Direct Sum of Invariant Subspaces and Its Characteristic Polynomial
📂Linear AlgebraDirect Sum of Invariant Subspaces and Its Characteristic Polynomial
Theorem
Let T:V→V be a linear transformation on a finite-dimensional vector space V above. Assume that V is the direct sum of the T-invariant subspaces Wi.
V=W1⊕W2⊕⋯⊕Wk
Let fi(t) be the characteristic polynomial of the restriction T∣Wi. Then, the characteristic polynomial of T, f(t), is as follows.
f(t)=f1(t)⋅f2(t)⋅⋯⋅fk(t)
Proof
Prove by mathematical induction.
When k=2, it holds.
Let β1,β2 be the ordered basis of W1,W2. And let’s say β=β1∪β2. Then, by the property of direct sums, β is the ordered basis of V.
Now, let’s say A=[T]β, B1=[TW1]β1, B2=[TW2]β2. Then, A is represented as the following block matrix.
A=[B1OOB2]
Here, let O be a zero matrix of appropriate size. Then, by the determinant of block matrices,
f(t)=det(A−tI)=det(B1−tI)det(B2−tI)=f1(t)⋅f2(t)
Assuming it holds when k−1≥2, it also holds when k.
Let V be the direct sum of the subspaces Wi.
V=W1⊕W2⊕⋯⊕Wk
Let W be the sum of Wi(1≤i≤k−1)s.
W=W1+W2+⋯+Wk−1
Then, W is T-invariant, and V=W⊕Wk holds. By the proof when k=2, if g(t) is the characteristic polynomial of T∣W, then f(t)=g(t)fk(t). In fact, W=W1⊕W2⊕⋯⊕Wk−1 holds and, by assumption, g(t)=f1(t)⋯fk−1(t). Therefore,
f(t)=g(t)fk(t)=f1(t)f2(t)⋯fk(t)
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