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There's something about the determinant which I don't get, so if someone could explain step-by-step how this is done, it would be much appreciated:

It's about finding eigenvalues via. the characteristic polynomial

  • Example $$K_A(\lambda)=A-\lambda E=\left( \begin{matrix}4 && -2 \\ 3 && -1\end{matrix} \right)- \left(\begin{matrix}\lambda && 0 \\ 0 && \lambda \end{matrix} \right)=\left( \begin{matrix} 4-\lambda && -2 \\ 3 && -1-\lambda \end{matrix} \right)$$

We then calculate: $K_A(\lambda)=det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]$

I don't get how you deduce that $det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\cdot(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$

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1 Answer 1

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You have a wrong minus sign:

$det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\color {red}-(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$

the correct determinat is

$det\big[\left( \begin{matrix}4-\lambda && -2 \\ 3 && -1-\lambda\end{matrix} \right) \big]=(4-\lambda)\cdot(-1-\lambda)-(-2)\cdot 3=\lambda^2-3\lambda+2$

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  • $\begingroup$ Ah yes I see - Thanks for correcting! $\endgroup$
    – Alex5207
    Nov 15, 2017 at 9:15
  • $\begingroup$ You are welcome! And after the correction what is your problem? Do you know the definition of determinant? en.wikipedia.org/wiki/Determinant#2_.C3.97_2_matrices $\endgroup$ Nov 15, 2017 at 9:30
  • $\begingroup$ Great! I totally forgot about the $$det\big[ \begin{matrix} a && b \\ c && d \end{matrix} \big] = ad - bc$$ Easy to see that's what happening here. Thank you! $\endgroup$
    – Alex5207
    Nov 15, 2017 at 9:44

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