By Jorge Rebaza

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Wq}9 and let x G V + W be arbitrary. Then x = v + w, for some v G V and w G W. Then, we can write x = (aivi H h apvp) + (fiiWi H h f3qwq), which implies that x lies in the spanofTiUT 2 . 39 are linearly independent. 37 is in a central concept in linear algebra given in the following definition. , vn} is said to be linearly independent if civi + . . + cnvn = 0 implies ci = • • • = c n = 0. , vn are linearly independent, none of them can be written as a combination of the others. (Otherwise, say ci ^ 0; then, we can write vi as the linear combination vi = — ^V2 — • • • — ^vn.
2. (Uniqueness) The system has at most one solution for every b if and only ifN(A) {o}. = Proof. Existence: We have already seen that for arbitrary x, the vector Ax G col(A). Then, for Ax — b to have a solution, b must lie in the same subspace. Uniqueness: IfN(A) ^ {0}, then besides x = 0, there is another solution to Ax = b, with 6 = 0, which is a contradiction. On the other hand, assuming N(A) = {0}, if there is a 6 for which Ax = b has more than one solution, that is Ax\ = b and Ax2 = b, with x\ 7^x2, then A(x\ — X2) = Ax\ — Ax2 — b — b = 0, which means that x\ — X2 — 0, or x\ = X2.
Thus, we can say that the col(^4) is also spanned by just the first two columns of A, and hence it has dimension two. 55, the column space of A is the subspace of vectors that can be obtained via linear combinations of the form "2" " 1 " 1 0 + C2 2 0 where ci, c2 are arbitrary scalars. The col(^4) is therefore a plane in M3. 55, we have expressed col(A) as a combination of the first two columns, but it can also be expressed as a combination of the last two, or the first and the last. In any case, there are only two linearly independent columns.