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Find all real numbers $x$ such that $S = \{\langle x, 2\rangle, \langle 1, x\rangle \}$ is not a basis for $\mathbb{R}^2$.

Should we find all numbers $x$ so that is a basis?

I see that $\operatorname{span} S = 2c_1\langle x, 1\rangle + c_2\langle 1, x\rangle$ for real $x$ so we show that this doesn't equal $\mathbb R^2$? How can we go about that?

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    if it's not a basis, then they must be multiples of one another: $\langle x,2\rangle=c\langle 1,x\rangle$. Solve for $c$, then for $x$. – 2017-01-19
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    @SteveD, I get $x = \pm \sqrt{2}$? But isnt there also a possibility that instead of being linearly dependent, the set just simply does not span $R^2$? – 2017-01-19
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    @Amad27 Right, the only way for two vectors in $\mathbb{R}^2$ to not span $\mathbb{R}^2$ is for them to be linearly dependent. – 2017-01-19
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    I'm the `\langle, \rangle` fairy, here to let you know that $\langle, \rangle$ plays nicer with $\TeX$ than $<, >$ does :) – 2017-01-19

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You can look for the values of $x\in\mathbb{R}$ such that $$ and $<1,x>$ are linearly dependent.

Let's try to see what happens if we suppose : $r\, + \,s\,<1,x>=<0,0>$

We get the following system :

$$\cases{xr+s=0\cr2r+xs=0}$$

Addind x times the first equation and (-1) times the second, we get $(x^2-2)r=0$.

If $x\not\in\{-\sqrt2,\sqrt2\}$, then $r=0$ and therefore $s=0$ too.

But if $x\in\{-\sqrt2,\sqrt2\}$, then we can choose whatever $r$ we want (and $s$ is computed according to the first equation) : the two vectors are linearly dependent in that last case.

You can also compute the determinant of the matrix :

$$A=\pmatrix{x & 2\cr 1 & x}$$

which is $\det(A)=x^2-2$. Hence $\det(A)=0\Leftrightarrow x\in\{-\sqrt2,\sqrt2\}$

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    I'm the `\langle, \rangle` fairy, here to let you know that $\langle, \rangle$ plays nicer with $\TeX$ than $<, >$ does :) – 2017-01-19
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    @Arthur That right ! Thanks for the hint – 2017-01-19
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If we have covered the results:

  1. $n$ linearly independent vectors on a vector space $V$ of dimenson $n$ form a basis of $V$
  2. The rank of a matrix is the same as the dimension of the space spanned by its rows

then, $$S\{(x,2),(1,x)\}\text{ is a basis of }\mathbb{R}^2\Leftrightarrow \text{rank }\begin{bmatrix}{x}&{2}\\{1}&{x}\end{bmatrix}=2\Leftrightarrow\det\begin{bmatrix}{x}&{2}\\{1}&{x}\end{bmatrix}=x^2-2\ne 0.$$ Hence, $$S\{(x,2),(1,x)\}\text{ is not a basis of }\mathbb{R}^2\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}.$$