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

MATH152 April 2011

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Question B 05 (b)
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 the other two eigenvalues, $\lambda _{2}$ and $\lambda _{3}$ of A.

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Hint
How can we relate eigenvalues to the determinant and trace of the matrix?

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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. Recall that our matrix is $A={\begin{bmatrix}3&1&-3\\-1&0&2\\1&0&0\end{bmatrix}}$ with one eigenvalue being, $\lambda _{1}=1$. Also recall that the product of the eigenvalues is equal to the determinant of the matrix and the sum is equal to the trace. We have that tr(A)=3 and det(A)=2. Therefore, ${\begin{aligned}{\textrm {det}}(A)&=\lambda _{1}\lambda _{2}\lambda _{3}=\lambda _{2}\lambda _{3}=2,\\{\textrm {tr}}(A)&=\lambda _{1}+\lambda _{2}+\lambda _{3}=1+\lambda _{2}+\lambda _{3}=3.\end{aligned}}$ Therefore we have that ${\begin{aligned}\lambda _{2}\lambda _{3}&=2,\\\lambda _{2}+\lambda _{3}&=2.\end{aligned}}$ We can rearrange for $\lambda _{3}$ to get $\lambda _{3}^{2}-2\lambda _{3}+2=0$ which has solution $\displaystyle \lambda _{3}=1\pm {}i$ where $i={\sqrt {-1}}$. If we were to plug this in to get $\lambda _{2}$ we'd get $\lambda _{2}=1\mp {}i.$ This is completely unsurprising since we know that complex eigenvalues occur in complex conjugate pairs. Therefore we take one of them for $\lambda _{2}$ and one for $\lambda _{3}$ to get that the three eigenvalues are ${\begin{aligned}\lambda _{1}&=1,\\\lambda _{2}&=1+i,\\\lambda _{3}&=1-i.\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If you are unfamiliar with the relationship between the determinant and trace of a matrix and its eigenvalues, then you can compute the remaining eigenvalues using the characteristic polynomial. That is, for the given matrix $A={\begin{bmatrix}3&1&-3\\-1&0&2\\1&0&0\end{bmatrix}},$ we can solve the polynomial equation $\det(\lambda I-A)=0$. We find that $|\lambda I-A|={\begin{vmatrix}\lambda -3&-1&3\\1&\lambda &-2\\-1&0&\lambda \end{vmatrix}}=\lambda ^{3}-3\lambda ^{2}+4\lambda -2=0.$ We already know that $\lambda _{1}=1$ is a solution, so we can long divide the polynomial by $(\lambda -1)$ to get an equation for the remaining eigenvalues. If we do this long division, we get ${\begin{matrix}\quad \lambda ^{2}-2\lambda +2\\\lambda -1{\overline {)\lambda ^{3}-3\lambda ^{2}+4\lambda -2}}\end{matrix}},$ so the other eigenvalues, $\lambda _{2}$ and $\lambda _{3}$, must solve the equation $\lambda ^{2}-2\lambda +2=0$, which is the same equation we found in Solution 1. Therefore we know that the solution is $\displaystyle \lambda =1\pm {}i,$ giving the three eigenvalues ${\begin{aligned}\lambda _{1}&=1,\\\lambda _{2}&=1+i,\\\lambda _{3}&=1-i.\end{aligned}}$