Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 05 (c)

MATH152 April 2011

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Question B 05 (c)
Consider the matrix below: $A={\begin{bmatrix}3&1&-3\\-1&0&2\\1&0&0\end{bmatrix}}$. It is known that $\lambda _{1}=1$ is an eigenvalue of A. Find eigenvectors associated with $\lambda _{2}$ and $\lambda _{3}$.

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Hint
What was the special relationship between $\lambda _{2}$ and $\lambda _{3}$. Does this relationship hold true for their eigenvectors?

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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. In part (b) we found that the other two eigenvalues $\lambda _{2}$ and $\lambda _{3}$ were complex conjugates, $1\pm {}i$. Therefore, the eigenvectors for $\lambda _{2}$ and $\lambda _{3}$ will also occur in complex conjugate pairs. It is then sufficient to just find one of the vectors and then conjugate it to find the other. We will solve the vector for $\lambda _{2}=1+i$. This means we are seeking a vector for which (A-$\lambda _{2}$I)x=0. $A-\lambda _{2}I=A-(1+i)I={\begin{bmatrix}2-i&1&-3\\-1&-1-i&2\\1&0&-1-i\end{bmatrix}}$. If we denote the vector x as [a,b,c] then we need to solve ${\begin{bmatrix}2-i&1&-3\\-1&-1-i&2\\1&0&-1-i\end{bmatrix}}{\begin{bmatrix}a\\b\\c\end{bmatrix}}={\begin{bmatrix}0\\0\\0\end{bmatrix}}$ which is equivalent to solving ${\begin{aligned}(2-i)a+b-3c&=0,\\-a+(-1-i)b+2c&=0,\\a+(-1-i)c=0.\end{aligned}}$ If we solve this we get a=(1+i)c and b=-ic for arbitrary c. We can set c=1 without loss of generality to get that the eigenvector for $\lambda _{2}$ is $\mathbf {x} _{2}={\begin{bmatrix}1+i\\-i\\1\end{bmatrix}}.$ This means automatically that the eigenvector for $\lambda _{3}$ is just the conjugate of the eigenvector for $\lambda _{2}$. Therefore, $\mathbf {x} _{3}={\overline {\mathbf {x} _{2}}}={\begin{bmatrix}1-i\\i\\1\end{bmatrix}}.$ If you still have trouble computing eigenvectors, it is a worthwhile exercise to actually compute $\mathbf {x} _{3}$ using the method above and comparing with this result.

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