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Either works when trying to find the eigenvalues, but only the former works when trying to find corresponding eigenvectors. I can understand how it makes a difference, but what I don't understand how one is supposed to "know" the former is the "correct" form, since it starts from here:

$AX = λX$

And from there, you can end up with either:

$AX - λX = 0$ or $λX - AX = 0$

And finally:

$(A-λI)X = 0$ or $(λI-A)X = 0$

And so should they not both be correct? Furthermore, once it's in that form, can you not multiply both sides by $-1$ to flip them?

I guess I'm overlooking some sort of rule of algebra when dealing with matrices. Let me know. Thanks.

EDIT

Here is an example:

$\pmatrix{5&3\\6&2}$ has an eigenvalue $-4$

If we do $(A-λI)X = 0$: $\pmatrix{9&3\\6&2}*X = 0$

This solves to $X = \pmatrix{-1\\3}$ (and multiples of it)

If we do $(λI-A)X = 0$: $\pmatrix{-9&3\\6&-2}*X = 0$

This solves to $X = \pmatrix{1\\3}$ (and multiples of it)

  • 0
    -4 is not an eigen value of the given matrix. The eigen values are 8, -1.2015-03-21

2 Answers 2

1

See this(explanation of eigenvalues and eigenvectors). Both equations should give you the corresponding eigenvectors. Perhaps your book just mentioned the first equation and you assumed that it couldn't be of the second form as well.

1

You usally use $(\lambda I-A)X$ to calculate the Eigenvalues because the polynom you will get is normalized in this case ( its highest degree coefficient is 1).

For Eigenvektors it totally doesn't matter which way you do it as you have

$(\lambda I-\vec A)\vec v=\vec0 \Leftrightarrow -(\lambda I-A)\vec v=\vec 0 \Leftrightarrow (A-\lambda I)\vec v=\vec 0$

  • 0
    Fine, I removed it because I agree that it could cause confusion :-)2011-02-03