- I want to 'implement' the zariski based proof for cayley hamilton in SAGE and show that it works by checking the computations scheme-theoretically.
- Let's work through the proof by hand. Take a 2x2 matrix
`[a, b; c, d]`

. - The charpoly is
`|[a-l; b; c; d-l]| = 0`

, which is`p(l) = (a-l)(d-l) - bc = 0`

- This simplified is
`p(l) = l^2 - (a + d) l + ad - bc = 0`

. - Now, let's plug in
`l = [a; b; c; d]`

to get the matrix eqn -
`[a;b;c;d]^2 - (a + d)[a;b;c;d] + [ad - bc; 0; 0; ad - bc] = 0`

. - The square is going to be
`[a^2 +]`

- Let
`X`

be the set of`(a, b, c, d)`

such that the matrices`[a;b;c;d]`

satisfy their only charpoly. - Consider the subset
`U`

of the set`(a, b, c, d)`

such that the matrix`[a;b;c;d]`

has distinct eigenvalues. - For any matrix with distinct eigenvalues, it is easy to show that they satisfy their charpoly.
- First see that diagonal matrices satisfy their charpoly by direct computation:
`[a;0;0;b]`

has eigenvalues`(a, b)`

. Charpoly is`l^2 - l(a + b) + ab`

. Plugging in the matrix, we get`[a^2;0;0;b^2] - [a(a+b);0;0;b(a+b)] + [ab;0;0;ab]`

which cancels out to`0`

. - Then note that similar matrices have equal charpoly, so start with
`|(λI - VAV')| = 0`

. rewrite as`(VλIV' - VAV') = 0`

, which is`V(λI - A)V' = 0`

, which is the same`λI - A = 0`

. - Thus, this means that a matrix with distinct eigenvalues, which is similar to a diagonal matrix (by change of basis), has a charpoly that satisfies cayley hamilton.
- Thus, the set of matrices with distinct eigenvalues,
`U`

is a subset of`X`

.

- However, it is not sufficient to show that the system of equations has an infinite set of solutions.
- For example,
`xy = 0`

has infinite solutions`(x=0, y=k)`

and`(x=l, y=0)`

, but that does not mean that it is identically zero. - This is in stark contrast to the 1D case, where a polynomial
`p(x) = 0`

having infinite zeroes means that it must be the zero polynomial. - Thus, we are forced to look deeper into the structure of solution sets of polynomials, and we need to come up with the notion of irreducibility.
- See that the space
`K^4`

is irreducible, where`K`

is the field from which we draw coefficients for our matrix.

- Next, we note that
`X`

is a closed subset of`k^4`

since it's defined by the zero set of the polynomial equations. - We note that
`U`

is an open subset of`k^4`

since it's defined as theof the discriminant of the charpoly! (ie, we want non-repeated roots)*non-zero set* - Also note that
`U`

is trivially non-empty, since it has eg. all the diagonal matrices with distinct eigenvalues. - So we have a closed subset
`X`

of`k^4`

, with a non-empty open subset`U`

inside it. - But now, note that the closure of
`U`

must lie in`X`

, since`X`

is a closed set, and the closure`U`

of the subset of a closed set must lie in`X`

. - Then see that since the space is irreducible, the closure of
`U`

(an open) must be the whole space. - This means that all matrices satisfy cayley hamilton!