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To my understanding characteristic polynomial can be expressed two different ways and when solving eigenvalues and eigenvectors should give same results. These two ways are:

$$ P_A(\lambda)=det(A-\lambda I) $$ $$ P_A(\lambda)=det(\lambda I -A) $$

So these two polynomials differ from one another by a sign $(-1)^n$.

We could test this with $2\times2$ matrix A: (because it's easy to compute)

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

$$ P_A(\lambda)=(A-\lambda I) $$ $$ P_A(\lambda)=det(\begin{bmatrix}a-\lambda & b \\ c & d - \lambda \end{bmatrix}) $$ $$ P_A(\lambda)= (a-\lambda)(d-\lambda)-bc $$ $$ ad + a(-\lambda) + (-\lambda)d + (-\lambda)(-\lambda)-bc $$ $$ \lambda^2 -\lambda(a+d) + ad - bc $$

Now we could do the same with second characteristic polynomial:

$$ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix} $$ $$ P_A(\lambda)=(\lambda I - A) $$ $$ P_A(\lambda)=det(\begin{bmatrix} \lambda - a & -b \\ -c & \lambda - d \end{bmatrix} $$ $$P_A(\lambda)=(\lambda-a)(\lambda-d)-(-b)(-c) $$ $$ P_A(\lambda)=\lambda\lambda+\lambda(-d)+(-a)\lambda + (-a)(-d)-(-b)(-c) $$ $$ P_A(\lambda)=\lambda^2-\lambda(a+d)+ad-bc$$

Now the problem is that this proof applies on 2x2 matrix but how about higher dimensions ? How do you prove this works in higher dimensions as well ? Does such situation exist when these would produce different results ?

If someone could provide some insight on this that would be much appreciated.

Thanks,

Tuki

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    $\begingroup$ You forgot to negate the off-diagonal entries. ;-) It is $\lambda I-A$, so all entries of $A$ have to multiplied by $-1$. $\endgroup$ – amsmath Oct 19 '17 at 21:18
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    $\begingroup$ To prove it in general, use the fact that $\det(cA)=c^n\det(A)$. $\endgroup$ – John Griffin Oct 19 '17 at 21:19
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    $\begingroup$ @Tuki: Have you seen Samuelson's Formula for the Characteristic Polynomial: mathworld.wolfram.com/CharacteristicPolynomial.html? Look at the 2x2, 3x3, 4x4, ... $\endgroup$ – Moo Oct 19 '17 at 21:20
  • $\begingroup$ @Moo no i am not aware of this. Thanks for providing this ! $\endgroup$ – Tuki Oct 19 '17 at 21:21
  • $\begingroup$ @amsmath i corrected this error on the post and this does not alter the outcome. $\endgroup$ – Tuki Oct 19 '17 at 21:54

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