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The set of all 2 x 2 matrices of the form

\begin{bmatrix} x & 1 \\ 1 & x \end{bmatrix}

Where each $x$ may be any scalar


I don't get why this doesn't close under addition textbook says thats the reason this isn't a vector space.

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closed as off-topic by Hurkyl, Mercy King, Claude Leibovici, Siong Thye Goh, k1.M Sep 10 '17 at 14:31

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    $\begingroup$ Try adding two of them. Any two of them, it doesn't matter which. $\endgroup$ – Hurkyl Sep 10 '17 at 3:02
  • $\begingroup$ Incidentally, the body of your post should be self-contained -- it's okay to duplicate information in the title, but nothing important should be only in the title. $\endgroup$ – Hurkyl Sep 10 '17 at 3:03
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Hint: $$\begin{bmatrix}x & 1\\1&x\end{bmatrix} + \begin{bmatrix}y & 1\\1&y\end{bmatrix}=\begin{bmatrix}x+y & \color{red}{2}\\\color{red}{2}&x+y\end{bmatrix}.$$

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    $\begingroup$ What does it mean to be closed under addition then? I don't get how this proves it $\endgroup$ – jimmy jimmy Sep 10 '17 at 3:05
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    $\begingroup$ @jimmyjimmy Let the set be $S$ whose elements are the matrices of the given form. The set is closed under addition if the sum of any two elements in $S$ results in another element in $S$. Does $\begin{bmatrix}x+y & \color{red}{2}\\\color{red}{2}&x+y\end{bmatrix}$ belong to $S$? $\endgroup$ – Math Lover Sep 10 '17 at 3:09
  • $\begingroup$ \begin{bmatrix} x & 1 \\ 1 & x \end{bmatrix} thats S? It has to stay that form, correct? Since adding them the new matrix arent in that form so that means it not closed under addition? Thats what I'm getting at $\endgroup$ – jimmy jimmy Sep 10 '17 at 3:16
  • $\begingroup$ @jimmyjimmy Yes, you got it right. $\endgroup$ – Math Lover Sep 10 '17 at 3:22

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