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I think so. If we take ##\mathbb R^2## as an example. ##\{(1,0), (0,1) \}## is the usual basis. Any other vector ##(x, y)## can be expressed in this basis:Salmone said:Hope the question is clear.
$$(x, y) = x(1, 0) + y(0,1)$$This can be rewritten as:
$$1(x, y) - x(1, 0) - y(0,1) = 0$$Which shows the linear dependence of ##(x, y), (1,0)## and ##(0, 1)##.
But, as long as ##y \ne 0## the vectors ##(x, y)## and ##(1, 0)## form a basis. And, as long as ##x \ne 0##, then vectors ##(x, y)## and ##(0, 1)## form a basis.
In ##\mathbb R^2## any two "different" vectors forms a basis. Checking linear independence is easy in ##\mathbb R^2##.
It's no so simple in higher dimensions, because you can have three "different" vectors that are linearly dependent. E.g.
$$(1, 2, 3), (2, 1, 5), (5, 4, 13)$$It takes a bit of work to determine whether they are linearly independent or not.