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.

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}

\mathcal{S}

,
Or say what is the space of

\mathcal{S}\cup\mathcal{U}

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

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.

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

(That is

\(6 + 6 = 3 + 9\)

)