# Thread: Space, basis, dimension, rank of Matrices

1. ## Space, basis, dimension, rank of Matrices

I'll start off easy.

How do you determine the dimension and a basis of the group of all 2x2 matrices over R. That would be $M_{2 \times 2}^R = \left\{\left(\begin{array}{cc}\alpha&\beta\\\lambd a&\delta\end{array}\right) | (\alpha, \beta, \lambda, \delta)\in R\right\}$

2. Originally Posted by jayshizwiz
I'll start off easy.

How do you determine the dimension and a basis of the group of all 2x2 matrices over R. That would be $M_{2 \times 2}^R = \left\{\left(\begin{array}{cc}\alpha&\beta\\\lambd a&\delta\end{array}\right) | (\alpha, \beta, \lambda, \delta)\in R\right\}$

Well, as it is, or should be, well known, $M_{n\times n}(\mathbb{R})\cong \mathbb{R}^{n^2}$ (isomorphism as vector spaces, among other things), so...

Tonio

3. Originally Posted by jayshizwiz
I'll start off easy.

How do you determine the dimension and a basis of the group of all 2x2 matrices over R. That would be $M_{2 \times 2}^R = \left\{\left(\begin{array}{cc}\alpha&\beta\\\lambd a&\delta\end{array}\right) | (\alpha, \beta, \lambda, \delta)\in R\right\}$
$\begin{pmatrix}\alpha & \beta \\ \gamma & \delta\end{pmatrix}$ $=\alpha \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}+ \beta\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}+ \gamma\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}+ \delta\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$.

It's that simple!

One letter wrong!

4. [QUOTE]LaTeX Error: Syntax error

It's that simple![QUOTE]

I wish it was that simple! (:

5. .
The extra 1 after the beta is a mistake, correct?
... $\beta \left(\begin{array}{cc}0&1\\0&0\end{array}\right)$

6. Yes, and thanks for catching that!

It should be
$
\begin{pmatrix}\alpha & \beta \\ \gamma & \delta\end{pmatrix}
$
$=\alpha \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}+ \beta\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}+ \gamma\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}+ \delta\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}
$

Now, it should be easy!