What are the eigenvalues of this matrix?

Question

empty · Answer

$$\begin{pmatrix}
3 &1 \ -1
 &3 
\end{pmatrix}$$

and 

$$\begin{pmatrix}
3 &1 \0 
 &3 
\end{pmatrix}$$

parthkohli · Answer

Cayley-Hamilton?$$|A - \lambda I |=0$$

parthkohli · Answer

Unless I'm missing something and those parentheses mean something else.

empty · Answer

I am sorta not interested in the answer to the first one, I'm using it to warm you up to the idea that the second supposedly defective matrix has two unique eigenvalues.

parthkohli · Answer

Ok, so I'm getting $\lambda = 3$ using the Cayley-Hamilton Theorem.$$\left | \begin{array}{l} { 3-\lambda }& 1 \ 0 & 3- \lambda  \end{array}\right | = (3 - \lambda)^2 =0$$

parthkohli · Answer

I really have zero idea about linear algebra. Sorry. :(

empty · Answer

Yeah, the first matrix has 2 complex eigenvalues and this somewhat similar one has only 1 eigenvalue? Here I claim that it actually has 2 eigenvalues after all, they're just not real nor complex, they're part of dual numbers. https://en.wikipedia.org/wiki/Dual_number

imqwerty · Answer

$$A=\left[\begin{matrix}3 & 1 \ 0 & 3\end{matrix}\right]$$
now $$determinant(A-\lambda I)=0$$
so $$\left[\begin{matrix}3-\lambda & 1 \ 0 & 3-\lambda\end{matrix}\right]=0$$$$(3-\lambda)^2=0$$so λ=3 is our eigen value

Av=3v
v is the eigen vector
Av=3v=>(A-6I)v=0
$$\left[\begin{matrix}3-3 & 1 \ 0 & 3-3\end{matrix}\right]\left(\begin{matrix}v_{1} \ v_{2}\end{matrix}\right)=0$$
$$0v_{1}+1v_{2}=0$$$$v_{2}=0$$and
$$0v_{1}+0v_{2}=0$$so v1 can be anything..
$$v=\left(\begin{matrix}v_{1} \ v_{2}\end{matrix}\right)$$v2=0
so v=1

imqwerty · Answer

wait we wanted eigen value? so it wuld be 3
there ws no need of that eigen vector..

imqwerty · Answer

but how is the second one having 2 values?

parthkohli · Answer

Exactly, how?

empty · Answer

The problem is that in the first matrix with complex eigenvalues we recognized that -1 has two square roots, however in the second matrix we didn't recognize that 0 has multiple roots in the dual numbers.

So just like in complex numbers we have $i^2=-1$ in the dual numbers we have $\varepsilon^2=0$

The charcteristic equation then becomes:

$$(3-\lambda)^2 = 0$$
$$3-\lambda = \pm k \varepsilon$$
where k is just some arbitrary real number to be determined.

So we have two eigenvalues:

$$\lambda_1 = 3+k\varepsilon$$
$$\lambda_2 = 3-k\varepsilon$$

See, and if we say wait a second what's happening here? Remember, the determinant of the matrix is the product of eigenvalues correct?

$$det(A)=9 = \lambda_1 \lambda_2 = (3+k\varepsilon)(3-k\varepsilon) = 3*3+k\varepsilon 3-k\varepsilon 3-k^2\varepsilon^2 = 9$$ since the middle terms cancel out by addition and the last one is 0 by multiplication.

What about the trace, that should also be the sum of eigenvalues, even easier:

$$tr(A) = 6 = \lambda_1+\lambda_2 = 3+k\varepsilon +3-k\varepsilon =6 $$

So if you're ok with complex eigenvalues then I see no reason why you can't be ok with dual eigenvalues as well, so we don't have to call them defective matrices anymore. :P

anonymous · Answer

if this were a matrix over the ring of dual numbers, then sure, but it's not so instead we just talk about multiplicity

anonymous · Answer

we like complex eigenvalues because the problem of guaranteeing a diagonalizable matrix has n eigenvalues is equivalent to the fundamental theorem of algebra by the fact that the determinant is precisely the polynomial that encodes the eigenvalues as roots

anonymous · Answer

the dual numbers are not a field let alone are they algebraically closed, so we're not interested in using them in place of the complex numbers for studying real matrices

anonymous · Answer

and as a side note, in the dual numbers we have *three* roots of zero: $$\varepsilon^2=(-\varepsilon)^2=0^2=0$$ why should we privilege $\pm\varepsilon$ here over $0$?

empty · Answer

Yeah, there are really going to be infinitely many since any real number could by multiplying $\varepsilon$. I think the reason to favor the + or - values is because really 0 can be seen as a special case of picking $k\varepsilon$ and $-k\varepsilon$ when k=0 they're I guess degenerate if that's the right word?

To be fair, I don't think dual numbers are the way to go, if anything it seems to me that complex numbers and dual numbers could be brought together under one roof,

$$\varepsilon = \begin{bmatrix}
0 &1 \ 
0 &0 
\end{bmatrix}$$

We're still able to deal with complex numbers as we have before since we can represent $i=\varepsilon - \varepsilon'$. 

Or will this not work for some reason?

anonymous · Answer

well, yes, we can put them *all* under one-roof using 2x2 real matrices, since this gives includes the (i.e. up to isomorphism) the complex, dual, and split-complex numbers as subrings

empty · Answer

@oldrin.bataku Cool, so this seems strange, or perhaps not quite right. This would seem to mean that all nxn matrices have a complete spectrum as long as you allow your numbers to also be essentially nxn matrices themselves. Where could I read more about this sort of stuff, can you give me any names of books or authors of this sort of stuff? It seems like at this point we're using matrices as entries of a matrix, which sorta makes stuff like the Cayley-Hamilton theorem maybe make more sense I'm not sure.