Math 502: Abstract Algebra

Homework 13

Lawrence Tyler Rush

<me@tylerlogic.com>

January 5, 2014

http://coursework.tylerlogic.com/courses/upenn/math502/homework13

1

We first lay out a helpful lemma. ¹

Lemma 1.1. Let V be a vector space over F where F is ℝ or ℂ. If S and T are hermitian and ST = TS, then there is an orthogonal basis for S and T.

Proof. Let λ be an eigenvalue of T and set W to be the eigenspace with respect to lambda. Then the commutativity of ST and TS gives us that

T Sw = ST w = S (λw) = λ(Sw )

for w ∈ W. In other words Sw ∈ W, implying that W is S-invariant. Now since T is hermitian, then W is 1-dimensional, implying that lambda is an eigenvalue for S and furthermore the eigenspace for S corresponding to λ is W. Since λ is arbitrary and both S and T are diagonalizable, then we know that there exists an orthogonal basis v₁, ⋅⋅⋅ ,v_n. __

Now for the main event. Let V be a vector space over F where F is ℝ or ℂ. Also let (⋅|⋅)₁ and (⋅|⋅)₂ be two distinct inner products on V . Due to Gram-schmitt, we can assume that (⋅|⋅)₁ is simply the standard inner product on V . So Then there exists hermitian, positive definite operators T ∈ End_F(V ) such that (x|y)₂ = (Tx|y)₁, and in the case of (⋅|⋅)₁, the hermitian operator corresponding to it is Id. Certainly T Id and IdT are commutative and T and Id are both hermitian. Hence Lemma 1.1 implies that we can find an orthogonal basis for the standard inner product, i.e. (⋅|⋅)₁, but because (x|y)₂ = (Tx|y)₁, then this will be othogonal with respect to both inner products.

2

(a)

(b)

Let w ∈ Ker(S - λId_W) then we have that (S - λId_W)w = 0, and so Sw = Tw = λw, which implies that w ∈ W(λ).

Now assume that w ∈ W(λ), then Sw = Tw = λw, which implies that w ∈ Ker(S - λId_W).

Hence Ker(S - λId_W) = W(λ).

(c)

We first factor S² - (λ -λ)S + λλId_W.

2 -- -- 2 -- -- S - (λ- λ)S +λ λIdW = S - λS - λS +-λλIdW = S(S - λ IdW)- λ (S - λIdW ) = (S - λ-IdW )(S - λ IdW )

First assume that w ∈ W(λ) + W(λ), then w = u + v where u ∈ W(λ) and v ∈ W(λ). But then u and v is killed by (S -λId_W) and (S - λId_W)w, respectively. Hence u + v = w is killed by (S -λId_W)(S - λId_W)w.

Now assume that w ∈ Ker(S² - (λ -λ)S + λλId_W) = Ker((S -λId_W)(S - λId_W)).

(d)

3

Define Q(v) as a quadratic form on ℝⁿ by

Q(x ,...,x ) = ∑ a xx 1 n 1≤i,j≤n ij i j

for some symmetric A = (a_ij) ∈ M_n(ℝ). Let λ₁ ≥ λ₂ ≥ ⋅⋅⋅ ≥ λ_n be the eigenvalues of A with possible multiplicity.

(a) Show that λ₁ = max{Q(v) | v ∈ ℝⁿ,(v|v) = 1}

Because A is symmetric, we can obtain an orthonormal basis β = {v₁,…,v_n} with respect to the dot product corresponding to the eigenvalues λ₁,…,λ_n. Then for any v ∈ ℝⁿ we have v = b₁v₁ +

+ b_nv_n which implies

Q (v) = (Av|v) = (b1λ1v1 + ⋅⋅⋅+ bnλnvn)|v) = (b λ v|v)+ ⋅⋅⋅+ (b λ v |v) 1 1 1 n n n = (b1λ1v1|b1v1) + ⋅⋅⋅+ (bnλnvn|bnvn) = b21λ1 + ⋅⋅⋅+ b2nλn

with the last two equalities coming to us by way of the orthonormality of β. Now assuming that (v|v) = 1, then

1 = (v|v) = (b1v1 + ⋅⋅⋅+ bnvn|b1v1 + ⋅⋅⋅+ bnvn) = b21 + ⋅⋅⋅+ b2n

again by the orthonormality of β. Hence λ₁ ≥ b₁²λ₁ + ⋅⋅⋅ + b_n²λ_n, i.e

n λ1 = max {Q(v) | v ∈ ℝ ,(v|v) = 1}

(b) Extra Credit

(c) Extra Credit

4 Extra Credit

5 Extra Credit

References

[HJE03] Friedberg S. H., Insel A. J., and Spence L. E. Linear Algebra. Prentice Hall, 2003.