Linear Algebra

400 Chapter Five. Similarityeach term comes from a rearrangement of the column numbers 1, . . . , n into anew order φ(1), . . . , φ(n). The upper right block Z 2 is all zeroes, so if a φ hasat least one of p + 1, . . . , n among its first p column numbers φ(1), . . . , φ(p)then the term arising from φ does not contribute to the sum because it is zero,e.g., if φ(1) = n then t 1,φ(1) t 2,φ(2) . . . t n,φ(n) = 0 · t 2,φ(2) . . . t n,φ(n) = 0.So the above formula reduces to a sum over all permutations with twohalves: any contributing φ is the composition of a φ 1 that rearranges only1, . . . , p and a φ 2 that rearranges only p + 1, . . . , p + q. Now, the distributivelaw and the fact that the signum of a composition is the product of the signumsgives that this( ∑|T 1 | · |T 2 | =)t 1,φ1 (1) · · · t p,φ1 (p) sgn(φ 1 )perms φ 1of 1,...,p(·∑)t p+1,φ2 (p+1) · · · t p+q,φ2 (p+q) sgn(φ 2 )perms φ 2of p+1,...,p+qequals |T| = ∑ contributing φ t 1,φ(1)t 2,φ(2) · · · t n,φ(n) sgn(φ).QED2.11 Example ∣ ∣∣∣∣∣∣∣∣ 2 0 0 01 2 0 0=2 00 0 3 0∣1 2∣ ·3 0∣0 3∣ = 360 0 0 3∣From Lemma 2.10 we conclude that if two subspaces are complementary andt invariant then t is one-to-one if and only if its restriction to each subspace isnonsingular.Now for the promised third, and final, step to the main result.2.12 Lemma If a linear transformation t: V → V has the characteristic polynomial(x − λ 1 ) p 1. . . (x − λ j ) p jthen (1) V = N ∞ (t − λ 1 ) ⊕ · · · ⊕ N ∞ (t − λ j ) and(2) dim(N ∞ (t − λ i )) = p i .Proof We start by proving that N ∞ (t−λ i )∩N ∞ (t−λ j ) = {⃗0} when i ≠ j. Thisshows that the bases for the generalized null spaces, when concatenated, form alinearly independent subset of the space V. We will then show that clause (2)holds because with that, since the degree p 1 + · · · + p k of the characteristicpolynomial equals the dimension of the space V, we will have proved clause (1)also.So consider N ∞ (t − λ i ) ∩ N ∞ (t − λ j ) when i ≠ j. By Lemma 2.8 bothN ∞ (t − λ i ) and N ∞ (t − λ j ) are t invariant. The intersection of t invariantsubspaces is t invariant and so the restriction of t to N ∞ (t − λ i ) ∩ N ∞ (t − λ j )is a linear transformation. Now, t − λ i is nilpotent on N ∞ (t − λ i ) and t − λ jis nilpotent on N ∞ (t − λ j ), so both t − λ i and t − λ j are nilpotent on theintersection. Therefore by Lemma 2.1 and the observation following it, if t has