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** source url Example text**

Suppose v1 , v2 , . . , vn are linearly dependent vectors in some vector space. i. Prove that there is some j ∈ {1, 2, 3, . . , n} such that vj ∈ span{v1 , v2 , . . , vj−1 , vj+1 , . . , vn }. ii. For j as in part i, prove that span{v1 , v2 , . . , vj , . . , vn } span{v1 , v2 , . . , vj−1 , vj+1 , . . , vn }. 9. Suppose u, v1 , v2 , . . , vn are vectors in some vector space. i. If u ∈ span {v1 , v2 , . . , vn }, prove that u, v1 , v2 , . . , vn are linearly dependent. ii. Suppose v1 , v2 , .

7. 8). 37 Let V be a vector space over a ﬁeld F. A subset U of V is a basis for V if U is a linearly independent set such that span U V. Bases are also characterized in the following way. 38 Let V be a vector space over a ﬁeld F, and let U be a nonempty subset of V. i. Suppose U is ﬁnite, say U {u1 , u2 , . . , un }, for some n ∈ N, with uj uk for j k. Then U is a basis for V if and only if 34 1. Background: Complex Numbers and Linear Algebra for each v ∈ V, there exist unique α1 , α2 , . . , αn ∈ F such that n v j 1 αj u j .

We claim that ker T {0}. To see this, suppose u ∈ U and T(u) 0. Then [T(u)]S is the 0 vector in Cn . So AT [u]R [T(u)]S 0. Multiplying this equation on the left by (AT )−1 gives [u]R (AT )−1 0 0. But [u]R 0 implies u 0. This proves the claim that ker T {0}. 7, T is 1 − 1. 8(v). Hence T is invertible. Another useful notion is the rank of a matrix. 56 Let A be an m × n matrix over C (or R). Let T be the linear transformation associated to A in the standard basis, that is, deﬁne T : Cn → Cm (or T : Rn → Rm ) by T(x) Ax.