Science:Math Exam Resources/Courses/MATH152/April 2022/Question B3 (b)/Solution 1

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 $ad-bc\neq 0$.