Science:Math Exam Resources/Courses/MATH152/April 2022/Question B3 (b)

MATH152 April 2022

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Question B3 (b)
This is a continuation of (a). (b) Find three linearly independent eigenvectors of $A$.

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
Based on the work in part (a), we are looking for vector solutions to the equations $$Av=-v\qquad {\text{and}}\qquad Av=5v.$$ Each vector equation above is a system of three linear equations. Can you try to solve these? Which one do you expect to have a two-dimensional solution space?

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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. We are looking for bases of the null-spaces of the matrices $$\left[{\begin{array}{ccc}0-\lambda &5&-5\\1&4-\lambda &1\\0&0&5-\lambda \end{array}}\right],$$ for $\lambda =-1,5$. Since the eigenvalue 5 is repeated, this is the value of $\lambda$ for which the matrix above has a 2-dimensional null-space. We apply row operations to find the null-spaces. For $\lambda =-1$, we have $$\left[{\begin{array}{ccc}1&5&-5\\1&5&1\\0&0&6\end{array}}\right]\xrightarrow {} \left[{\begin{array}{ccc}1&5&-5\\0&0&6\\0&0&6\end{array}}\right],$$ so the null-space is the span of the vector $v_{1}=\left[{\begin{array}{c}-5\\1\\0\end{array}}\right]$; on other words, $v_{1}$ is an eigenvector corresponding to the eigenvalue -1. For the eigenvalue 5, the row operations yield $$\left[{\begin{array}{ccc}-5&5&-5\\1&-1&1\\0&0&0\end{array}}\right]\xrightarrow {} \left[{\begin{array}{ccc}1&-1&1\\0&0&0\\0&0&0\end{array}}\right],$$ so the null space contains the linearly independent vectors $$v_{2}=\left[{\begin{array}{c}1\\1\\0\end{array}}\right]\qquad {\text{and}}\qquad v_{3}=\left[{\begin{array}{c}0\\1\\1\end{array}}\right],$$ for example. By the rank-nullity theorem, we know that the null-space is 2 dimensional, so $v_{2},v_{3}$ span the null-space; but this is besides the point. The vectors $v_{1},v_{2},v_{3}$ are three linearly independent eigenvectors of $A$. Note that $v_{2},v_{3}$ are far from unique; another choice is $$v_{2}'=\left[{\begin{array}{c}1\\1\\0\end{array}}\right]\qquad {\text{and}}\qquad v_{3}'=\left[{\begin{array}{c}1\\0\\-1\end{array}}\right],$$ or, more generally, any $v_{2}'=av_{2}+bv_{3},\ v_{3}'=cv_{2}+dv_{3}$, where $ad-bc\neq 0$.