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The equation $(A−λI)x=0$ has a nontrivial solution (a solution where $x≠0$) if and only if $det(A−λI)=0$.

Why is that?

How can this be proven?

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    $\begingroup$ Forget the $-\lambda I$. (It is a red herring.) Do you know how to show that $A x = 0$ has a non-trivial solution iff $\det A = 0$? $\endgroup$ – Kenny Wong May 19 '17 at 20:31
  • $\begingroup$ @KennyWong: I might have chosen a different letter for the matrix in the simplification, to avoid confusion between the two problems. $\endgroup$ – Hurkyl May 19 '17 at 20:33
  • $\begingroup$ @KennyWong I don't know how to show that. $\endgroup$ – octavian May 19 '17 at 20:42
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    $\begingroup$ $\det A$ being non-zero is the same as $A$ being invertible. This knowledge is usually assumed when discussing eigenvalues. $\endgroup$ – M. Winter May 19 '17 at 20:45
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    $\begingroup$ @octavian Do you already know that $\det(M) \ne 0 \iff M$ is invertible? $\endgroup$ – user137731 May 19 '17 at 20:45
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If $\det (A-\lambda I) \neq 0$, then it has an inverse and so the equation has solution $x=(A-\lambda I)^{-1} 0 = 0 $ as its only solution. So in order for any other solution to exist (a non-trivial one, that is) $A-\lambda I$ can't have an inverse. Therefore its determinant is $0$.

Reverse: If $det (A-\lambda I) = 0$ then it has less than full rank. So when you row reduce, you get at least one row of zeros. So the solution has at least one free variable. You can pick the value of the free variable as you please, specifically not $0$, and get a non-trivial solution.

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  • $\begingroup$ Got it. What about the reverse implication? $\endgroup$ – octavian May 19 '17 at 20:49
  • $\begingroup$ See edits. Word. $\endgroup$ – B. Goddard May 19 '17 at 20:53

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