Assum T: R^m to R^n is a matrix transformation with matrix A.Prove that if the column

babeeb0oL 2020-11-30 Answered

Assum T: $R_{m}$ to $R_{n}$ is a matrix transformation with matrix A. Prove that if the columns of A are linearly independent, then T is one to one (i.e injective). (Hint: Remember that matrix transformations satisfy the linearity properties.
Linearity Properties:
If A is a matrix, v and w are vectors and c is a scalar then
$A 0 = 0$
$A (c v) = c A v$
$A (v + w) = A v + A w$

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casincal

Answered 2020-12-01 Author has 82 answers

Proof:
Let T be defined as

T (v) = A v

Take

v = [\begin{matrix} v_{1} \\ v_{2} \\ . \\ . \\ v_{m} \end{matrix}] \in R_{m} a n d A = [\begin{matrix} a_{11} & a_{12} & . & . & a_{1 m} \\ a_{21} & a_{22} & . & . & a_{2 m} \\ . & . & . & . & . \\ . & . & . & . & . \\ a_{n 1} & a_{n 2} & . & . & a_{n m} \end{matrix}] .

Further obtain the result as follows:

T (v) = A v

= [\begin{matrix} a_{11} & a_{12} & . & . & a_{1 m} \\ a_{21} & a_{22} & . & . & a_{2 m} \\ . & . & . & . & . \\ . & . & . & . & . \\ a_{n 1} & a_{n 2} & . & . & a_{n m} \end{matrix}] [\begin{matrix} v_{1} \\ v_{2} \\ . \\ . \\ v_{m} \end{matrix}]

[\begin{matrix} a_{11} v_{1} & + & a_{12} v_{2} & + \dots + & a_{1 m} v_{m} \\ a_{21} v_{1} & + & a_{22} v_{2} & + \dots + & a_{2 m} v_{m} \\ . & . & . & . & . \\ . & . & . & . & . \\ a_{n 1} v_{1} & + & a_{n 2} v_{2} & + \dots + & a_{n m} v_{m} \end{matrix}]

As the colums of A are linearly idependent, no rows Av will be equal. Thus, the vector obtained is different than one another in

R_{n} .

Hence, if the columns of A are linearly independent, then T is one to one.

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