# MA 3065 Study Guide - Quiz Guide: Rayleigh Quotient, Symmetric Matrix, Fundamental SolutionExam

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**preview**shows page 1. to view the full**5 pages of the document.**Math 5588 – Homework 7 Solutions

Recall a matrix A= (aij )∈Rn×nis symmetric if aij =aji, so that AT=A. We say a

symmetric matrix Ais positive deﬁnite, written A≥0, if

vTAv =

n

X

i=1

n

X

j=1

aijvivj≥0for all v∈Rn.

We write A≤Bwhenever B−A≥0.

1. A real symmetric matrix A∈Rn×nhas nreal eigenvalues λ1≤λ2≤ · · · ≤ λn, counted

with multiplicity. Show that Ais positive deﬁnite (A≥0) if and only if λ1≥0. [Hint:

Recall that

λ1= min

v6=0

vTAv

vTv.

The right hand side above is called a Rayleigh quotient.]

Solution. Ais positive deﬁnite if and only if vTAv ≥0for all v∈Rn, which holds if and

only if

vTAv

vTv≥0for all v6= 0.

By the Raleigh quotient interpretation of λ1,Ais positive deﬁnite if and only if λ1≥

0.

2. Let A∈R2×2be a symmetric matrix. Show that Ais positive deﬁnite (A≥0) if and

only if

det(A)≥0and Trace(A)≥0.(1)

Solution. We have det(A) = λ1λ2and Trace(A) = λ1+λ2, where λ1≤λ2are the real

eigenvalues of A.

If Ais positive deﬁnite then λ1, λ2≥0, and so (1) holds. Conversely, if (1) holds, then

λ1λ2≥0and λ1+λ2≥0. It follows that λ2≥0. If λ2= 0 then clearly λ1≥0. If

λ2>0, then if follows from λ1λ2≥0that λ1≥0. Hence λ1≥0and by problem 1 Ais

positive deﬁnite.

3. (a) Let A, B ∈Rn×nbe diagonal matrices. Show that A≤Bif and only if aii ≤bii

for all i∈ {1, . . . , n}.

Solution. We have A≤Bif and only if

n

X

i=1

n

X

j=1

(bij −aij)vivj≥0

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for all v∈Rn. Since Aand Bare diagonal, A≤Bif and only if

n

X

i=1

(bii −aii)v2

i≥0

for all v∈Rn. Choosing v=ekwe get that A≤Bif and only if bkk −akk ≥0for

all k.

(b) Give an example of diagonal matrices A, B ∈Rn×nfor which neither A≤Bnor

B≤Ahold.

Solution. A=diag(1,0) and B=diag(0,1).

4. Consider a nonlinear PDE in the form

H(x, ∇u, ∇2u) + F(x, ∇u) = 0.(2)

Assume that His linear in ∇2u, that is,

λH(x, p, A) + H(x, p, B) = H(x, p, λA +B)

for any λ, A, B. Such a PDE is called quasilinear. Show that (2) is elliptic if and only if

A≤0 =⇒H(x, p, A)≥0.

Solution. If (2) is elliptic, then

A≤B=⇒H(x, p, B) + F(x, p)≤H(x, p, A) + F(x, p).

Since His linear in the Hessian, H(x, p, 0) = 0. Setting B= 0 we have

A≤0 =⇒F(x, p)≤H(x, p, A) + F(x, p)

or

A≤0 =⇒H(x, p, A)≥0.

Conversely, assume

A≤0 =⇒H(x, p, A)≥0.

Let A≤B. Then A−B≤0and so

H(x, p, A −B)≥0.

Since His linear

H(x, p, A)≥H(x, p, B).

Adding F(x, p)to both sides we have

H(x, p, A) + F(x, p)≥H(x, p, B) + F(x, p).

Hence (2) is elliptic.

2

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