Page 339 - 35Linear Algebra

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10. (i) We say that the vectors {v 1 , v 2 , . . . v n } are linearly independent if there
2
n
1
2
1
exist no constants c , c , . . . c (not all vanishing) such that c v 1 + c v 2 +
n
· · · + c v n = 0. Alternatively, we can require that there is no non-trivial
2
1
1
n
2
solution for scalars c , c , . . . , c to the linear system c v 1 + c v 2 + · · · +
n
c v n = 0. (ii) We say that these vectors span a vector space V if the set
n
2
2
1
1
n
span{v 1 , v 2 , . . . v n } = {c v 1 + c v 2 + · · · + c v n : c , c , . . . c ∈ R} = V . (iii)
We call {v 1 , v 2 , . . . v n } a basis for V if {v 1 , v 2 , . . . v n } are linearly independent
and span{v 1 , v 2 , . . . v n } = V .
3
For u, v, w to be a basis for R , we firstly need (the spanning requirement)
 
x
that any vector   can be written as a linear combination of u, v and w
y
z
       
−1 4 10 x
y
5
c 1  −4  + c 2   + c 3  7  =   .
3 0 h + 3 z
The linear independence requirement implies that when x = y = z = 0, the
3
2
1
only solution to the above system is c = c = c = 0. But the above system
in matrix language reads
1
     
−1 4 10 c x
−4 5 7 c = y .
   2   
3 0 h + 3 c 3 z
Both requirements mean that the matrix on the left hand side must be
invertible, so we examine its determinant
 
−1 4 10
det  −4 5 7  = −4 · (−4 · (h + 3) − 7 · 3) + 5 · (−1 · (h + 3) − 10 · 3)
3 0 h + 3
= 11(h − 3) ·
Hence we obtain a basis whenever h 6= 3.

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