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Direct Sum of Invariant Subspaces and Its Characteristic Polynomial 📂Linear Algebra

Direct Sum of Invariant Subspaces and Its Characteristic Polynomial

Theorem1

Let T:VVT : V \to V be a linear transformation on a finite-dimensional vector space VV above. Assume that VV is the direct sum of the TT-invariant subspaces WiW_{i}.

V=W1W2Wk V = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k}

Let fi(t)f_{i}(t) be the characteristic polynomial of the restriction TWiT|_{W_{i}}. Then, the characteristic polynomial of TT, f(t)f(t), is as follows.

f(t)=f1(t)f2(t)fk(t) f(t) = f_{1}(t) \cdot f_{2}(t) \cdot \cdots \cdot f_{k}(t)

Proof

Prove by mathematical induction.

  • When k=2k=2, it holds.

    Let β1,β2\beta_{1}, \beta_{2} be the ordered basis of W1,W2W_{1}, W_{2}. And let’s say β=β1β2\beta = \beta_{1} \cup \beta_{2}. Then, by the property of direct sums, β\beta is the ordered basis of VV.

    Now, let’s say A=[T]βA = \begin{bmatrix} T \end{bmatrix}_{\beta}, B1=[TW1]β1B_{1} = \begin{bmatrix} T_{W_{1}} \end{bmatrix}_{\beta_{1}}, B2=[TW2]β2B_{2} = \begin{bmatrix} T_{W_{2}} \end{bmatrix}_{\beta_{2}}. Then, AA is represented as the following block matrix. A=[B1OOB2] A = \begin{bmatrix} B_{1} & O \\ O & B_{2} \end{bmatrix} Here, let OO be a zero matrix of appropriate size. Then, by the determinant of block matrices, f(t)=det(AtI)=det(B1tI)det(B2tI)=f1(t)f2(t) f(t) = \det(A - tI) = \det(B_{1} - tI) \det(B_{2} - tI) = f_{1}(t) \cdot f_{2}(t)

  • Assuming it holds when k12k-1 \ge 2, it also holds when kk.

    Let VV be the direct sum of the subspaces WiW_{i}. V=W1W2Wk V = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k} Let WW be the sum of Wi(1ik1)W_{i}(1\le i \le k-1)s. W=W1+W2++Wk1 W = W_{1} + W_{2} + \cdots + W_{k-1} Then, WW is TT-invariant, and V=WWkV = W \oplus W_{k} holds. By the proof when k=2k=2, if g(t)g(t) is the characteristic polynomial of TWT|_{W}, then f(t)=g(t)fk(t)f(t) = g(t)f_{k}(t). In fact, W=W1W2Wk1W = W_{1} \oplus W_{2} \oplus \cdots \oplus W_{k-1} holds and, by assumption, g(t)=f1(t)fk1(t)g(t) = f_{1}(t) \cdots f_{k-1}(t). Therefore, f(t)=g(t)fk(t)=f1(t)f2(t)fk(t) f(t) = g(t)f_{k}(t) = f_{1}(t) f_{2}(t) \cdots f_{k}(t)


  1. Stephen H. Friedberg, Linear Algebra (4th Edition, 2002), p319-320 ↩︎