Science:Math Exam Resources/Courses/MATH221/April 2009/Question 12 (f)

MATH221 April 2009

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Question 12 (f)
If T: $R^{5}\rightarrow R^{4}$ is a linear transformation, and if $\displaystyle v_{1},v_{2},v_{3}$ are vectors in $R^{5}$ such that $\displaystyle T(v_{1}),T(v_{2}),T(v_{3})$ are linearly independent, then $\displaystyle v_{1},v_{2},v_{3}$ must be linearly independent.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The answer is true. Suppose that $a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3}=0,$ where $a_{1},a_{2},a_{3}\in \mathbb {R}$. We want to show that $a_{1}=a_{2}=a_{3}=0$. Applying T to the above equation, we find $T(a_{1}v_{1}+a_{2}v_{2}+a_{3}v_{3})=a_{1}T(v_{1})+a_{2}T(v_{2})+a_{3}T(v_{3})=0.$ Since we are given that $T(v_{1}),T(v_{2})$, and $T(v_{3})$ are linearly independent, the above equality forces $a_{1}=a_{2}=a_{3}=0$, which is what we wanted to show. Therefore, $v_{1},v_{2},v_{3}$ are linearly independent.

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