Given that $G=\left\{ \left(\begin{array}{cc} a & -a\\ b & c \end{array}\right):a,b,c\in\mathbb{R}\right\} $ and $H=\left\{ \left(\begin{array}{cc} x & y\\ z & -z \end{array}\right):x,y,z\in R\right\} $ are matrices in $\mathbb M_2(\mathbb{R)}$.

(1) Find a basis for $G$.

(2) Find a basis for $H$

(3) Find a basis for $G\cap H$

For the first two of these could be argue that since both of the matrices

are subsets of $\mathbb M_2(\mathbb R)$that a basis for each of the matrices would be the standard basis for

$\mathbb M_2(\mathbb R)$ ie $\left\{ \left(\begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)\left(\begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right)\right\} $?.

For the intersection of $G$ and $H$

I am unsure what a basis would be ?

  • $\begingroup$ Can you find a basis for $\{(a,-a,b,c)\colon a,b,c\in \mathbb R\}$ and see why your answer is wrong? $\endgroup$ – Git Gud Jul 23 '14 at 10:02
  • $\begingroup$ I thought the standard basis would suffice ? $\endgroup$ – Massin Jul 23 '14 at 10:03
  • $\begingroup$ Are you stating that the standard basis is a basis of $\{(a,-a,b,c)\colon a,b,c\in \mathbb R\}$? $\endgroup$ – Git Gud Jul 23 '14 at 10:05
  • $\begingroup$ Yes thats what i would have thought. $\endgroup$ – Massin Jul 23 '14 at 10:10
  • $\begingroup$ OK, this is helpful. You don't know what a basis is. Do you know the difference between a basis and a set of generators? $\endgroup$ – Git Gud Jul 23 '14 at 10:10

Hint: How many independent parameters do you have for each set?

I'll complete it if you need more help.

First edit:

For example, for $G$, you can write each element in $G$ like this $$ \left( \begin{array}{cc} a & -a\\ b & c \end{array}\right) = a \left( \begin{array}{cc} 1 & -1\\ 0 & 0 \end{array}\right)+b \left( \begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right)+ c \left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right)$$

Therefore, the set $\{\left( \begin{array}{cc} 1 & -1\\ 0 & 0 \end{array}\right),\left( \begin{array}{cc} 0 & 0\\ 1 & 0 \end{array}\right),\left( \begin{array}{cc} 0 & 0\\ 0 & 1 \end{array}\right) \}$ spans $G$. You can show that these matrices are also linearly independent, it's easy to see that. Therefore they form a basis for $G$.

Can you do the same for $H$ and find a basis for it?

Can you see what kind of matrices will lie in $H \cap G$?

Second edit:

Well, if something lies in both $G$ and $H$, then it must have the forms of elements in both sets.

So, if $A$ is in $G \cap H$ we can find $a,b,c,x,y,z \in \mathbb{R}$ such that:

$$A= \left( \begin{array}{cc} a & -a\\ b & c \end{array}\right)=\left( \begin{array}{cc} x & y\\ z & -z \end{array}\right)$$

Since the two matrices are equal, we realize that $A$ must look like this:

$$A=\left( \begin{array}{cc} x & -x\\ z & -z \end{array}\right)$$


$$G \cap H = \{ \left( \begin{array}{cc} x & -x\\ z & -z \end{array}\right): x,z \in \mathbb{R} \}$$

Can you find a basis for $G \cap H$ now?

  • $\begingroup$ Do we only need two ? . $\endgroup$ – Massin Jul 23 '14 at 10:12
  • $\begingroup$ @Massin: Well, for $G$ you need three parameters, for $H$ you need $3$ parameters as well. For their intersection, generally speaking, sometimes you can use the formula $\dim(G+H)=\dim G+\dim H−\dim(G \cap H)$, but sometimes you need to find their intersection first and then see how many parameters you need to describe the elements in $G \cap H$. In this particular problem, following the second approach is easy. If I need to explain more clearly, tell me. $\endgroup$ – math.n00b Jul 23 '14 at 10:30
  • $\begingroup$ Howcould we tell the matrix in H intersection G $\endgroup$ – Massin Jul 23 '14 at 10:30

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