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

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Filmmaking-Sound
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MA 3065
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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 A0, if
vTAv =
n
X
i=1
n
X
j=1
aijvivj0for all vRn.
We write ABwhenever BA0.
1. A real symmetric matrix ARn×nhas nreal eigenvalues λ1λ2 · · · λn, counted
with multiplicity. Show that Ais positive definite (A0) if and only if λ10. [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 vRn, which holds if and
only if
vTAv
vTv0for all v6= 0.
By the Raleigh quotient interpretation of λ1,Ais positive definite if and only if λ1
0.
2. Let AR2×2be a symmetric matrix. Show that Ais positive definite (A0) if and
only if
det(A)0and Trace(A)0.(1)
Solution. We have det(A) = λ1λ2and Trace(A) = λ1+λ2, where λ1λ2are the real
eigenvalues of A.
If Ais positive definite then λ1, λ20, and so (1) holds. Conversely, if (1) holds, then
λ1λ20and λ1+λ20. It follows that λ20. If λ2= 0 then clearly λ10. If
λ2>0, then if follows from λ1λ20that λ10. Hence λ10and by problem 1 Ais
positive definite.
3. (a) Let A, B Rn×nbe diagonal matrices. Show that ABif and only if aii bii
for all i∈ {1, . . . , n}.
Solution. We have ABif and only if
n
X
i=1
n
X
j=1
(bij aij)vivj0
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for all vRn. Since Aand Bare diagonal, ABif and only if
n
X
i=1
(bii aii)v2
i0
for all vRn. Choosing v=ekwe get that ABif and only if bkk akk 0for
all k.
(b) Give an example of diagonal matrices A, B Rn×nfor which neither ABnor
BAhold.
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
A0 =H(x, p, A)0.
Solution. If (2) is elliptic, then
AB=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
A0 =F(x, p)H(x, p, A) + F(x, p)
or
A0 =H(x, p, A)0.
Conversely, assume
A0 =H(x, p, A)0.
Let AB. Then AB0and 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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