Science:Math Exam Resources/Courses/MATH221/December 2008/Question 10 (a)

MATH221 December 2008

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Question 10 (a)
True or False (give a proof if true, or provide a counterexample if false): Suppose that $v_{1},\ldots ,v_{n}$ are a basis for $\mathbb {R} ^{n}$, and A is an invertible n x n matrix. Then the vectors $Av_{1},\ldots ,Av_{n}$ are also a basis for $\mathbb {R} ^{n}$.

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
The invertibility of A is important.

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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 statement is true. To prove that $Av_{1},\ldots ,Av_{n}$ is a basis for $\mathbb {R} ^{n}$, we actually only need to verify that $Av_{1},\ldots ,Av_{n}$ are linearly independent vectors. (Why? Recall that for a finite-dimensional vector space V, a linearly independent list of vectors with length equal to the dimension of V is a spanning list, and hence a basis. In our case $V=\mathbb {R} ^{n}$ and dimV = n.) To prove linear independence, assume $c_{1}Av_{1}+\ldots +c_{n}Av_{n}=0,$ where $c_{1},\ldots ,c_{n}\in \mathbb {R}$. We want to show that $c_{1}=\ldots =c_{n}=0$. To begin, write the above equation as $A(c_{1}v_{1}+\ldots +c_{n}v_{n})=0.$ We are given that $A$ is invertible, so apply $A^{-1}$ to the above equation to find $c_{1}v_{1}+\ldots +c_{n}v_{n}=A^{-1}A(c_{1}v_{1}+\ldots +c_{n}v_{n})=A^{-1}0=0.$ Now the vectors $v_{1},\ldots ,v_{n}$ are linearly independent because they are a basis. This means that the only solution to $c_{1}v_{1}+\ldots +c_{n}v_{n}=0$ is $c_{1}=\ldots =c_{n}=0$, which is exactly what we wanted to show. Therefore $Av_{1},\ldots ,Av_{n}$ is a linearly independent list of vectors with length $n={\text{dim}}\mathbb {R} ^{n}$, and thus a basis.

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