be G

So, Using Replacement Theorem

\(\displaystyle{\left|{L}\right|}\le{\left|{G}\right|}\)

So, \(\displaystyle{\left|{L}\right|}\le{n}={\left|{B}\right|}\)

Hence, \(\displaystyle{\left|{S}\right|}={4}{>}\dim{\left({R}^{{3}}\right)}={3}\)

So,

\(\displaystyle{S}={\left\lbrace{\left({1},{4},{6}\right)},{\left({1},{5},{8}\right)},{\left({2},{1},{1}\right)},{\left({0},{1},{0}\right)}\right\rbrace}\) is not linearly independent.