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I need to show that the following set of vectors are linearly dependent over $\mathbb C$. $(1-i, i)$ and $(2, -1+i)$.

If l multiply the first vector by $a+bi$ and the second vector by $c+di$, l will obtain four equations after combining the real part and imaginary parts together. Their sum will be equal to zero , l will get 4 equations. So if l write them into an augmented matrice and show that the rank is less than 4 (it is 2), hence prove that $a$, $b$, $c$, $d$ are not equal to zero or that the solution is non trivial. Will this procedure be correct ?

emacs drives me nuts
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Aristotle Stagiritis
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Yes, but what about simply checking whether or not $\frac2{1-i}=\frac{-1+i}i$? The answer is affirmative (both quotients are equal to $1+i$) and therefore the vectors are linearly dependent.

What I am using here is the fact that two vectors are linearly dependent if and only if one of them is a multiple of the other one.

José Carlos Santos
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You can create a matrix whose columns are the above vectors and compute the determinant. If the determinant if non-zero, they are linearly independent and if it is zero they are linearly dependent. Alternatively you can use Gaussian elimination (here: Linear Independence of a set of Complex Vectors)

Alessio K
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  • I haven't used the determinant method, what is the reasoning behind that ? – Aristotle Stagiritis Feb 14 '20 at 09:01
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    If the determinant is zero then the column vectors are linearly dependent because one vector will be a multiple of another or when you reduce the matrix to row echelon form, there will be a zero row. (https://math.stackexchange.com/questions/499321/why-is-the-determinant-zero-iff-the-column-vectors-are-linearly-dependent) – Alessio K Feb 14 '20 at 09:06