Operation on subspaces

Say $\mathcal{S}$ denotes the space of all $3\times 3$ symmetric matrices.
Say $\mathcal{U}$ denotes the space of all $3\times 3$ lower triangular matrices.

• $\mathcal{S}\cap\mathcal{U}$ What about all $3\times 3$ matrices who are lower triangular AND symmetric.
  We are asking for space contain by both $\mathcal{U}$ AND $\mathcal{S}$, OR say what is the space of $\mathcal{S}\cap\mathcal{U}$.
  And it's all $3\times 3$ Diagonal matrices.
  And what is the dimension of all $3\times 3$ Diagonal matrices.

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Space of all $3\times 3$ diagonal matrices is a subspace of all $3\times 3$ matrices.
It has $3$ independent entities for a $3\times 3$ matrix.
So it's dimension is $3$.

• $\mathcal{S}\cup\mathcal{U}$ What about matrices who are lower triangular OR symmetric.
  We are asking for space contain by both $\mathcal{U}$ OR $\mathcal{S}$,
  Or say what is the space of $\mathcal{S}\cup\mathcal{U}$.

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But all $3\times 3$ lower triangular OR symmetric matrices do not form a subspace of all $3\times 3$ matrices.

But Why?
Let's say $\mathcal{Q}=\mathcal{S}\cup\mathcal{U}$
Consider two matrices,

$$\text{ (symmetric) } A= \begin{bmatrix} \color{red}{1} & \color{red}{1} & 0 \\ \color{red}{1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$$\text{ (lower triangular) }B= \begin{bmatrix} 0 & 0 & 0 \\ \color{red}{-1} & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

$A\in\mathcal{Q}$ and $B\in\mathcal{Q}$ so for $\mathcal{Q}$ to be a vector space, linear combination of $A$ and $B$ must $\in\mathcal{Q}$.

$$A+B= \begin{bmatrix} \color{red}{1} & \color{red}{1} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

And $A+B$ is neither lower triangular and nor symmetric.
So $A+B\notin \mathcal{Q}$.
So $\mathcal{Q}$ is not a vector space.

• $\mathcal{S} + \mathcal{U}$ Here we took any matrix in $\mathcal{S}$ (say $A$) and any matrix in $\mathcal{U}$ (say$B$), and add them.

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You can say it's some sort to linear combinations of space $\mathcal{S}$ and space $\mathcal{U}$.
Say you have a matrix $A\in\mathcal{S}$ and a matrix $B\in\mathcal{U}$.

• If $A\in\mathcal{S}\Rightarrow \alpha\cdot A\in \mathcal{S}$ because $\mathcal{S}$ is a vector space
• If $B\in\mathcal{U}\Rightarrow \beta\cdot B\in \mathcal{U}$ because $\mathcal{U}$ is a vector space And $\alpha\in\mathbb{R}$ and $\beta\in\mathbb{R}$
  And we are looking at $\alpha\cdot A + \beta \cdot B;\quad$ $\forall A\in\mathcal{S}$ and $\forall B\in\mathcal{U}$.
  And it's a linear combinations in between elements of $\mathcal{S}$ and $\mathcal{U}$.

$A\in\mathcal{S}$ and $B\in\mathcal{U}$ and we are looking for $A+B$

$$\begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}} & \color{green}{\mathbb{R}} & \mathbb{R} \\ \end{bmatrix} + \begin{bmatrix} \mathbb{R} & 0 & 0 \\ \mathbb{R} & \mathbb{R} & 0 \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix} = \begin{bmatrix} 2\mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}}+ \mathbb{R} & 2\mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}}+ \mathbb{R} & \color{green}{\mathbb{R}}+ \mathbb{R} & 2\mathbb{R} \\ \end{bmatrix}$$

equivalently,

$$\begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}} & \color{green}{\mathbb{R}} & \mathbb{R} \\ \end{bmatrix} + \begin{bmatrix} \mathbb{R} & 0 & 0 \\ \mathbb{R} & \mathbb{R} & 0 \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix} = \begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \mathbb{R} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix}$$

now there is no boundation so,

$$\begin{bmatrix} \mathbb{R} & \color{red}{\mathbb{R}} & \color{blue}{\mathbb{R}} \\ \color{red}{\mathbb{R}} & \mathbb{R} & \color{green}{\mathbb{R}} \\ \color{blue}{\mathbb{R}} & \color{green}{\mathbb{R}} & \mathbb{R} \\ \end{bmatrix} + \begin{bmatrix} \mathbb{R} & 0 & 0 \\ \mathbb{R} & \mathbb{R} & 0 \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix} = \begin{bmatrix} \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \mathbb{R} & \mathbb{R} & \mathbb{R} \\ \end{bmatrix}$$

Here we have $9$ independent entities for a $3\times 3$ matrix.
So it has $9$ Basis and $9$ Dimensions.

We can also see it by a beautiful formula.

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$\text{dim}(\mathcal{S}) + \text{dim}(\mathcal{U}) = \text{dim}(\mathcal{S}\cap\mathcal{U}) + \text{dim}(\mathcal{S} + \mathcal{U})$

(That is $6 + 6 = 3 + 9$)