Science:Math Exam Resources/Courses/MATH307/April 2010/Question 04 (b)/Solution 1

Knowing $A_{1}$ has an eigenvalue of 1 we need to solve the equation:

$A_{1}x=\lambda x$ where $\lambda$ = $1$

${\begin{bmatrix}a&b\\1&0\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}={\begin{bmatrix}x_{1}\\x_{0}\end{bmatrix}}$

making it into a system of linear equations we get:

1) $ax_{1}+bx_{0}$ = $x_{1}$

2) $x_{1}$ = $x_{0}$

using 2) we can substitute into 1) so that we are only working with one variable,

$ax_{0}+bx_{0}$ = $x_{0}$

$x_{0}$($a$ $+$ $b$) $=$ $x_{0}$

$a$ $+$ $b$ $=$ $1$

now that we see how $a$ and $b$ have to relate, let’s find the other eigenvalue of $A_{1}$

solve for either $a$ or $b$ and input back into $A_{1}$ and then do our standard $det(A_{1}$ $-$ $\lambda I)$ $=$ $0$

I’m going to continue after having solved for $b$

so our new $A_{1}$ = ${\begin{bmatrix}a&1-a\\1&0\end{bmatrix}}$

$det(A_{1}-\lambda I)=det({\begin{bmatrix}a-\lambda &1-a\\1&-\lambda \end{bmatrix}})=(a-\lambda )(-\lambda )-(1-a)=0$

we end up obtaining the equation: $\lambda ^{2}$ $-$ $\lambda a$ $+$ ($a$ $-$ $1$) $=$ $0$

using the quadratic formula with a = $1$, b = $-a$ and c = $a$ $-$ $1$, our result is:

$\lambda ={\frac {a\pm {\sqrt {a^{2}-4(a-1)}}}{2}}={\frac {a\pm {\sqrt {a^{2}-4a+4}}}{2}}={\frac {a\pm {\sqrt {(a-2)^{2}}}}{2}}={\frac {a\pm (a-2)}{2}}$

$\lambda$ $=$ $1$, $a$ $-$ $1$

we knew about the eigenvalue of $1$ already,

and so the other eigenvalue of $A_{1}$ is $a$ $-$ $1$ or, equivalently, $-b$