Question

A is 2x2 matrix and its eigenvalue mu with eigenvector v. P(A)= x^2+ 5x. How do we show that v is the eigenvector and what is its related eigenvalue?

Answer 1

\[A\vec{v}=\mu \vec{v}\]

Answer 2

do we not do anything with P(A)? like with it being equal to x^2+5x?

Answer 3

Well, need to get something useable out of it. Just pointing out one of the ways to prove an eigenvalue/eigenvector.

Answer 4

I do not think this is very explanatory..

Answer 5

Well, what does \(P(A)= x^2+ 5x.\) tell you about A?

Answer 6

Thats what I do not get... I do not know how it is connected to the question and how to use the equation... bec I know the general formula to descrive P(A)..

Answer 7

P(A)= A^2+5A..

Answer 8

Yah, I'm not seeing a reference to that in what I have. Sorry I do not think I can help clarify it. Which I could!

Answer 9

I am making a correction-- P(A)= A^2+5A not to x^2+5x

Answer 10

@KingGeorge

Answer 11

Well, if P(x)=x^2+5x, then the roots of this are the eigenvalues. Since \(x^2+5x=x(x+5)\), the eigenvalues are \(0,-5\).

At least that's what it is if I remember my definitions of P(A) correctly.

Answer 12

Do you have actual numbers for A and v?

Answer 13

That is what I suspected, opened my book.... and found no reference. Looked online, found no reference... but I have learned that Linear is not as rigorously standard as say Algebra 1, so there is more room for different notations, etc.

Answer 14

no we do not.. it is just asking How do we show that v is the eigenvector...

Answer 15

how you found eigenvalue made so much sense.. but just is there anyway that I am supposed to connect it to showing that v is the eigenvector?

Answer 16

Without knowing any values for v or A, I don't know how to continue.

Answer 17

\[A\vec{v}=\mu \vec{v}\implies A\vec{v}=0 \vec{v} \text{ or }A\vec{v}=-5 \vec{v}\]but it relies on having other values.... as KG points out.

Answer 18

Found a reference to the characteristic polynomial that I had not seen before. It is related to the Cayley–Hamilton theorem.
\[p(A)=A^2-(a+d)A+(ad-bc)I_2=\left[ \begin{matrix} 0&0\\0&0 \end{matrix} \right]\]

Answer 19

With \(P(A)= x^2+ 5x\) the determinant would be 0 if I match it up with that form. And \(a+d=-5\). Don't know if that gets it any furthur.

Answer 20

That's a good observation. But a+d=5, not -5.

Answer 21

Note the negative in front.

Answer 22

Oh. Right. I accounted for one too many negatives.

Answer 23

\[A^2=
\left[ \begin{matrix} a&b\\c&d \end{matrix} \right]
\left[ \begin{matrix} a&b\\c&d \end{matrix} \right]\\
\implies A^2=
\left[ \begin{matrix} a^2+bc&ab+bd\\ca+dc&cb+d^2\end{matrix} \right]\\
\implies A^2=
\left[ \begin{matrix} a^2+bc&-5b\\-5c&cb+d^2\end{matrix} \right]\]

Answer 24

And the determinant of that should still be 0.

Answer 25

Then if we add 5A, and use the Cayley-Hamilton theorem, we also find that \(a^2+5a+bc=d^2+5d+bc=0\).

Answer 26

\((a^2+bc)(cb+d^2)-25=0\)

Answer 27

So two equations that both =0. Too bad Melisac went offline. I think we finally got somewhere. Not far, but something.

Answer 28

If we assume A is diagonal we could get a little farther, but I'm not sure how that would change things.

Answer 29

det[A]=0... does that preclude diagonalizable?

Answer 30

It has to be at least linearly dependent or a row/col of 0 or something like that.

Answer 31

Oops. I meant upper/lower-triangular. But if that's the case, then at least one row or one column must be 0.

Answer 32

But in any case, since A has distinct eigenvalues, it should be diagonalizable to \[\begin{bmatrix}0&0\\0&-5\end{bmatrix}.\](This would be the Jordan canonical form).

Answer 33

AH HA! Yah, you came to what I was just thinking! There has to be a -5 in either a or d, it can not have a value in both!

Answer 34

I don't know if that's necessarily true yet, but it's certainly similar to such a matrix, so we might be able to work with that.

Answer 35

We know it is 2x2 and it has ... ah, yah... they could be a multiple... so one possibility, if they are 0 in one row or colume would mean -5 in only one, but the other possibility, say row 2 is 1/2 row a, leaves that open.

Answer 36

For example \[\begin{bmatrix}1&3\\-2&-6\end{bmatrix}\]would be such a matrix. However, the column vectors are not linearly independent.

Answer 37

yah.\[\left[ \begin{matrix} -10&10\\-5&5 \end{matrix} \right]\]Looks possible too.So lots of options.

Answer 38

Well, this has been interesting, but I've got to sleep now.

Answer 39

yah, we had a good stab at it. I hope Melisac finds something useful in here.