Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 03 (c)

MATH152 April 2016

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[hide] Question B 03 (c)
Consider the $3\times 3$ matrix $A=\left[{\begin{array}{ccc}2&2&-1\\0&0&1\\0&-4&5\end{array}}\right].$ (c) Find a basis of eigenvectors of $A$.

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[show] Hint
If $\lambda$ and $v$ be eigenvalue and eigenvector of a matrix $A$ (respectively), then we have $(A-\lambda I)v=0.$

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[show] Solution

From part (1), we have the corresponding eigenvector to $\lambda _{1}=1$ as $v_{1}=\left[{\begin{array}{c}-1\\1\\1\end{array}}\right].$ For the eigenvector $v_{2}$ corresponding to the eigenvalue $\lambda _{2}=2$, we solve $(A-2I)v_{2}=\left[{\begin{array}{ccc}2-2&2&-1\\0&0-2&1\\0&-4&5-2\end{array}}\right]v_{2}=\left[{\begin{array}{ccc}0\\0\\0\end{array}}\right].$ Let now $v_{2}=\left[{\begin{array}{ccc}a\\b\\c\end{array}}\right]$. Then, we have $\left[{\begin{array}{ccc}0&2&-1\\0&-2&1\\0&-4&3\end{array}}\right]\left[{\begin{array}{ccc}a\\b\\c\end{array}}\right]=\left[{\begin{array}{ccc}2b-c\\-2b+c\\-4b+3c\end{array}}\right]=\left[{\begin{array}{ccc}0\\0\\0\end{array}}\right].$ This means that $2b=c$ and $3c=4b.$ So, we have $b=c=0.$ On the other hand, $a$ could be any real number. So, let $a=1$ to get $v_{2}=a\left[{\begin{array}{ccc}1\\0\\0\end{array}}\right]=\left[{\begin{array}{ccc}1\\0\\0\end{array}}\right].$ Similarly, for the eigenvector $v_{3}$ corresponding to the eigenvalue $\lambda _{3}=4$, putting $v_{3}=\left[{\begin{array}{ccc}a\\b\\c\end{array}}\right]$, we solve $(A-4I)v_{3}=\left[{\begin{array}{ccc}-2&2&-1\\0&-4&1\\0&-4&1\end{array}}\right]\left[{\begin{array}{ccc}a\\b\\c\end{array}}\right]=\left[{\begin{array}{ccc}-2a+2b-c\\-4b+c\\-4b+c\end{array}}\right]=\left[{\begin{array}{ccc}0\\0\\0\end{array}}\right].$ This means that $c=4b$ and hence $a=-b$. Thus, putting $b=1$ for simplicity, we have $v_{3}=\left[{\begin{array}{ccc}a\\b\\c\end{array}}\right]=b\left[{\begin{array}{ccc}-1\\1\\4\end{array}}\right]=\left[{\begin{array}{ccc}-1\\1\\4\end{array}}\right].$ Therefore, the basis of eigenvector of A is the set $\color {blue}\left\{\left[{\begin{array}{c}-1\\1\\1\end{array}}\right],\left[{\begin{array}{ccc}1\\0\\0\end{array}}\right],\left[{\begin{array}{ccc}-1\\1\\4\end{array}}\right]\right\}$