# Linear Algebra by Larry Smith (auth.)

By Larry Smith (auth.)

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**Example text**

Let E be the set of vectors E = {p(x) Isuch that degree p(x) is at most 1} in (#' ilR). Then E is a linearly dependent set of vectors because the vectors 1, x, 1 + x belong to E and 1(1 + x) + (-1)x + (-1)(1) = O. Definition. A set of vectors E that is not linearly dependent is said to be linearly independent. EXAMPLE 3. Let E be the vectors {(1, 1, 1), (0, 1, 1), (0, 0, 1)} 33 5: Linear independence and dependence in [R3. Then E is a linearly independent set of vectors. In order to prove this suppose to the contrary that E is linearly dependent.

So let E 1/'. Then (x, y, z) (*) x +y+Z = O. We wish to write (x, y, z) = a(1, -1,0) + b(O, 1, -1) = (a,b - a, -b) so we must have x = a, y = b - a, Z = -b. To solve for a and b use the first and last equations giving b = -z. a = x, But is this consistent with the middle equation? Substituting gives y= - z - x which recalling (*) we see is valid precisely for the vectors of 1/'. Thus if E 1/' then (x, y, z) (x, y, z) = x(l, -1,0) - z(O, 1, -1), and therefore the vectors (1, -1,0), (0, 1, -1) span 1/'.

Thus ... , An} for 1JIt. 9) o as required. EXAMPLE 15. The vectors (1,1,1), (1,1,0), (1,0,0) are a basis for 1R3. 9) we need only check that the three vectors are linearly independent. If PROOF. aj(1, 1, 1) + a 2 (1, 1,0) + ail, 0, 0) = (0,0,0) 49 6: Bases and finite-dimensional vector spaces then so al al + a2 + a3 = + a2 = 0 0 =0 and al = a2 = a3 = O. So they are linearly independent. EXAMPLE 16. The vectors (1, 1, - 2), (0, 3, - 3) are a basis for the subspace "Y = {(x, y, z)lx + y + z = O}.