Let $V = M^{2\times 2}(\bf F),$

$$W_1 =\left\{\begin{bmatrix}a & b \\c & a\end{bmatrix}\in V\;:\; a, b, c\in F\right\}$$


$$W_2 =\left\{ \begin{bmatrix}0 & a \\-a & b\end{bmatrix}\in V\;:\; a, b, \in F\right\}$$

Prove that $W_1$ and $W_2$ are subspaces of $V$ and find the dimensions of $\,W_1\,,\, W_2\,,\, W_1+W_2\,,\, W_1\cap W_2\,$.

My attempt: Clearly, $W_1$ is of dimension $3$ since it has three independent components, and $W_2$ is of dimension $2$ since it only has $2$. However, does this mean $W_1+W_2$ will have $\dim = 3$ since there will be three independents in total? How do I prove that?

  • $\begingroup$ Matrices cannot be spaces (unless it is the zero matrix), so surely you meant $\,W_1\,,\,W_2\,$ sets of all the spaces with the required conditions. I edited your question to clear this out. $\endgroup$ – DonAntonio Sep 28 '12 at 3:13
  • $\begingroup$ Pick a basis for $V$. See whether, for each $A$ in the basis, you can find $B$ in $W_1$ and $C$ in $W_2$ such that $B+C=A$. If you can, you have proved the dimension of $W_1+W_2$ is 4. $\endgroup$ – Gerry Myerson Sep 28 '12 at 3:16
  • $\begingroup$ @DonAntonio You can't say a group of matrices forms a subspace? $\endgroup$ – CodyBugstein Sep 28 '12 at 3:33
  • $\begingroup$ You meant a set of matrices? Yes, you can say that, yet that is not what you wrote in your question's heading. $\endgroup$ – DonAntonio Sep 28 '12 at 3:35

Since $\,\dim (W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)\,$ , it is probably wiser to try to find first the dimension of the intersection:

$$A=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}\in W_1\cap W_2\Longleftrightarrow 0=\alpha=\delta\,\,,\,\,\beta=-\gamma$$

and from the above clearly the dimension of the intersection is $\,1\,$ . Now complete the answer.

Added: $\,W_1\,$ is closed under multiplication by scalar:

$$k\begin{pmatrix}a&b\\c&a\end{pmatrix}:=\begin{pmatrix} ka&kb\\kc&ka\end{pmatrix}$$

and we can easily see the right hand matrix belongs to $\,W_1\,$ as its main diagonals' elements are equal, as required from any element in $\,W_1\,$

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  • $\begingroup$ How did you find the intersection? I don't see any intersection between them. $\endgroup$ – CodyBugstein Sep 28 '12 at 3:21
  • $\begingroup$ Can you explain this more? I don't understand what A is. $\endgroup$ – CodyBugstein Sep 28 '12 at 3:25
  • $\begingroup$ As I showed: first, it must be $\,\alpha=0\,$ since $\,A\in W_2\,$, but also it must be $\,\alpha=\delta\,$ since $\,A\in W_1\,$. The rest follows. $\endgroup$ – DonAntonio Sep 28 '12 at 3:25
  • $\begingroup$ So then why don't you get an intersection of dimension 2? According to what you're telling me, W1 intersection W2 has two base matrices - [0 & -b \\ b & 0] and [a & 0 \\ 0 & a]] $\endgroup$ – CodyBugstein Sep 28 '12 at 3:32
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    $\begingroup$ A basis could be $$\begin{pmatrix}0&1\\\!\!\!-1&0\end{pmatrix}$$ $\endgroup$ – DonAntonio Sep 28 '12 at 3:58

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