Find spectral information that survives coordinates

Try

For A=[[3,4],[0,−2]], calculate det(tI−A)=(t−3)(t+2)=t²−t−6. The roots are its eigenvalues, their sum is the trace 1, and their product is the determinant −6. The entry 4 affects eigenvectors but not this polynomial. Change it to zero and predict which spectral information remains unchanged. This gives a first warning: the characteristic polynomial is an invariant, but it does not completely identify the operator.

Predict

Compare diag(3,3) with B=[[3,1],[0,3]]. Both have characteristic polynomial (t−3)². In the first case every vector is an eigenvector; in the second, solving (B−3I)v=0 forces the second coordinate to zero. Predict whether knowing all eigenvalues with repetitions determines diagonalizability. These two matrices show why an additional count—the dimension of each eigenspace—is necessary.

Experiment

Create a table listing matrix, characteristic polynomial, eigenspace dimension and trace for those two examples. Add the rotation R=[[0,−1],[1,0]]. Over R its polynomial t²+1 has no root and no eigenvector. Over C it has roots i and −i. Solve (R−iI)v=0 to get an eigenvector (1,−i), and check it directly. The arithmetic of the matrix has not changed, but the allowed scalar field has changed the available decompositions.

Understand

For an n-dimensional operator T, define χ_T(t)=det(tI−A), where A is any coordinate matrix. If B=S⁻¹AS, then tI−B=S⁻¹(tI−A)S, so determinant multiplicativity proves the definition is basis-independent. A scalar λ is an eigenvalue exactly when χ_T(λ)=0, because singularity of λI−A is equivalent to a nonzero kernel. Algebraic multiplicity is the root multiplicity in χ_T; geometric multiplicity is dim ker(T−λI). The latter is at most the former: start a basis with an eigenspace basis, giving a block triangular matrix whose first diagonal block is λI. Its determinant contributes that many factors t−λ.

Apply

If χ_T splits into linear factors over F, T can be triangularized over F. Choose an eigenvector, pass to the induced operator on the quotient by its line, and use induction. The quotient characteristic polynomial is a factor of χ_T by the block determinant formula, so it also splits. Lifting a quotient triangularizing basis completes the proof. Over C the fundamental theorem of algebra supplies splitting; over R it may fail. Explain why this argument produces a triangular basis but not necessarily an eigenbasis: lifted quotient eigenvectors may acquire components in the previously chosen invariant subspace.

Use the following comparison to diagnose what the polynomial omits.

Matrix	Characteristic polynomial	Dimension of the 3-eigenspace	Diagonalizable?
diag(3,3)	(t−3)²	2	Yes
[[3,1],[0,3]]	(t−3)²	1	No

Explain the final column by counting an eigenbasis, not by inspecting the repeated root alone. The off-diagonal entry changes the kernel equation even though determinant expansion does not see it.

Trace as a structural test: cyclicity and commutators

Trace is the sum of the diagonal entries of a square matrix. It is useful when a proposed matrix identity would force an impossible equality of these sums. First prove the rule rather than assuming that matrices commute.

Let A have m rows and n columns, and B have n rows and m columns, over a field F. Both AB and BA are square, although their sizes can differ. Expanding their diagonal entries gives

tr(AB)=Σ from i=1 to m Σ from j=1 to n AᵢⱼBⱼᵢ =Σ from j=1 to n Σ from i=1 to m BⱼᵢAᵢⱼ=tr(BA).

The sums are finite, so their order can be exchanged; field scalars commute, even though the matrices generally do not. For an invertible n-by-n S this proves tr(S⁻¹AS)=tr((S⁻¹A)S)=tr(S(S⁻¹A))=tr(A). Thus trace is an operator invariant, independently of any assumption that the characteristic polynomial splits.

For compatible square matrices, repeated grouping gives tr(ABC)=tr(CAB)=tr(BCA). It does not permit arbitrary rearrangement. For example, take A=diag(1,0), B=[[0,1],[0,0]] and C=[[0,0],[1,0]]. Then BC=diag(1,0), CB=diag(0,1), so tr(ABC)=1 but tr(ACB)=0. Check each product before making a claim about its trace.

The commutator of square A,B is AB−BA: it measures their failure to commute. Linearity and cyclicity give tr(AB−BA)=0. If AB−BA=I on a vector space of positive finite dimension n, we must therefore have n·1_F=0, where 1_F is the field’s multiplicative identity and the left side is its sum with itself n times. Over R or C, or any field of characteristic zero, this is impossible. The dimension assumption matters; this is a finite matrix argument.

In characteristic p, where p is prime, that necessary condition only says p divides n. The obstruction can actually disappear. On a p-dimensional space with basis e₀,…,e_(p−1), define maps X and D by X eⱼ=e_(j+1) for j before the final index, X e_(p−1)=0, D e₀=0 and D eⱼ=j e_(j−1) for j≥1. Integers in these formulas mean their field values. For j before the final index,

(DX−XD)eⱼ=((j+1)−j)eⱼ=eⱼ.

For the final basis vector, (DX−XD)e_(p−1)=−(p−1)e_(p−1)=e_(p−1), since p·1_F=0. Equality on a basis proves DX−XD=I. Its trace is p·1_F=0, fully consistent with cyclicity. The familiar characteristic-zero conclusion cannot be transported to a new field without checking its arithmetic.

Practice before reading the check. For A=[[1,2,0],[0,1,1]] and B=[[1,0],[0,1],[1,1]], calculate AB and BA and verify their traces despite their different sizes. Then write the two-by-two D and X above over F₂, calculate their commutator, and explain why the same displayed matrices over R do not give I. Finally repair the statement “equal traces imply AB=BA.”

Check the products and the field boundary

AB=[[1,2],[1,2]] has trace 3. BA=[[1,2,0],[0,1,1],[1,3,1]] also has trace 3, although one matrix is two-by-two and the other three-by-three. In basis e₀,e₁, D=[[0,1],[0,0]] and X=[[0,0],[1,0]]. Their commutator is diag(1,−1); this equals I over F₂ because −1=1, but not over R. The same real matrices also refute the last statement: DX and XD both have trace 1, yet they are different diagonal matrices. Trace gives a necessary scalar test for an identity, not a reconstruction of the entire matrix.

Continue

Triangular form makes trace and determinant visible as the sum and product of diagonal eigenvalues, counted with multiplicity. The coefficient identities remain valid over any field by determinant expansion, even when the roots do not belong to that field. Next, the minimal polynomial will record how far an operator is from being annihilated by simpler factors. This unit works in finite dimension; infinite-dimensional spectral theory is not a direct extension of a determinant calculation.

Review Linear Algebra I when a prerequisite needs practice. Continue to Find the shortest polynomial relation an operator obeys.