5
$\begingroup$

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)

  • 1
    Uhm... Can you give an example when one of the equations work but not the other?2011-02-03
  • 3
    Joey: Your reasoning about their equivalence is correct; they are.2011-02-03
  • 0
    What makes you think that $(\lambda I - A)x=0$ doesn't work? Are you responding to something that someone told you, or are you reporting that when you tried solving the equation it somehow didn't give you eigenvectors?2011-02-03
  • 0
    I have edited my post with an example2011-02-03
  • 4
    In your example both your equations involving $A - \lambda I$ and $\lambda I - A$ contain errors (careful with the signs and subtractions!).2011-02-03
  • 0
    Oh I see. I forgot to make the rest of A negative. Thanks!2011-02-03
  • 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
    Your last sentence does not make sense. You could wish to erase it, since the preceding part of your answer is correct and to the point.2011-02-03
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    I tried to edit it to make more sense. What I want to say is that the vector $v'$ you get when you swap is exactly the negative of the eigenvector $v$ you get by doing it the other way around.2011-02-03
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    Which is meaningless and exactly why I think you should delete this sentence. If $v$ is an eigenvector in the sense that $(\lambda I -A)v=0$ (1), then $v$ is also an eigenvector in the sense that $(A-\lambda I)v=0$ (2), and vice versa! To introduce a new vector $v'$ to replace $v$, such as $v'=-v$, as if $v$ was OK for (1) but not for (2) and, presumably, $v'$ was OK for (2) but not for (1), just confuses things. Linearity is a powerful ally, my friend...2011-02-03
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    Fine, I removed it because I agree that it could cause confusion :-)2011-02-03