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

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

and

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

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?

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    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.2012-09-28
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    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 4.2012-09-28
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    @DonAntonio You can't say a group of matrices forms a subspace?2012-09-28
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    You meant a set of matrices? Yes, you can say that, yet that is not what you wrote in your question's heading.2012-09-28

1 Answers 1

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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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    How did you find the intersection? I don't see any intersection between them.2012-09-28
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    Can you explain this more? I don't understand what A is.2012-09-28
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    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.2012-09-28
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    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]]2012-09-28
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    This is not making sense to me at all. The way I understand it, $W1$ has a basis of 3 matrices, $W2$ has a basis of 2. The intersection of $W1$ and $W2$ should be the common base matrices W1 and W2 have - and I can't find any. I'm completely lost with the matrix you just posted above, I have no idea where that comes in.2012-09-28
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    I can try to help you understand this, but I can't do anything about something so basic not making sense to you: the matrix A is *assumed* to belong to the intersection of $\,W_1\,,\,W_2\,$ , so it **must** fulfill both sets' characterizations, right? Because it belongs to $\,W_1\,$ both elements on its main diagonal must be equal, but since it *also* belongs to $\,W_2\,$ the entry 11 on the main diagonal **must** be zero, so in fact the whole main diagonal of A must be zeros....what isn't clear here?!2012-09-28
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    So why did you put a 1 on the main diagonal??2012-09-28
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    Because I made a stupid typing mistake. I shall fix this at once, but the important point here is whether you understood what's going on here or not.2012-09-28
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    A basis could be $$\begin{pmatrix}0&1\\\!\!\!-1&0\end{pmatrix}$$2012-09-28
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    Yes now I've got it. The intersection of $W1$ and $W2$ is the matrix subspace that satisfies the properties of $W1$ and $W2$, and that subspace has a dimension of 1.2012-09-28
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    Yup, that's right.2012-09-28
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    I'm stuck on one more thing though; how do I show that W1 is a subspace? I tried closing it under multiplication but I can't prove it's within W1. Can you start a new answer thread with a little help? Thanks alot for all this!2012-09-28
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    I'll add some hints to this in my original answer. Wait for it.2012-09-28