# Thread: How to find a basis for this

1. ## How to find a basis for this

We have a vectorspace (is it a vectorspace)$\displaystyle R^{2x2}$ we have a subspace $\displaystyle U \subseteq R^{2x2}$ where U is the amount of symmetric matrices, this means that the matrix $\displaystyle A {2x2}$
belongs to $\displaystyle U$ if and only if $\displaystyle A=A^T$

Then I have to find a basis. I would say that:

$\displaystyle E_1=\begin{bmatrix}1&0 \\ 0 & 0 \end{bmatrix} , E_2=\begin{bmatrix}0&1 \\ 0 & 0 \end{bmatrix} , E_3=\begin{bmatrix}0&0 \\ 1 & 0 \end{bmatrix} , E_4=\begin{bmatrix}0&0 \\ 0 & 1 \end{bmatrix}$

Is a basis.

Is that true and how can I show that?

Furthermore is it true that the dimension of $\displaystyle U$ is $\displaystyle 4$?

2. Hi,
Originally Posted by lo2
We have a vectorspace (is it a vectorspace)$\displaystyle R^{2x2}$ we have a subspace $\displaystyle U \subseteq R^{2x2}$ where U is the amount of symmetric matrices, this means that the matrix $\displaystyle A {2x2}$
belongs to $\displaystyle U$ if and only if $\displaystyle A=A^T$

Then I have to find a basis.
A basis of which vector space ? $\displaystyle \mathbb{R}^{2\times 2}$ or $\displaystyle U$ ? (what you've done is correct if the answer to the former question is $\displaystyle \mathbb{R}^{2\times 2}$, otherwise it's wrong since $\displaystyle E_2\not \in U$)

Furthermore is it true that the dimension of $\displaystyle U$ is $\displaystyle 4$?
No. 4 is the dimension of $\displaystyle \mathbb{R}^{2\times 2}$. As $\displaystyle U\subset\mathbb{R}^{2\times 2}$, saying that $\displaystyle \dim U = 4 = \dim \mathbb{R}^{2\times 2}$ amounts to saying that $\displaystyle U=\mathbb{R}^{2\times 2}$... It can't be true since not every $\displaystyle 2\times 2$ matrix is symmetric.

3. Originally Posted by flyingsquirrel
Hi,

A basis of which vector space ? $\displaystyle \mathbb{R}^{2\times 2}$ or $\displaystyle U$ ? (what you've done is correct if the answer to the former question is $\displaystyle \mathbb{R}^{2\times 2}$, otherwise it's wrong since $\displaystyle E_2\not \in U$)
A basis for $\displaystyle U$. So I guess I have not found the proper basis.

Originally Posted by flyingsquirrel
No. 4 is the dimension of $\displaystyle \mathbb{R}^{2\times 2}$. As $\displaystyle U\subset\mathbb{R}^{2\times 2}$, saying that $\displaystyle \dim U = 4 = \dim \mathbb{R}^{2\times 2}$ amounts to saying that $\displaystyle U=\mathbb{R}^{2\times 2}$... It can't be true since not every $\displaystyle 2\times 2$ matrix is symmetric.
Yeah that makes sense. I can see that. But what is the dimension then?

4. Originally Posted by lo2
A basis for $\displaystyle U$. So I guess I have not found the proper basis.
Let $\displaystyle A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ be a (real) symmetric matrix. One has $\displaystyle A=A^T \Longleftrightarrow b=c$ hence every symmetric matrix looks like $\displaystyle A=\begin{pmatrix}a & b \\ b & d \end{pmatrix}$ where $\displaystyle a,b$ and $\displaystyle d$ are real numbers. That should help you find a basis.

But what is the dimension then?
How many "vectors" are there in the basis of $\displaystyle U$ you've found ?

5. Originally Posted by flyingsquirrel
Let $\displaystyle A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ be a (real) symmetric matrix. One has $\displaystyle A=A^T \Longleftrightarrow b=c$ hence every symmetric matrix looks like $\displaystyle A=\begin{pmatrix}a & b \\ b & d \end{pmatrix}$ where $\displaystyle a,b$ and $\displaystyle d$ are real numbers. That should help you find a basis.
I guess:

$\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}$

Originally Posted by flyingsquirrel
How many "vectors" are there in the basis of $\displaystyle U$ you've found ?
3.

6. Originally Posted by lo2
I guess:

$\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}$
That's it. Now to show that it is a basis you have to say that the three matrices are linearly independent and you have to show that

$\displaystyle \left\{\begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix} , \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} , \begin{pmatrix}0 &0 \\ 0 & 1 \end{pmatrix}\right\}$

is a spanning set of $\displaystyle U$ (that's more or less what I did in my previous post).

3.
I agree.

7. Thank you for your help!

But how can you show that $\displaystyle span\{E_1, E_2, E_3\}=V$?

8. Originally Posted by lo2
But how can you show that $\displaystyle span\{E_1, E_2, E_3\}=V$?
By showing that $\displaystyle U\subseteq S$ and $\displaystyle S\subseteq U$ where $\displaystyle S = \mathrm{span}\left(\left\{ E_1=\left(\begin{smallmatrix} 1 & 0\\ 0 & 0 \end{smallmatrix}\right), E_2=\left(\begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix}\right), E_3=\left(\begin{smallmatrix} 0 & 0\\ 0 & 1 \end{smallmatrix}\right)\right\} \right)$.

Remember that every real, symmetric matrix can be written as $\displaystyle \left(\begin{smallmatrix}a & b \\ b & d \end{smallmatrix}\right)$ where $\displaystyle a,b$ and $\displaystyle d$ are real numbers.