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If $V$ is vector space of all $2\times 2$ matrices over $R$ having subspaces $W_1=\begin{pmatrix}a&-a\\c&d\end{pmatrix}$ and $W_2=\begin{pmatrix}a&b\\-a&d\end{pmatrix}$. Then what is the $\dim(W_1+W_2)$? It is clear that $\dim(W_1)=\dim(W_2)=3$. Is it valid if I write $W_1+W_2=\begin{pmatrix}2a&-a+b\\-a+c&2d\end{pmatrix}$? So, it will have dim $4$. Is my way correct?

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2 Answers 2

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I am doing it in the alternate way,$\dim(W_1+W_2)=\dim W_1+ \dim W_2-\dim W_1 \cap W_2=3+3-2=4$

Let $x \in W_1 \cap W_2$, then $x \in W_1 $ and $x \in W_2$. So, $x=\begin{pmatrix}a_1&-a_1\\c_1&d_1\end{pmatrix}$ and $x=\begin{pmatrix}a_2&b_2\\a_2&d_2\end{pmatrix}$. so, $a_1=a_2=a(say),b_2=-a_1,c_1=a_2,d_1=d_2=d(say)$. So $x$ is of the form $x=\begin{pmatrix}a&-a\\a&d\end{pmatrix}$

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  • $\begingroup$ How do you calculate $dim(W_1\cap W_2)=2$? $\endgroup$
    – Believer
    Dec 31, 2017 at 10:51
  • $\begingroup$ @omkarGirkar I have added the reason, please check. $\endgroup$
    – user464147
    Dec 31, 2017 at 11:05
  • $\begingroup$ Thanks$\ddot\smile$ $\endgroup$
    – Believer
    Dec 31, 2017 at 18:02
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You are doing it wrong; when adding $W_1$ and $W_2$ you must use distinct letters;$$W_1+W_2=\left\{\begin{pmatrix}a+a'&-a+b\\c-a'&d+d'\end{pmatrix}\,\middle|\,a,b,c,d,a',d'\in\mathbb{R}\right\}.$$But your conclusion is correct: $W_1+W_2$ is the space of all $2\times2$ matrices and therefore its dimension is $4$.

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  • $\begingroup$ A small typo... $\endgroup$
    – user371838
    Dec 31, 2017 at 10:49
  • $\begingroup$ @Rohan Thank you for the edition. $\endgroup$ Dec 31, 2017 at 10:58

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