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I'll write it again but the users below don't seem to understand ... THIS IS NOT A DUPLICATE . PLEASE READ MY QUESTION THOROUGLY.

I need to find a matrix with characteristic equation given by

λ² − λ − 1 = 0

I know that Since the polynomial p(λ) = λ² − λ − 1 has degree 2,so we look for a 2 × 2 matrix.

I let A be any 2 × 2 matrix, where a, b, c, d ∈ R.

A=$$ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $$

The characteristic equation of A is:

p(λ) = det(A − λId) = λ² − (a + d)λ + ad − bc

And here is what I don't get . On the answer sheet is says that this polynomial coincides with λ² − λ − 1 if and only if:

−(a + d) = −1 and ad − bc = −1.

I have dyscalculia and even the things that might seem simple are hard for me to understand. Can somebody please explain where they are getting

−(a + d) = −1 and ad − bc = −1

and why , for example, they are not setting λ²=-1? What does that mean?

Please be as simple and clear as possible! I really appreciate the help ! Thanks guys

Edit: please do not mark this as a duplicate. This is a UNIQUE question in which i am trying to understand ONE of the stages to determine the characteristic equation .

BM97
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  • Look at https://en.wikibooks.org/wiki/Control_Systems/Standard_Forms#Observable_Canonical_Form for example. – copper.hat Dec 10 '18 at 17:11
  • No ! It is not a duplicate ! I am asking a specific question reguarding my problem ! please check thoroughly both questions before writing down random things. – BM97 Dec 10 '18 at 17:20
  • Yes i have, and no one appears to answer my question . All the solutions i've seen on this site take a different approach on answering the question . I have one type of method on answering the question which i really haven't seen here and i would really really like it if one of you guys could help me out ? I mean you the time to mark my question as a duplicate but not answer it? – BM97 Dec 10 '18 at 17:24
  • I mean is is that hard to answer? – BM97 Dec 10 '18 at 17:32
  • Is it possible...just possible ... that the highly-experienced users here are more adept at recognizing a duplicate than you are? Also: I think you may have forgotten exactly who cares about this question. One answer: not the people you insult as having offered you insufficient or inadequate free help. You ask "Is it that hard to answer?" The answer (in my case) is "No, it's not that hard. But I'm choosing not to." – John Hughes Dec 10 '18 at 18:01
  • Check below :)) It wasn't that hard after all – BM97 Dec 10 '18 at 18:14

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You have $$\lambda^2-\lambda-1=\lambda^2\color{red}{-1}\cdot\lambda+\color{blue}{(-1)},$$ you agree? Now the number which is multiplied by $\lambda$ is $\color{red}{-1}$, the number which is just added is $\color{blue}{-1}$.

In $$\lambda^2\color{red}{-(a+d)}\cdot\lambda+ \color{blue}{ad-bc}$$ the number which is multiplied by $\lambda$ is $\color{red}{-(a+d)}$ and the number which is just added is $\color{blue}{ad-bc}$.

From here $$\color{red}{-1}=\color{red}{-(a+d)}\qquad\text{and}\quad\color{blue}{-1}=\color{blue}{ad-bc}. $$

Michael Hoppe
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