Dimension and Basis

Basis of all $3\times 3$ matrices

Our matrix space is something like,

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$$\begin{bmatrix} \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix}$$

We have $9$ independent entities for every $3\times 3$ matrices.
So there are $9$ basis,

note

$$\begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & \color{red}{1} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix}$$

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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?

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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,

note

$$\begin{bmatrix} \mathbb{R} & 0 & 0 \\ \mathbb{R} & \mathbb{R} & 0 \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix}$$

We have $6$ independent entities for a $3\times 3$ matrix.
So there are $6$ basis,

note

$$\begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix}$$

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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?

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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,

note

$$\begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}} & \color{green}{\mathbb{R}} & \mathbb{R} \\ \end{bmatrix}$$

We have $6$ independent entities for a $3\times 3$ matrix.
So there are $6$ basis,

note

$$\begin{bmatrix} \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ \color{red}{1} & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & \color{red}{1} & 0 \\ \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \color{red}{1} \\ \end{bmatrix}$$

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We can get any Symmetric matrix by the linear combination of these matrices

What is the Dimension of all $3\times 3$ Symmetric matrix?

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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 $6$.