a Prove that if v1 vk is a linearly dependent list of ve

a) Prove that if v1, . . . , vk is a linearly dependent list of vectors in R n , and T: R n R m is a linear transformation, then T(v1), . . . , T(vk) is also linearly dependent.

b) Prove that if S: R n R m and T: R m R p are linear transformations, then the composite transformation T S (defined by T S(x) = T(S(x))) is also a linear transformation. Prove that if T S is one-to-one, then S is one-to-one as well. Prove that if T and S are both one-to-one, then T S is one-to-one.

c) Prove that if B is an m × n matrix with linearly independent columns and A is a p × m matrix with linearly independent columns, then the columns of AB are also linearly independent. Prove that if B is m × n, A is p × m, and AB has linearly independent columns, then the columns of B are also linearly independent. How is this related to part b)?

Solution

Given that v1,v2,v3...vn are linearly dependent.

Hence it is possible to represent atleast one vi say vm as a linear combination of other n-1 vector

v_m = av1+bv2 +cv3+......pv_m-1+qv_m+1+...

Tranform Vm now

T(v_m) = T(av1+bv2 +cv3+......pv_m-1+qv_m+1+...)

₌ aT(v1)+bT(v2)+cT(v3)+....+pT(v_m-1)+qT(v_m+1)+...) (due to linearlity of T)

i.e. it directly follows that T(vm) can be represented as a linear combination of other n-1 vectors of vi.
So they are linearly dependent.

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Given that  S: R n R m and T: R m R p are linear transformations.

T S (x+y) = T{S(x)+S(y)} since S is linear

= T(s1+s2)

= T(s1)+T(s2) (since T is linear)

= T(S(x))+T(S(y))

= T0S (x) + T0S (y)

Follows that ToS is also linear

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Let T0S (x) be one to one

Then T0S (x)=T0S (y) gives x =y

T{S(x)} = T{S(y}} implies x =y

Or S(x) = S(y) implies x=y hence S is also one to one.

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If T and S both are one to one

then ToS(x) = ToS(y)

implies

ToS(x-y) =0

x=y hence one to one.

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A is a p × m matrix with linearly independent columns

B is a m × n matrix with linearly independent columns

The matrix multiplication of AB is nothing but multiplying each row of matrix A with each column of matrix B.

Since B is columnwise linearly independent, the matrix AB got columns as a linear combination of linearly independent columns of B.

So it follows that Columns of AB are linearly independent.

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Conversely if AB is linearly independent, AB can be said to have columns as

ToS where S is for columns of B and T for rows of A

Since by part B, it follows that when ToS is one to one, T is one to one.

Here we have unique combination of linear combinations B also has.

Hence B is columnwise linearly independent.