Quiz 10 Solutions, Math 246, Professor David Levermore
Tuesday, 24 November 2009
(1) [4] Let A=๎3 1
โ1 1๎. Compute etA.
Solution. The characteristic polynomial is p(z) = z2โ4z+ 4 = (zโ2)2. Hence,
etA=e2t๎I+ (Aโ2I)t๎
=e2t๎๎1 0
0 1๎+๎1 1
โ1โ1๎t๎
=e2t๎1 + t t
โt1โt๎.
(2) [3] A=๎1 1
3โ1๎has eigenvalues โ2 and 2. Find an eigenvector for each eigenvalue.
Solution: One has A+ 2I=๎3 1
3 1๎and Aโ2I=๎โ1 1
3โ3๎.
The eigenvectors v1associated with the eigenvalue โ2 satisfy (A+ 2I)v1= 0. You
can either solve this system or simply read-off from a nonzero column of Aโ2Ithat
these have the form
v1=ฮฑ1๎โ1
3๎for some ฮฑ16= 0 .
The eigenvectors v2associated with the eigenvalue 2 satisfy (Aโ2I)v2= 0. You
can either solve this system or simply read-off from a nonzero column of A+ 2Ithat
these have the form
v2=ฮฑ2๎1
1๎for some ฮฑ26= 0 .
(3) [3] A real 2 ร2 matrix Ahas the eigenpair ๎โ1 + i3,๎2
โi๎๎. Diagonalize A.
Solution: Because Ais real, a second eigenpair will be the complex conjugate of
the given eigenpair. If you use the eigenpairs
๎โ1 + i3,๎2
โi๎๎ ,๎โ1โi3,๎2
i๎๎ ,
then set
V=๎2 2
โi i๎,D=๎โ1 + i3 0
0โ1โi3๎.
Because det(V) = 2 ยทiโ(โi)ยท2 = 4i, you obtain the diagonalization
A=VDVโ1=๎2 2
โi i๎๎โ1 + i3 0
0โ1โi3๎1
4i๎iโ2
i2๎.