Recall: Basis Let V be a vector space over a field F. A set of S of vectors in V is snid to be a basis of V if
(i) S is linearly independent in V, and

(ii) S generates V ie. linear span of \(S, L(S) = V\)

Theorem: Let V be a vector space of dimension n over a field F. Then any linearly independent set of n vectors of V is a basis of V.

\(M_{22}\) is the vector space of all \(2 \times 2\) real matrices.

Let \(S = \left\{\left(\begin{array}{c}21\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 21\end{array}\right), \left(\begin{array}{c}3-1\\ 00\end{array}\right),\left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

We have to show that 5 is a basis of \(M_{22}\)

Let \(\alpha_{1} = \left\{\left(\begin{array}{c}21\\ 00\end{array}\right),\alpha_{2}= \left(\begin{array}{c}00\\ 21\end{array}\right),\alpha_{3}= \left(\begin{array}{c}3-1\\ 00\end{array}\right),\alpha_{4}=\left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

Let us consider the relation

\(c_{1} \alpha_{1} + c_{2} \alpha_{2} + c_{3} \alpha_{3} + c_{4} \alpha_{4} = 0\)

where \(c_{1}, c_{2}, c_{3}, c_{4} \in RR.\)

Then we have \(c_{1} = \left(\begin{array}{c}21\\ 00\end{array}\right),+c_{2} \left(\begin{array}{c}00\\ 21\end{array}\right),+c_{3} \left(\begin{array}{c}3-1\\ 00\end{array}\right),+c_{4}\left(\begin{array}{c}00\\ 31\end{array}\right)=\left(\begin{array}{c}00\\ 00\end{array}\right) \Rightarrow \left(\begin{array}{c}2c_{1}+3c_{3}c_{1}-c_{3}\\ 2c_{2}+3c_{3}c_{2}+c_{4}\end{array}\right)=\left(\begin{array}{c}00\\00\end{array}\right)\)

Fron the above equation we get,

\(2c_{1} + 3c_{3} = 0\)

\(c_{1} - c_{3} = 0\)

\(2c_{2} + 3c_{4} = 0\)

\(c_{2} + c_{4} = 0\)

Solving these equation we get, \(c_{1} = 0, c_{2} = 0, c_{3} = 0, c_{4} = 0\)

This proves that the set \({\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}}\) is lineary independent.

Now, we know that \(M_{22}\) is a vector space of dimension 4 and its standard

basis is \left\{\left(\begin{array}{c}10\\00\end{array}\right),\left(\begin{array}{c}01\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 10\end{array}\right), \left(\begin{array}{c}00\\ 01\end{array}\right)\right\}\)

As the dimension of \(M_{22}\ is\ 4\ and\ {\alpha_{1}, \alpha_{2}, \alpha_{2}, \alpha_{4}}\) is a linearly independent

set containing 4 vectors of \(M_{22},\)

therefore the set \({\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}}\)

is a basis of \(M_22\).

Result:

\(\left\{\left(\begin{array}{c}21\\00\end{array}\right),\left(\begin{array}{c}00\\ 21\end{array}\right), \left(\begin{array}{c}3-1\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

is a basis for \(M_{22}\)

(ii) S generates V ie. linear span of \(S, L(S) = V\)

Theorem: Let V be a vector space of dimension n over a field F. Then any linearly independent set of n vectors of V is a basis of V.

\(M_{22}\) is the vector space of all \(2 \times 2\) real matrices.

Let \(S = \left\{\left(\begin{array}{c}21\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 21\end{array}\right), \left(\begin{array}{c}3-1\\ 00\end{array}\right),\left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

We have to show that 5 is a basis of \(M_{22}\)

Let \(\alpha_{1} = \left\{\left(\begin{array}{c}21\\ 00\end{array}\right),\alpha_{2}= \left(\begin{array}{c}00\\ 21\end{array}\right),\alpha_{3}= \left(\begin{array}{c}3-1\\ 00\end{array}\right),\alpha_{4}=\left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

Let us consider the relation

\(c_{1} \alpha_{1} + c_{2} \alpha_{2} + c_{3} \alpha_{3} + c_{4} \alpha_{4} = 0\)

where \(c_{1}, c_{2}, c_{3}, c_{4} \in RR.\)

Then we have \(c_{1} = \left(\begin{array}{c}21\\ 00\end{array}\right),+c_{2} \left(\begin{array}{c}00\\ 21\end{array}\right),+c_{3} \left(\begin{array}{c}3-1\\ 00\end{array}\right),+c_{4}\left(\begin{array}{c}00\\ 31\end{array}\right)=\left(\begin{array}{c}00\\ 00\end{array}\right) \Rightarrow \left(\begin{array}{c}2c_{1}+3c_{3}c_{1}-c_{3}\\ 2c_{2}+3c_{3}c_{2}+c_{4}\end{array}\right)=\left(\begin{array}{c}00\\00\end{array}\right)\)

Fron the above equation we get,

\(2c_{1} + 3c_{3} = 0\)

\(c_{1} - c_{3} = 0\)

\(2c_{2} + 3c_{4} = 0\)

\(c_{2} + c_{4} = 0\)

Solving these equation we get, \(c_{1} = 0, c_{2} = 0, c_{3} = 0, c_{4} = 0\)

This proves that the set \({\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}}\) is lineary independent.

Now, we know that \(M_{22}\) is a vector space of dimension 4 and its standard

basis is \left\{\left(\begin{array}{c}10\\00\end{array}\right),\left(\begin{array}{c}01\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 10\end{array}\right), \left(\begin{array}{c}00\\ 01\end{array}\right)\right\}\)

As the dimension of \(M_{22}\ is\ 4\ and\ {\alpha_{1}, \alpha_{2}, \alpha_{2}, \alpha_{4}}\) is a linearly independent

set containing 4 vectors of \(M_{22},\)

therefore the set \({\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}}\)

is a basis of \(M_22\).

Result:

\(\left\{\left(\begin{array}{c}21\\00\end{array}\right),\left(\begin{array}{c}00\\ 21\end{array}\right), \left(\begin{array}{c}3-1\\ 00\end{array}\right), \left(\begin{array}{c}00\\ 31\end{array}\right)\right\}\)

is a basis for \(M_{22}\)