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I have $(\lambda - a)(\lambda - d)-bc = 0$ which can also be written as $\lambda^2 - \lambda(a+d) + (ad - bc) = 0$. This quadratic equation can be solved by finding the roots $\lambda_1$ and $\lambda_2$. With the roots, I have been told that $\lambda^2 - \lambda(a+d) + (ad - bc) = 0$ can be written as $(\lambda - \lambda_1)(\lambda - \lambda_2) = 0$ - why?

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    $\begingroup$ You said $(\lambda - a)(\lambda - d)-bc = 0$ can be written as $\lambda^2 - \lambda(a+b) + (ad - bc) = 0$ but shouldn't it be $-\lambda(a+d)$? $\endgroup$ – ArsenBerk Jun 6 '19 at 11:36
  • $\begingroup$ Yes, of course. Thanks! $\endgroup$ – Fac Pam Jun 6 '19 at 11:43
  • $\begingroup$ Is your question related to chromatic numbers? Because you are used $\lambda$ as variable. $\endgroup$ – BarzanHayati Jun 6 '19 at 12:48
  • $\begingroup$ It's obvious, you could write an equation with two roots in two forms. So your question is not clear. $\endgroup$ – BarzanHayati Jun 6 '19 at 12:50
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If $x=\alpha$ is a root of a polynomial $p(x)$, you can always write $p(x)$ as $$ p(x) = (x-\alpha)\cdot q(x), $$ where $q(x)$ is a polynomial with degree $\deg(q)=\deg(p)-1$.

In your example $p(\lambda) = (\lambda - a)(\lambda - d)-bc$ is a polynomial of degree $2$. Since it has roots $\lambda_1$, $\lambda_2$ you can write is as $$ p(\lambda) = (\lambda-\lambda_1)(\lambda-\lambda_2)\cdot q(\lambda) $$ where $q(\lambda)$ has degree $0$, that is, is a constant. However, since the coefficient of $\lambda^2$ in $p(\lambda)$ is $1$ we must have $q(\lambda)=1$ so that $$ p(\lambda) = (\lambda-\lambda_1)(\lambda-\lambda_2). $$

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Hint: Expanding $$(\lambda-a)(\lambda-d)-bc=0$$ we get $$\lambda^2-a\lambda-d\lambda+ad-bc=0$$ this can be written as $$\lambda-\lambda(a+d)+ad-bc=0$$ using the quadratic formula we obtain $$\lambda_{1,2}=\frac{a+d}{2}\pm\frac{1}{2}\sqrt{(a-d)^2+4bc}$$ now you can write $$(\lambda-\lambda_1)(\lambda-\lambda_2)=0$$ It is the same like above, called the theorem of Vieta.

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  • $\begingroup$ What does this have to do with the roots $\lambda_1$ and $\lambda_2$? $\endgroup$ – Fac Pam Jun 6 '19 at 11:31
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    $\begingroup$ I still do not understand why you can get from $$\lambda_{1,2}=\frac{a+d}{2}\pm\frac{1}{2}\sqrt{(a-d)^2+4bc}$$ to $$(\lambda-\lambda_1)(\lambda-\lambda_2)=0$$ $\endgroup$ – Fac Pam Jun 6 '19 at 11:39
  • $\begingroup$ @FacPam See the BiFactor Theorem. and the higher degree generalization of the Factor Theorem that I prove there. $\endgroup$ – Bill Dubuque Jun 6 '19 at 17:04
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Referring to Vieta's theorem: $$\lambda^2 - \lambda(a+b) + (ad - bc) = 0 \Rightarrow \begin{cases}\lambda_1+\lambda_2=a+b\\ \lambda_1\lambda_2=ad-bc\end{cases}$$ and: $$(\lambda - \lambda_1)(\lambda - \lambda_2) = 0 \Rightarrow \lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2=0$$ Do you see the relationship?

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Have a close look at the polynomial $p_α(λ) = α(\lambda - λ_1)(\lambda - λ_2)$, $α\ne0$. Its roots are really obvious: $p_α(λ) = 0$ if and only if at least one of the factors $λ-λ_1$ and $λ-λ_2$ is zero. So this polynomials have the same roots as your polynomial $p(λ) = (\lambda - a)(\lambda - d)-bc$. The correct value of $α$ is the leading coefficient of $p(λ) = 1\cdotλ^2−λ(a+d)+(ad−bc)$, that is $α=1$.

When you really want to understand the mathematical background why this factorization works you need the Fundamental Theorem of Algebra but this may be beyond your skills.

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