Canonical Forms Linear Algebra Notes

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$Math 121 Midterm Exam Version A Spring 2012 Instructor: Lang Don ...$

2. This non-zero element e ∈ V above is called a characterisitic vector of T associated to λ. A characterisitic vector is also known an eigen vector. 3. Write N(λ) = {v ∈ V : T (v) = λv}. Then N(λ) is a subspace of V and is said to be the characterisitic space or eigen space of T associated to λ. 2.2 (Lemma.) Let V be a vector space over a field F and T ∈ L(V, V ) Then T is singular if and only if det(T ) = 0. Proof. (⇒): We have T (e1) = 0 for some e1 ∈ V with e1 = 0. We can extend e1 to a basis e1, e2, . . . , en of V. Let A be the matrix of T with respect to this basis. Since, T (e1) = 0, the first row of A is zero. Therefore, det(T ) = det(A) = 0. So, this implication is established. (⇐): Suppose det(T ) = 0. Let e1, e2, . . . , en be a basis of V and A be the matrix of T with respect to this basis. So det(T ) = det(A) = 0. Therefore, A is not invertible. Hence AX = 0 for some non-zero column ⎛ ⎞ X = ⎜ ⎝ c1 c2 . . . cn ⎟ ⎠ . Write v = c1e1 + c2e2 + · · · + cnen. Since not all ci is zero, v = 0. Also, T (v) = ciT (ei) ⎛ ⎞ ⎛ ⎞ ⎜ = (T (e1), T (e2), . . . , T (en)) ⎜ ⎝ So, T is singular. c1 c2 . . . cn ⎟ ⎠ = (e1, ⎜ e2, . . . , en)A ⎜ ⎝ The following are some equivalent conditions. 2 c1 c2 . . . cn ⎟ ⎠ = 0..
2.3 (Theorem.) Let V be a vector space over a field F and T ∈ L(V, V ) be linear operator. Let λ ∈ F be a scalar. Then the following are equivalent: 1. λ is a characteristic value of T. 2. The operator T − λI is singular (or is not invertible). 3. det(T − λI) = 0. Proof. ((1) ⇒ (2)): We have (T − λI)(e) = 0 for some e ∈ V with e = 0. So, T − λI is singular and (2) is established. ((2) ⇒ (1)): Since T − λI is singular, we have (T − λI)(e) = 0 for some e ∈ V with e = 0. Therefore, (1) is established. ((2) ⇔ (3)): Immediate from the above lemma. 2.4 (Definition.) Let A ∈ Mn(F) be an n × n matrix with entries in a field F. 1. A scalar λ ∈ F is said to be a characteristic value of A if det(A − λIn) = 0. Equivalently, λ ∈ F is said to be a characteristic value of A if the matrix (A − λIn) is not invertible. 2. The monic polynomial det(XI − A) is said to be the characteristic polynomial of A. Therefore, characteristic values of A are the roots of the characteristic polynomial of A. 2.5 (Lemma.) Let A, B ∈ Mn(F) be two n × n matrices with entries in a field F. If A and B are similar, then they have some characteristic polynomials. Proof. Suppose A, B are similar matrices. Then A = P BP −1 . The characteristic polynomial of A = det(XI−A) = det(XI−P BP −1 ) = det(P (XI − B)P −1 ) = det(XI − B) = the characteristic polynomial of B. 2.6 (Definitions and Facts) Let V be a vector space over a field F and T ∈ L(V, V ) be linear operator. 3
Page 1: 1 Introduction Canonical Forms Line
Page 5 and 6: 2.2 Decomposition of V 2.8 (Definit
Page 7 and 8: Since Ei is a basis, λij = 0 for a
Page 9 and 10: 2.15 (Theorem) Suppose V is vector
Page 11 and 12: 4. Therefore, ann(T ) = F[X]p(X) wh
Page 13 and 14: (T (e1), T (e2), . . . , T (en)) =
Page 15 and 16: 1. Let p be the characteristic poly
Page 17 and 18: (⇐): Now assume that the MMP p fa
Page 19 and 20: V2 V = W2 ⊕ V2 T2 V2 pr T V =
Page 21 and 22: (⇐): We we asssume that p(X) = (X
Page 23 and 24: 6 Direct Sum Part of this section w
Page 25 and 26: 7 Invariant Direct Sums This sectio
Page 27 and 28: Also, for i = j note that p | hihj.

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