Try
Apply the matrix J with rows (0,−1),(1,0) to a coordinate pair: J(x,y)=(−y,x). Over the real numbers this is a quarter-turn. That geometric description does not supply a Euclidean angle for vectors over a finite field. Here we ask an algebraic question: is there a nonzero vector whose image is a scalar multiple of itself?
Such a vector v is an eigenvector, and its scalar λ is an eigenvalue: Jv=λv. Both must use the field currently named. Reducing the same integer entries modulo 3, 5 or 2 gives different scalar systems. Before calculating, predict whether an eigenvector over one of these fields must remain an eigenvector over the others. Reduction between different prime characteristics is not an inclusion of fields.
For a refresher, return to eigenvalues and eigenspaces. No complex-number geometry is required for the calculations below.
Predict
Write the equations −y=λx and x=λy. If y=0, the second equation forces x=0, which is not an eigenvector. Otherwise substitute x=λy into the first equation and divide by the nonzero y. The necessary condition is λ²=−1. It is also sufficient: whenever λ²=−1, the vector (λ,1) satisfies J(λ,1)=(−1,λ)=λ(λ,1).
The determinant calculation gives the same condition: det(xI−J)=x²+1. In this polynomial, x is a formal variable, not a selected vector coordinate. List the squares in F3 and F5 yourself before opening the comparison.
Compare the square tables
In F3 the squares of 0,1,2 are 0,1,1; none equals −1=2. In F5 the squares of 0,1,2,3,4 are 0,1,4,4,1, so the roots are 2 and 3. In F2, −1=1 and the only root is 1. The number of roots, not the appearance of the integer matrix, controls this calculation.
Experiment
Complete this finite arithmetic investigation on paper. Multiply each proposed vector by J in the stated field, then compare with λv. Every displayed equality is a field equality, so reduce both coordinates after multiplication.
| Field | Proposed λ and v | Jv | λv | Conclusion |
|---|
| F3 | λ=1, v=(1,1) | (2,1) | (1,1) | Not an eigenvector for λ=1 |
| F5 | λ=2, v=(2,1) | (4,2) | (4,2) | Eigenvector |
| F5 | λ=3, v=(3,1) | (4,3) | (4,3) | Eigenvector |
| F2 | λ=1, v=(1,1) | (1,1) | (1,1) | One eigenvector direction |
Over F5 put the two successful vectors in the columns of P. Its determinant is 2−3=4, a nonzero field element, so those columns form a basis. The equation JP=P diag(2,3) now proves P⁻¹JP=diag(2,3). Changing coordinates has converted the operator into two independent scalar multiplications.
Do not infer the same result over F2 merely because its characteristic polynomial splits. There J−I has two identical rows (1,1), so its rank is one and its kernel is the line spanned by (1,1). There is only one independent eigenvector direction, while a basis of the plane needs two. An explicit generalized vector makes the failure visible: for w=(1,0) and v=(1,1), (J−I)w=v and (J−I)v=0. In the ordered basis v,w, J has columns (1,0),(1,1), a non-diagonal upper triangular matrix.
Understand
Why are distinct eigenvalues enough? Suppose eigenvectors with distinct eigenvalues had a nontrivial linear dependence. Choose one with the fewest participating vectors, all coefficients nonzero. A single nonzero vector cannot be dependent, so there are at least two. Apply A−λI, where λ belongs to the last participating vector. Its term disappears. Every other coefficient is multiplied by a nonzero difference of eigenvalues, leaving a shorter nontrivial dependence: a contradiction. Thus eigenvectors for distinct eigenvalues are independent over any field.
If the characteristic polynomial of an n-dimensional operator has n distinct roots in its field, each root makes A−λI singular and supplies a nonzero eigenvector. Independence gives an eigenbasis. This is a sufficient criterion, not a characterization of every diagonalizable matrix. The identity has a repeated root and is already diagonal; a nonzero off-diagonal Jordan block can have the same polynomial without enough eigenvectors.
We can retain characteristic three while adding new scalars. Define F9 as the quotient F3[t]/(t²+1), and write i for the class of t. Every element has a unique form a+bi with a,b in F3, and i²=−1. Quotient arithmetic gives associative addition and multiplication. To verify this quotient is a field rather than assuming it, compute
(a+bi)(a−bi)=a²+b².
If exactly one of a,b is nonzero, the right side is 1; if both are nonzero it is 2. Thus it never vanishes for a nonzero element. Its inverse is (a−bi)/(a²+b²). We have constructed a nine-element field containing F3 as the elements with b=0.
Over this extension J has roots i and −i and eigenvectors (i,1),(−i,1). Their determinant 2i is nonzero, so they diagonalize J. The old claim was “no eigenvectors with coordinates in F3”; it did not prohibit vectors with new coordinates. This is extension of scalars. Moving to F5 instead changes characteristic: no unital field embedding F3→F5 can preserve 1+1+1=0 because that sum is nonzero in F5.
Apply
Investigate a changed matrix U with rows (1,1),(0,1) over F3. Its characteristic polynomial is (x−1)², while its eigenvector equation forces the second coordinate to be zero. Therefore U is not diagonalizable over F3. Extending to any field containing F3 still leaves the same rank-one equation, so adding roots cannot fix this particular obstruction: the roots were present already.
Write N=U−I. Since N²=0, multiplication gives U²=I+2N and U³=I+3N=I in characteristic three. Check the first two powers are not I. This operator has finite order three but no eigenbasis. Consequently “some positive power equals I” is not by itself a valid diagonalization argument in positive characteristic.
For a second changed action, count rather than solve again. The F5 plane has 24 nonzero vectors, partitioned into six lines of four nonzero vectors each. Exactly two of these lines are J-invariant, because an invariant line must be an eigenline. In the F9 plane there are ten lines, each containing eight nonzero vectors; J still has only two eigenlines. Explain why sixteen eigenvectors in F9 do not mean sixteen eigenvalues. This links scalar arithmetic to the counting lesson without treating all vector directions as interchangeable.
Continue
For every spectral claim record the field, the roots that belong to it, the dimensions of their eigenspaces, and the invertibility of any proposed eigenvector matrix. A characteristic polynomial alone does not settle the repeated-root case. Distinguish changing the characteristic, adjoining roots within the same characteristic, and merely changing coordinates over a fixed field.
For more extension-field arithmetic, return to finite fields. The present result needs only the explicit nine-element construction above; a general classification of finite fields is additional study.
