Q.4. Prove that, if V is a finite dimensional vector space over F and if U1, U2, ... Um belongs to V are linearly independent, then we can find vectors Um+1, Um+2, ..., Um+r in V such that {ui | 1 less than equal to i less than equal to m + r } is a basis of V.
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Answers
legth of PQ=8cm
length of OQ = 6cm
PQ is tangent
therefore ∠OQP=90
so by apply pythoguras theorem
we can find OP
OP² =OQ²+PQ ²
OP² =6² +8²
OP=10cm
Step-by-step explanation:
Suppose U is a subspace of V and S ∈ L(U, W). Choose a basis u1, ..., um
of U. Then u1, ..., um is a linearly independent list of vectors in V, and so can
be extended to a basis u1, ..., um, v1, ..., vn of V (by Proposition 2.33 ). Using
Proposition 3.5, we know that there exists a unique linear map T ∈ L(V, W) such
that
T ui = Sui for all i ∈ {1, 2, ..., m}
T vj = 0 for all j ∈ {1, 2, ...n}
Now we are going to prove T u = Su for all u ∈ U.
For any u ∈ U, u can be written as a1u1 + ... + amum, since S ∈ L(U, W),
Su = a1Su1 + a2Su2 + ... + amSum (see Definition 3.2 ).
Since T ∈ L(V, W), we have
T u = T(a1u1 + ... + amum)
= a1T u1 + a2T u2 + ... + amT um
= a1Su1 + a2Su2 + ... + amSum
= Su
Therefore we have T u = Su for all u ∈ U, so we have proved that every linear
map on a subspace of V can be extended to a linear map on V .