Question

Eigenvalues and eigenvectors problem?

Answer 1

This is the answer key, but I am having problems understanding it

Answer 2

How can you immediately tell that the eigenvalues are 1,2,3 without having to calculate det(A-xI)?

Answer 3

Look... linear algebra isn't my cup of tea, but just because it's unreasonably hard for me doesn't mean it has to be for you too...

Yes, I can tell the eigenvalues near-instantly, because it's a(n upper) triangular matrix, and the determinant of an upper triangular matrix is just the product of the diagonal entries...

Just you try calculating that determinant... but let me demonstrate anyway ^_^
\[\Large \left[\begin{matrix}1&4&-16\\0&2&3\\0&0&3\end{matrix}\right]-\left[\begin{matrix}x&0&0\\0&x&0\\0&0&x\end{matrix}\right]\]

Let's subtract...

Answer 4

We get:
\[\Large \left[\begin{matrix}1-x&4&-16\\0&2-x &3\\0&0&3-x\end{matrix}\right]\]
And the determinant of this matrix is
\[\Large (1-x)(2-x)(3-x)\]
Setting this to be equal to zero (to find the roots, and therefore the eigenvalues), can you now see why it's relatively straightforward, why the eigenvalues are 1,2,and 3?

Answer 5

Oh my gosh, thanks so much! I get it, thanks! However, I don't get the next part about eigenvectors. When I try calculating the eigenvector based on the eigenvalue 1, I somehow do not end up with the same answer as the answer key. Besides, the answer key seems to use a method I do not recognise involving i, j, k unit vectors?

Answer 6

I do not recognise it neither.

Want to show me how you got your eigenvectors?

Answer 7

I take it you tried this:
\[\Large \left[\begin{matrix}1&4&-16\\0&2&3\\0&0&3\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right]=\left[\begin{matrix}a\\b\\c\end{matrix}\right]\]
And then just solved the system?

Answer 8

Okay, I got it, thanks! It turns out I made an arithmetic error. However, for the eigenvalue 2, I can't seem to get why the 'b' value is 1?

Answer 9

Let's try...
\[\Large \left[\begin{matrix}1&4&-16\\0&2&3\\0&0&3\end{matrix}\right]\left[\begin{matrix}a\\b\\c\end{matrix}\right]=\left[\begin{matrix}2a\\2b\\2c\end{matrix}\right]\]

This is it, right?

Answer 10

Yes, so the equations would be:
a+4b-16c = 2a ---> a-4b+16c = 0
2b+3c = 2b ---> 3c = 0
3c = 2c
That would mean c = 0, and then how can I get the b?

Answer 11

Magic. LOL
Let me see... how do I put this...

Answer 12

Okay, from the second equation, that is...
2b + 3c = 2b

It doesn't seem very helpful, yes?
All it tells you is that
b = b.

Right?

Answer 13

Yes, doesn't that mean all values of b will satisfy the equation?

Answer 14

Not quite.
What about the first equation?
a + 4b - 16c = 2a

c = 0, so
a + 4b = 2a

Or in other words...
a = 4b

Catch me so far?

Answer 15

Hmm, so why can't b be 0? That would mean a = 0, too and \[\left[\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right]\] would be an eigenvector? I'm confused....

Answer 16

Zero vectors are never eigenvectors, *dear* (LOL, can I call you something more appropriate, like your nickname, perhaps?)

Because... any matrix times the zero vector always gives the zero vector, it's kind of a freak-case, really :D

Answer 17

Ohhh, okay I get it! Thanks so much, you're a savior!
Hmm, what about the A^n part? Does this have anything to do at all with diagonalisation?

Answer 18

Hang on, do you understand why b = 1 now?

Answer 19

Any scalar multiple of an eigenvector is also an eigenvector, right?

So it's almost like b can take any value, and it can, but the side-effect is that a must ALWAYS be four times the value of b.

Answer 20

Oh, and I'm kind of at my limit when it comes to diagonalizable matrices and stuff... better get a professional for that :D
(sorry T.T)

Answer 21

Okay, thanks a lot! That really helped me!

Answer 22

No problem. Sorry I couldn't be of more help... but look around, the expert's probably just around the corner ^_^.

Answer 23

***
Hmm, what about the A^n part? Does this have anything to do at all with diagonalisation?
***

the diagonalization is
\[A= PDP ^{-1}\]
where P is the matrix whose columns are the eigenvectors
and D is a diagonal matrix whose entries have the corresponding eigenvalues

Answer 24

notice that
\[ A^2= A\cdot A = \\= PD(P ^{-1}P)DP ^{-1}\\= PDIDP ^{-1} = PDDP ^{-1} \\= PD^2P^{-1}\]

Answer 25

or in general
\[ A^n = PD^nP^{-1} \]

Answer 26

Thanks so much! I get it now!