$V$ is vector space of real $n \times n$ matrices and $W$ is a set of elements of $A$ with trace equal to zero. How can we find the basis of $W$?
I have seen a few questions previously posted including one that was identical with the exception that it considered only $2 \times 2$ matrices.
I have tried to use the logic derived from them and worked out the following solution:
The dimension of the basis of $W$ will be $n-1$. The basis will have $n \times n$ matrices of form
Mi = { a11 = 1, all elements except ai +1 i + 1 =0, ai +1 i + 1 = -1} . So M1 = {a11 = 1, a22 = -1, rest all elements = 0}
Please correct me if I'm wrong.
Best Answer
Your answer is partially correct. For instance
$$ A=\left( {\begin{array}{*{20}{c}} 2 & 0 & 1 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \\ \end{array}} \right) \in M_3 $$
is not in the span of your basis because $a_{1,3} \ne 0$ but $trace(A)=0$. When $n=2$ your subspace is
$$\begin{align} M_2 &=\left\{ \left( {\begin{array}{*{20}{c}} a & b \\ c & d \\ \end{array}} \right) | a,b,c,d \in \mathbb{R},a+d=0 \right\} \\ &=\left\{ \left( {\begin{array}{*{20}{c}} a & b \\ c & {-a} \\ \end{array}} \right) | a,b,c \in \mathbb{R} \right\} \\ &=\left\{ a \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {-1} \\ \end{array}} \right) + b\left( {\begin{array}{*{20}{c}} 0 & 1 \\ 0 & 0 \\ \end{array}} \right)+ c\left( {\begin{array}{*{20}{c}} 0 & 0 \\ 1 & 0 \\ \end{array}} \right)| a,b,c \in \mathbb{R} \right\} \end{align}$$
so a a natural basis for $M_2$ is
$$B=\left\langle \left( {\begin{array}{*{20}{c}} 1 & 0 \\ 0 & {-1} \\ \end{array}} \right), \left( {\begin{array}{*{20}{c}} 0 & 1 \\ 0 & 0 \\ \end{array}} \right), \left( {\begin{array}{*{20}{c}} 0 & 0 \\ 1 & 0 \\ \end{array}} \right) \right\rangle.$$
So $dim(M_2)=3=2*2-1$. To extend above idea, we have for $n \in \mathbb{N}$
$$\begin{align} M_n &=\left\{ \left( {\begin{array}{*{20}{c}} {{a_{1,1}}} & \cdots & {{a_{1,n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n,1}}} & \cdots & {{a_{n,n}}} \\ \end{array}} \right) | \sum\limits_{j=1}^{n}{a_{j,j}}=0 \right\} \\ &=\left\{ \left( {\begin{array}{*{20}{c}} {{a_{1,1}}} & \cdots & {{a_{1,n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n,1}}} & \cdots & {{a_{n,n}}} \\ \end{array}} \right) | a_{n,n}=-\sum\limits_{j=1}^{n-1}{a_{j,j}} \right\} \\ &=\left\{ \left( {\begin{array}{*{20}{c}} {{a_{1,1}}} & \cdots & {{a_{1,n}}} \\ \vdots & \ddots & \vdots \\ {{a_{n,1}}} & \cdots & {{\sum\limits_{j=1}^{n-1}{-a_{j,j}}}} \\ \end{array}} \right) \right\} \\ &=\left\{ {{a_{1,1}}\left( {\begin{array}{*{20}{c}} 1 & 0 & \cdots & 0 \\ 0 & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & 0 \\ 0 & \ldots & 0 & { - 1} \\ \end{array}} \right) + {a_{1,2}}\left( {\begin{array}{*{20}{c}} 0 & 1 & 0 & 0 \\ 0 & 0 & \vdots & \vdots \\ \vdots & \vdots & \vdots & 0 \\ 0 & \ldots & 0 & { - 1} \\ \end{array}} \right) + \cdots + {a_{n,n - 1}}\left( {\begin{array}{*{20}{c}} 0 & 0 & \cdots & 0 \\ 0 & \vdots & \vdots & \vdots \\ \vdots & \vdots & 0 & 0 \\ 0 & \vdots & 1 & { - 1} \\ \end{array}} \right)} \right\} \end{align} $$
and a natural basis for $M_n$ is
$$B={\left\langle {\left( {\begin{array}{*{20}{c}} 1 & 0 & \cdots & 0 \\ 0 & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & 0 \\ 0 & \ldots & 0 & { - 1} \\ \end{array}} \right),\left( {\begin{array}{*{20}{c}} 0 & 1 & 0 & 0 \\ 0 & 0 & \vdots & \vdots \\ \vdots & \vdots & \vdots & 0 \\ 0 & \ldots & 0 & { - 1} \\ \end{array}} \right), \cdots ,\left( {\begin{array}{*{20}{c}} 0 & 0 & \cdots & 0 \\ 0 & \vdots & \vdots & \vdots \\ \vdots & \vdots & 0 & 0 \\ 0 & \vdots & 1 & { - 1} \\ \end{array}} \right)} \right\rangle } $$
hence $dim(M_n)=n^2 -1$.