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Let $V(R)=M$ of order $2$

$$W_1=\left\{ \begin{pmatrix} a & b \\ 0 & 0 \\ \end{pmatrix} : a ,b \in R\right\}$$

$$W_2=\left\{ \begin{pmatrix} a & 0 \\ c & 0 \\ \end{pmatrix} : a ,c \in R \right\}$$

$$W_1+W_2=\left\{\begin{pmatrix} 2a & b \\ c & 0 \\ \end{pmatrix} : a ,b,c \in R\right\}$$

I am trying to find out the dimension of $W_1+W_2$ using the definition: The number of elements in the basis for $V$ is called dimension of $V$.

Now we have vectors $v_1=(2a,c),v_2=(b,0)$

both are linearly independent and each have two elements in it so dimension is $2$. Is this the correct way to find dimension ?

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  • $\begingroup$ For $W_i$, do you intend to type them in set notation but having difficulty displaying them? $\endgroup$ – Siong Thye Goh Jan 13 '19 at 11:24
  • $\begingroup$ What is $W_2 $? $\endgroup$ – Bernard Jan 13 '19 at 11:26
  • $\begingroup$ @SiongThyeGoh I didnt 't get you. I am just unable to find dimension. $\endgroup$ – Daman Jan 13 '19 at 11:28
  • $\begingroup$ @Bernard typo check now $\endgroup$ – Daman Jan 13 '19 at 11:29
  • $\begingroup$ is $W_1$ a subspace? or a particular matrix? what do you mean by dimension of a particular matrix if it is a particular matrix? $\endgroup$ – Siong Thye Goh Jan 13 '19 at 11:32
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Check that $$\left\{\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} \right\}$$

is a basis for $W_1+W_2$.

Hopefully you can state the dimension correctly.

The question is not talking about column space of a particular matrix.

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  • $\begingroup$ oh so I was calculated column space which $2$ here. $\endgroup$ – Daman Jan 13 '19 at 11:39
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Note that$$W_1+W_2=\left\{\begin{pmatrix}a+a'&b\\c&0\end{pmatrix}\,\middle|\,a,a',b,c\in\mathbb{R}\right\}.$$It's dimension is three, because$$W_1+W_2=\operatorname{span}\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix},\begin{pmatrix}0&0\\1&0\end{pmatrix}\right\}$$and the previous set is linearly independent.

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  • $\begingroup$ What does previous set is linearly independent means ? $\endgroup$ – Daman Jan 13 '19 at 11:40
  • $\begingroup$ Why it is $a+a^{'}$ not $2a$ $\endgroup$ – Daman Jan 13 '19 at 11:44
  • $\begingroup$ It means that$$a\begin{pmatrix}1&0\\0&0\end{pmatrix}+b\begin{pmatrix}0&1\\0&0\end{pmatrix}+c\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}\iff a=b=c=0.$$ $\endgroup$ – José Carlos Santos Jan 13 '19 at 11:46
  • $\begingroup$ Because the $a$ from an element of $W_1$ doesn't have to be the same $a$ that was used in an element of $W_2$. $\endgroup$ – José Carlos Santos Jan 13 '19 at 11:49

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