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What is the difference between solving $\det(xI- A) = 0$ and $\det(A-\lambda I) = 0$ to find eigenvalues of a Matrix $A$?

  1. Is the only difference that the first equation will give you the characteristic polynomial?
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2 Answers

up vote 3 down vote accepted

Both are the same. There is no difference since $$\det(xI - A) = (-1)^{n} \det(A-xI)$$ where $n$ is the size of your square matrix $A$.

Typically you would prefer to have the leading coefficient i.e. the coefficient of $x^n$ in your characteristic polynomial to be positive, in which case $\det(xI - A)$ will always give you a polynomial with the leading coefficient positive, whereas $\det(A - xI)$ will have its leading coefficient positive when $n$ is even and the leading coefficient will be negative when $n$ is odd.

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Solving both are the one and the same. If you consider both $\lambda$ and $x$ as variables, both will give you the characteristic equation. Because, for any matrix $B$, if $\det{(B)}=0$, then $\det{(-1*B)}=0$.

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