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sets {$x_1, e_2, e_3, e_4$}, {$e_1, x_2, e_3, e_4$} and {$e_1, e_2, x_3, e_4$} are basis but {$x_1, x_2, x_3, e_4$} is not a basis?

I think the answer is no, but what if $x_4$ was a linear combination of some $x_1, x_2, or x_3$.

Is it then possible?

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  • $\begingroup$ By the way, the notion of standard basis implies that is usually called the standard basis. $\endgroup$ – Hagen von Eitzen Oct 22 '14 at 5:33
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Write $\def\\#1{{\bf#1}}\\x_1=a\\e_1+b\\e_2+c\\e_3+d\\e_4$. We use the following theorem:

Let $S$ be an independent set. Then $S\cup\{\\y\}$ is independent if and only if $\\y$ is not in the span of $S$.

In your question it is given that $\{\\x_1,\\e_2,\\e_3,\\e_4\}$ is independent. And it is clear that $\{\\e_2,\\e_3,\\e_4\}$ is independent. So by the theorem, $\\x_1$ is not in ${\rm span}(\\e_2,\\e_3,\\e_4)$; that is, the first component of $\\x_1$ is not zero.

Similarly, the second component of $\\x_2$ is not zero, and the third component of $\\x_3$ is not zero. So the question is, can you find such vectors for which $\{\\x_1,\\x_2,\\x_3,\\e_4\}$ is dependent? And it's not too hard to see that this is the case when, for example, $$\\x_1=(1,0,1,0)\ ,\quad \\x_2=(0,1,1,0)\ ,\quad \\x_3=(1,1,2,0)\ .$$

So the answer is yes, the required set can exist.

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Yes.

Let $x_1 = x_2 = e_1+e_2, x_3 = e_3$. Then $|\{x_1,x_2,x_3,e_4\}| = 3$, so it cannot be a basis.

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