Science:Math Exam Resources/Courses/MATH152/April 2015/Question B 2 (c)

MATH152 April 2015

• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QA 9 • QA 10 • QA 11 • QA 12 • QA 13 • QA 14 • QA 15 • QA 16 • QA 17 • QA 18 • QA 19 • QA 20 • QA 21 • QA 22 • QA 23 • QA 24 • QA 25 • QA 26 • QA 27 • QA 28 • QA 29 • QA 30 • QB 1(a) • QB 1(b) • QB 1(c) • QB 2(a) • QB 2(b) • QB 2(c) • QB 3(a) • QB 3(b) • QB 3(c) • QB 4(a) • QB 4(b) • QB 4(c) • QB 5(a) • QB 5(b) • QB 5(c) • QB 5(d) • QB 6(a) • QB 6(b) • QB 6(c) •

Other MATH152 Exams

• April 2016 • April 2015 • April 2014 • April 2022 • April 2011 • April 2010 • April 2012 • April 2013 • April 2017 •

Question B 2 (c)
Let $A$ be the matrix ${\begin{bmatrix}4&-1&7\\0&3&0\\1&2&-2\end{bmatrix}}$ (c) How many linearly independent eigenvectors does $A$ have? Justify briefly.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Don't try to find other eigenvectors. Note that we have three distinct eigenvalues; what does that mean?

Hint 2

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. In fact, eigenvectors corresponding to distinct eigenvalues are linearly independent. Let’s prove it. Suppose that $\mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}$ are eigenvectors corresponding to distinct eigenvalues $\lambda _{1},\lambda _{2},\dots ,\lambda _{n}$ , respectively. Let $m$ be the largest integer for which $\mathbf {v} _{1},\dots ,\mathbf {v} _{m}$ are linearly independent. If we had $m<n$ , we could write $\mathbf {v} _{m+1}=\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}$ for some $c_{i}$ not all zero. Now $A\mathbf {v} _{m+1}=\lambda _{m+1}\mathbf {v} _{m+1}\implies A\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}=\sum _{i=1}^{m}c_{i}\mathbf {\lambda } _{i}\mathbf {v} _{i}=\lambda _{m+1}\sum _{i=1}^{m}c_{i}\mathbf {v} _{i}$ , whence $\sum _{i=1}^{m}c_{i}\mathbf {(} \lambda _{i}-\lambda _{m+1})\mathbf {v} _{i}=\mathbf {0}$ . As the $\lambda _{i}$ are distinct, this implies (by linear independence) that $c_{1}=\cdots =c_{n}=0$ , so $\mathbf {v} _{m+1}=\mathbf {0}$ , contradicting the assumption that $\mathbf {v} _{m+1}$ was an eigenvector. We conclude that $m=n$ , so all the eigenvectors are linearly independent. Each of the 3 eigenvalues found in part (b) has a corresponding eigenvector, and by the above, they are linearly independent. (Clearly, there cannot be more linearly independent eigenvectors, since the dimension of the matrix is 3.) Thus, the answer is ${\textstyle \color {blue}3.}$