Solution. Let T: R" → R™ be a linear transformation and let {V₁,..., vp} be aset of linearly independent vectors in R". Suppose that {T(v₁),...,T(v₂)} isinearly dependent. Then, there are real scalars c₁,..., Cp, not all zero, such thatC₁T (V₁) ++ cpT(vp) = 0. • Given that I is linear, what can you do with the left hand side of the centeredequation above? Once you figure out what I mean, you'll end up with anequation of the form T(...) = 0, with justification.Can you now argue that there is a nonzero vector that I sends to zero? Hereyou will use the fact that {V₁,..., Vp} is a linearly independent set (what isthe ONLY solution to the equation ₁V₁ ++ pvp = 0?).• Why does the fact that I sends a non-zero vector to the zero vector implythat I' is not one-to-one?|

Given that is a linear transformation.Suppose that is the set of linearly independent vectors such…

Answered: olution. Let T: R" →→R" be a linear…

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter6: Vector Spaces

Section6.4: Linear Transformations

Problem 24EQ

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Solution. Let T: R" → R™ be a linear transformation and let {V₁,..., vp} be a
set of linearly independent vectors in R". Suppose that {T(v₁),...,T(v₂)} is
inearly dependent. Then, there are real scalars c₁,..., Cp, not all zero, such that
C₁T (V₁) +
+ cpT(vp) = 0.

• Given that I is linear, what can you do with the left hand side of the centered
equation above? Once you figure out what I mean, you'll end up with an
equation of the form T(...) = 0, with justification.
Can you now argue that there is a nonzero vector that I sends to zero? Here
you will use the fact that {V₁,..., Vp} is a linearly independent set (what is
the ONLY solution to the equation ₁V₁ ++ pvp = 0?).
• Why does the fact that I sends a non-zero vector to the zero vector imply
that I' is not one-to-one?|

Expert Solution

Step 1: Introduction

Given that $T colon straight real numbers to the power of n rightwards arrow straight real numbers to the power of m$ is a linear transformation.

Suppose that $open curly brackets v subscript 1 comma v subscript 2 comma horizontal ellipsis comma v subscript p close curly brackets$ is the set of linearly independent vectors such that $open curly brackets T open parentheses v subscript 1 close parentheses comma T open parentheses v subscript 2 close parentheses comma horizontal ellipsis comma T open parentheses v subscript p close parentheses close curly brackets$ is linearly dependent.

We need to prove that $T$ is not one-one.

We know that $T open parentheses 0 subscript straight real numbers to the power of n end subscript close parentheses equals 0 subscript straight real numbers to the power of m end subscript$ for zero vector $0 subscript straight real numbers to the power of n end subscript element of straight real numbers to the power of n$ and $0 subscript straight real numbers to the power of m end subscript element of straight real numbers to the power of n$ .

We know that, if $T$ is a linear transformation, then $T open parentheses c u plus v close parentheses equals c T open parentheses u close parentheses plus T open parentheses v close parentheses$ , where $u comma v$ are vectors and $c$ is scalar.

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olution. Let T: R" →→R" be a linear transformation and let {v₁,..., vp} be a et of linearly independent vectors in R". Suppose that {T(v₁),...,T(v₂)} is hearly dependent. Then, there are real scalars c₁,..., Cp, not all zero, such that C₁T(v₁) ++ cpT(vp) = 0.