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Show that the vectors $X_{1}=(1, 1+i, i), X_{2}=(i, -i, 1-i), X_{3}=(0,1-2i,2-i)$ in $\mathbb C^3$ are linearly independent over the field of real numbers but are linearly dependent over the field of complex numbers.

My Solution- I could easily prove the first part i.e. the vectors are linearly independent over R but in the second part I proceeded as follows and somehow couldn't get the desired result.

I tried showing linear dependence by finding a complex number $(a+ib)$ such that one vector is a multiple of the other i.e. $X_{1}=(a+ib)X_{2}$. Also, $X_{1} + (a+ib) X_{2}=X_{3}$ couldn't give me the result.

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    Welcome to Math SX! Just compute the determinant. – 2017-02-14
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    Thank you ! Can you elaborate a bit. I think am missing something – 2017-02-14
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    $n$ vectors in $K^n$ ($K$ a field) are linearly independent if and only if their determinant is $\ne 0$. Now a determinant of order $3$ is easy to calculate. – 2017-02-14
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    @Bernard Oh ! Thanks, I just forgot that. Just one more doubt, would be highly thankful if you can answer that. I am a bit confused about proving Linear Independence by rref. Conventionally, I think the vectors are arranged in columns and then reduced. But at times my Professor, also arranges them in a row and then gets a rref. Does arranging the vectors either in columns or rows make any difference. – 2017-02-14
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    @Bernard As mentioned, those vectors have to be arithmetical ($K^n$), but over reals, those given vectors are not arithmetical. Computing the determinant can decide the (in)dependence only in case of $\mathbb{C}$-dependence. (Heuristically - determinant should not depend on the field you're working in.) – 2017-02-14
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    @user18900: It makes no difference as to the *rank* of a system of vectors. If you want to extract a *maximal system* of linearly independent vectors from a given system of vectors, it's easier to do it with *row* vectors, as elementary operations on rows the correspond to lear combinations of the given vectors, so that it's easier to give an interpretation. – 2017-02-14
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    @pepa.dvorak: I purposely gave only a general criterion. Of course, over $\mathbf R$, one has to convert the vectors in $\mathbf C^3$ to vectors in $\mathbf R^3$ and compute an echelon form of the resulting $6\times 3$ matrix (or compute the $3\times 3$ minors of the matrix). – 2017-02-14

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You need to find complex numbers $a,b,c$ such that $$ aX_1 + bX_2 + cX_3 = 0. $$ Assuming for a moment that $a$ is non-zero, we can divide each coefficient by it, and thus choose $a = 1$. Now, since the first entry of $X_3$ is 0, we must have that $1 + bi = 0$, hence $b = i$. Can you now figure out $c$?

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    Thanks ! One general query about proving linear independence: While reducing the matrix to rref, does arranging the vectors in columns or in rows make any difference. – 2017-02-14
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    @user18900, no, it does not. The rank of a matrix is equal to the rank of its transpose. – 2017-02-14
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    Yup, that's what I thought but while going through a Schaum's series book there's something called Casting out Algorithm, in which the author finds the basis by arranging the vectors in columns and then counts the rows which have pivots, thereby getting the basis. Do you have any idea about this ? Am really sorry to bother you but I was getting little confused. – 2017-02-14
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    @user18900, I am nor familiar with the algorithm. Perhaps the author is interested in more than just the rank, e.g. is trying to compute the image, or a subset of the domain on which it is an isomorphism. But I can't know without context. – 2017-02-15