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I'm not sure how many of you are familiar with Khan Academy, but the videos on their website have been a tremendous help in guiding me through my Linear Algebra course. Anyways, I had a question because there seems to be a fundamental contradiction in the lessons on the website. (One in particular is here: http://www.khanacademy.org/video/linear-algebra--example-solving-for-the-eigenvalues-of-a-2x2-matrix?playlist=Linear%20Algebra) The author of the video states that $\lambda$ is an eigenvalue of A if and only if $det(\lambda I_n - A) = 0$. However, my linear algebra textbook as well as Wikipedia both state that the eigenvalues of A are the solutions $\lambda$ to the equation $det(A - \lambda I_n) = 0$. As far as I'm aware, these two equations are not equivalent. I know it's always possible that Khan Academy may have made a mistake, but this seems unlikely because the instructor has used this equation in many different examples in many different videos.

Is there any case where either equation may be necessary? Is this just a mistake on behalf of Khan Academy? I'm a little confused.

Thanks!

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I am not familar with Khan Academy. But $det(\lambda I_n -A)=0$ is equivalent to $det(A-\lambda I_n)=0$ because $det(\lambda I_n -A)=(-1)^n det(A-\lambda I_n)$

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  • $\begingroup$ Ah okay. As expected, it is because of my limited understanding of determinants. Thanks! $\endgroup$ – dkaranovich Apr 11 '11 at 8:08
  • $\begingroup$ More over, the $\det (\lambda I -A)$ saves a $(-1)^n$ factor in most of the formulas. (Note: I am not of the Khan Academy). $\endgroup$ – AlainD May 30 '13 at 19:30

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