How to find a basis for this

• Oct 25th 2008, 07:30 AM
lo2
How to find a basis for this
We have a vectorspace (is it a vectorspace) $R^{2x2}$ we have a subspace $U \subseteq R^{2x2}$ where U is the amount of symmetric matrices, this means that the matrix $A {2x2}$
belongs to $U$ if and only if $A=A^T$

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

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

Is a basis.

Is that true and how can I show that?

Furthermore is it true that the dimension of $U$ is $4$?
• Oct 25th 2008, 07:44 AM
flyingsquirrel
Hi,
Quote:

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

Then I have to find a basis.

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

Quote:

Furthermore is it true that the dimension of $U$ is $4$?
No. 4 is the dimension of $\mathbb{R}^{2\times 2}$. As $U\subset\mathbb{R}^{2\times 2}$, saying that $\dim U = 4 = \dim \mathbb{R}^{2\times 2}$ amounts to saying that $U=\mathbb{R}^{2\times 2}$... It can't be true since not every $2\times 2$ matrix is symmetric.
• Oct 25th 2008, 07:50 AM
lo2
Quote:

Originally Posted by flyingsquirrel
Hi,

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

A basis for $U$. So I guess I have not found the proper basis.

Quote:

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

Yeah that makes sense. I can see that. But what is the dimension then?
• Oct 25th 2008, 08:03 AM
flyingsquirrel
Quote:

Originally Posted by lo2
A basis for $U$. So I guess I have not found the proper basis.

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

Quote:

But what is the dimension then?
How many "vectors" are there in the basis of $U$ you've found ?
• Oct 25th 2008, 08:07 AM
lo2
Quote:

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

I guess:

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

Quote:

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

3.
• Oct 25th 2008, 08:20 AM
flyingsquirrel
Quote:

Originally Posted by lo2
I guess:

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

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

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

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

Quote:

3.
I agree.
• Oct 25th 2008, 08:47 AM
lo2

But how can you show that $span\{E_1, E_2, E_3\}=V$?
• Oct 25th 2008, 10:18 AM
flyingsquirrel
Quote:

Originally Posted by lo2
But how can you show that $span\{E_1, E_2, E_3\}=V$?

By showing that $U\subseteq S$ and $S\subseteq U$ where $S = \mathrm{span}\left(\left\{
E_1=\left(\begin{smallmatrix}
1 & 0\\
0 & 0
\end{smallmatrix}\right),
E_2=\left(\begin{smallmatrix}
0 & 1\\
1 & 0
\end{smallmatrix}\right),
E_3=\left(\begin{smallmatrix}
0 & 0\\
0 & 1
\end{smallmatrix}\right)\right\}
\right)$
.

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