Dimension and Basis
Basis of all \(3\times 3\) matrices
Our matrix space is something like, We have\[ \begin{bmatrix} \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix} \]
\(9\)
independent entities for every \(3\times 3\)
matrices.So there are
\(9\)
basis,\[ \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & \color{red}{1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \]\[ \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ 0 & 0 & 0 \\ \end{bmatrix} \]\[ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix} \]
We can get any\(3\times 3\)matrix by the linear combination of these\(9\)matrices
What is the Dimension of all
\(3\times 3\)
matrix?There are\(9\)basis and by there linear combination we can get all other matrices.
So Dimension of all\(3\times 3\)matrices is\(9\).
Basis of all \(3\times 3\) Lower Triangular matrices
Our matrix space is something like,We have\[ \begin{bmatrix} \mathbb{R} & 0 & 0 \\ \mathbb{R} & \mathbb{R} & 0 \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix} \]
\(6\)
independent entities for a \(3\times 3\)
matrix.So there are
\(6\)
basis,\[ \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \]\[ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix} \]
We can get any Lower Triangular matrix by the linear combination of these matrices
What is the Dimension of all
\(3\times 3\)
Lower Triangular matrix?There are\(6\)basis and by there linear combination we can get all other Lower Triangular matrices.
So Dimension of all\(3\times 3\)Lower Triangular matrices is\(6\).
Basis of all \(3\times 3\) Symmetric matrices
Our matrix space is something like,We have\[ \begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}} & \color{green}{\mathbb{R}} & \mathbb{R} \\ \end{bmatrix} \]
\(6\)
independent entities for a \(3\times 3\)
matrix.So there are
\(6\)
basis,\[ \begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \]\[ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix} \]
We can get any Symmetric matrix by the linear combination of these matrices
What is the Dimension of all
\(3\times 3\)
Symmetric matrix?There are\(6\)basis and by there linear combination we can get all other Symmetric matrices.
So Dimension of all\(3\times 3\)Symmetric matrices is\(6\).