Answer to Question #115063 in Linear Algebra for Sourav mondal

Question

how that the set of all 3 x 3 upper
triangular matrices with entries from R is a
vector space over R under usual addition
and scalar multiplication of matrices. Find a
basis of this vector space.

Answer 1

We will verify all the axiom which qualify the set of all $3 \times 3$ upper triangular matrix forms a vector space over $\mathbb{R}$ ,

Let's define $M:=\bigg\{ \begin{pmatrix} a & b & c\\ 0& d&e\\ 0&0&f \end{pmatrix} : \forall a,b,c,d,e,f \in \mathbb{R} \bigg\}$ . Now, let for any $A , B ,C \in M$ such that $A = \begin{pmatrix} a_1 & b_1 & c_1\\ 0& d_1&e_1\\ 0&0&f_1 \end{pmatrix}$ , $B=\begin{pmatrix} a_2 & b_2& c_2\\ 0& d_2&e_2\\ 0&0&f_2 \end{pmatrix}$ ,$C=\begin{pmatrix} a_3 & b_3 & c_3\\ 0& d_3&e_3\\ 0&0&f_3 \end{pmatrix}$ . We will prove the first four axioms i.e $M$ is an Abelian group under addition.

$1. A+B =\begin{pmatrix} a_1+a_2 & b_1+b_2 & c_1+c_2\\ 0& d_1+d_2&e_1+e_2\\ 0&0&f_1+f_2 \end{pmatrix}=B+A$ $\in M$

$2.(A+B)+C = \begin{pmatrix} a_1+a_2+a_3 & b_1+b_2+ b_3& c_1+c_2+c_3\\ 0& d_1+d_2+d_3&e_1+e_2+e_3\\ 0&0&f_1+f_2+f_3 \end{pmatrix}=A+(B+C)$

$3.A+O =O+A=A$ ,where $O$ null matrix.

$4. A +(-A)=(-A)+A=O$

Now, we will show that multiplicative and distributive axioms holds true,

$5.O\cdot A = O \: (clearly, trivial\: multiplication)$

$6. for, 1 \in \mathbb{R}, 1\cdot A =A$

$7. \forall \alpha, \beta \in \mathbb{R} \: and \: A \in M, \: \alpha(\beta \cdot A)=\alpha \begin{pmatrix} \beta a_1 & \beta b_1 & \beta c_1\\ 0& \beta d_1&\beta e_1\\ 0&0&\beta f_1 \end{pmatrix}$

$=\begin{pmatrix} \alpha \beta a_1 & \alpha\beta b_1 & \alpha\beta c_1\\ 0&\alpha \beta d_1&\alpha\beta e_1\\ 0&0&\alpha\beta f_1 \end{pmatrix}=(\alpha\cdot \beta)A$

$8.$ for any $\alpha \in \mathbb{R} \& A,B \in M$ ,

$\alpha(A+B)= \begin{pmatrix} \alpha (a_1+a_2) & \alpha( b_1+b_2 )& \alpha( c_1+c_2)\\ 0& \alpha(d_1+d_2)& \alpha(e_1+e_2)\\ 0&0& \alpha(f_1+f_2) \end{pmatrix}$ $=$  $\alpha A + \alpha B$

$9.$ for any $\alpha, \beta \in \mathbb{R}$ and $A \in M$ ,

$( \alpha + \beta)A =\begin{pmatrix} ( \alpha + \beta)a_1 & ( \alpha + \beta)b_1 &( \alpha + \beta) c_1\\ 0& ( \alpha + \beta)d_1&( \alpha + \beta)e_1\\ 0&0&( \alpha + \beta)f_1 \end{pmatrix}$ $=\alpha A + \beta A$

Therefore, the set $M$ satisfy all the axioms of vector space,hence $M$ is a vector space.

One basis of $M$ is follows,

$B=$ $\bigg\{$ $\begin{pmatrix} 1 & 0 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}$ ,$\begin{pmatrix} 0 & 1 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}$ ,$\begin{pmatrix} 0 & 0 & 1\\ 0& 0&0\\ 0&0&0 \end{pmatrix}$ ,$\begin{pmatrix} 0 & 0 & 0\\ 0& 1&0\\ 0&0&0 \end{pmatrix}$ ,$\begin{pmatrix} 0 & 0 & 0\\ 0& 0&1\\ 0&0&0 \end{pmatrix}$ ,$\begin{pmatrix} 0 & 0 & 0\\ 0& 0&0\\ 0&0&1 \end{pmatrix}$ $\bigg \}$

proof:

Let, for any $a,b,c,d,e,f \in \mathbb{R}$ , $a\begin{pmatrix} 1 & 0 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+b\begin{pmatrix} 0 & 1 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+c\begin{pmatrix} 0 & 0 & 1\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+$

$d\begin{pmatrix} 0 & 0 & 0\\ 0& 1&0\\ 0&0&0 \end{pmatrix}+e\begin{pmatrix} 0 & 0 & 0\\ 0& 0&1\\ 0&0&0 \end{pmatrix}+f\begin{pmatrix} 0 & 0 & 0\\ 0& 0&0\\ 0&0&1 \end{pmatrix}=\begin{pmatrix} a & b & c\\ 0& d&e\\ 0&0&f \end{pmatrix} \in M$ .

Hence, $B$ spans $M$ .

Now, we will check the linear independence of $B$ .

let for any $a,b,c,d,e,f \in \mathbb{R},$ $a\begin{pmatrix} 1 & 0 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+b\begin{pmatrix} 0 & 1 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+c\begin{pmatrix} 0 & 0 & 1\\ 0& 0&0\\ 0&0&0 \end{pmatrix}+$

$d\begin{pmatrix} 0 & 0 & 0\\ 0& 1&0\\ 0&0&0 \end{pmatrix}+e\begin{pmatrix} 0 & 0 & 0\\ 0& 0&1\\ 0&0&0 \end{pmatrix}+f\begin{pmatrix} 0 & 0 & 0\\ 0& 0&0\\ 0&0&1 \end{pmatrix}=\begin{pmatrix} a & b & c\\ 0& d&e\\ 0&0&f \end{pmatrix} =$ $\begin{pmatrix} 0 & 0 & 0\\ 0& 0&0\\ 0&0&0 \end{pmatrix}$ ,

$\implies a=b=c=d=e=f=0$ , Hence, $B$ satisfy the linear independence.Thus $B$ is a basis of $M$ .

Therefore, we are done.