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(1) Suppose T ∈ L(V )
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If 9 is an eigenvalue of T 2 , then T 2 v = 9v for some v ̸= 0
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Thus, (T − 3I)v = 0 or (T + 3I) = 0, so
λ = 3 or λ = −3
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It follows that T 2 u = 9u, so 9 is an eigenvalue of T 2
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Find all eigenvalues and eigenvectors of T
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T (v) = λv ⇒

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x1 + x2 + x3 = λx1
x1 + x2 + x3 = λx2
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
x1 + x2 + x3 = λx3
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(1 − λ)x1 + x2 + x3 = 0
x1 + (1 − λ)x2 + x3 = 0
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
x1 + x2 + (1 − λ)x3 = 0
For λ = 3, the system simplifies to −2x1 + x2 = x3 = 0, which gives the
eigenvector v = (1, 1, 1)
For λ = 0, any vector satisfying x1 +x2 +x3 = 0 is an eigenvector, indicating
an eigenspace of dimension 2
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(3) Suppose F = C, T ∈ L(V ), p ∈ P (C), and α ∈ C
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If T v = λv, then (T k )v = (λk )v for some integer k ≥ 0
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+ an tn , we have:
p(T )v = (a0 I + a1 T +
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+ an (λn )v = p(λ)v
Thus, p(λ) is an eigenvalue of p(T ) with eigenvalue u ̸= 0
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Express p(t) − α as a product of its roots ri :
p(t) − α = c(t − r1 )(t − r2 ) · · · (t − rm )
Thus, (p(T ) − αI)u = c(T − r1 I)(T − r2 I) · · · (T − rm I)u = 0
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(4) Suppose T ∈ L(F 2 ) is defined by T (w, z) = (−z, w)
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Observe T ’s action on (w, z): T (w, z) = (−z, w) ⇒ T 2 (w, z)
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This yields that T 2 (w, z) = −I(w, z), where I is the
identity of F 2
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(5) Suppose L ∈ L(V ) and p ∈ P (F )
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Let mL be the minimal polynomial L with deg(mL ) = n
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Then, when
we apply L, we get p(L) = mL (L)q(L) + r(L)
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Given anything produced by the division algorithm is used, there
exists a single r(t) for each p(t) given an mL (t)
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Find the minimal polynomial of T −1

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