Today § 5.1 : Eigenvalues
of § 5.2 : Characteristic Polynomial
Next : Review for Midterm 2
Reminders
My Math Lab Homework # 6 : Due Tomorrow by ltispm
Midterm # 2 : This Wednesday 8- 10pmRom / seat assign .
& practicemidterms posted on course webpage .
Given an nxn matrix A,
an eigenvector is a nonzero
vector v. e- R" with the property that
A v. = tvfor Some scalar XEIR
,called the eigen value
.
• The set of eigenvectors of A for a given eigen value X←
Without s .
B equal to Nul ( ATI ) .
So it B a subspace
of Rn,
called the eigen space for X.
The
eigenvectors are the nonzero vectors in the eigenspahe .
• Typically hand to find the eigenvalues of a matrix ;
once known,
finding the eigenspab is routine.
E 9.Let
A=|§y 6g ) .
2 isaneigenvaluefrlt ;
find abasB for the eigenspad .
Tkeymspalefoz is Null A . ZI ) ←dim ⇒
restitute .tk#Ef 't :#→
k¥31www.t#.sffxM*fxtYtxskB.rHit.HDt 17
KIM?ftEµMl ban .
Eg.
As Hl )
tnde9moahes.pj.YgyAfiIftoHIeHiftssl-fyty.ydeta.oznetwrtble@yjdetthII.o
.
A- It fold , Null :D :{ wyd }espan{ tell dm=t
E.g .
�1� = Idf ) B .v=I± = t foany ten.
↳
Kouyate,
and
its eigenspevis 1122.
ldm = 2)bbasis { ldfl :B
Theorem. Eigenvectors with distinct eigenvalues are
linearly independent .
( Case of zdstnetegenuames
xp Atm )
v. w.
to
Asian ttwgnw Assume { x. d) are Independent.
b W. = Cy for some
A±=µw↳i¥YIa⇒.!
*
So µw.=X± i.e. ( µ. 7) a =o
¥ ¥- Impossible .
Corollary If A is an nxn matrix with n distinct
eigenvahes ,
then there is a basis of Rn
consisting of eigenvectors of A.
D= { s,
I,
.. ; In } basis of e. vectors forA. c)
×' k t Se every meter t.tk
"
can beexpanded as a lm .
Gmb.
I B.
[ Axles ftp.YD⇐ * . Hagey
y =Y,
gtx.at -- - txn.tn
.
=/"
oh, ?yn|µ§n| Ad = x,
As + KAI t - - txnb.vn= X
,X
,9 + ×2X↳ht - .
. tktn.vn
ask.fi#.=fI*d=fgEnleidWhat does that really mean ? D-
II. . ,b eigenvectors
P=[gb . .. b.| l invertible)
AP - AH , g. " b) =[ Ay Ahn . A± )
Ap =p ,=lhbX¥ ... and
i. AMPDM = 14.4 - b) ("
ftp./=PDA2=(PDPY(PDpD=PDP@DPt=PD2p"
Definition :nxn matrices A and B are called similar
IF there B an invertible nxw matrix P with the property
A = PBP-
1 B=PtA@ Q =P- '
i. Q- ' =P↳
.
: AP - PB → i. FHP ' B\B = a apt
• we just saw that, if A has all distinct eigen values
,
then it B similar to a diagonal matrix.
That's important ;
more on that next time.
• Similarity is net the same as now equivalent .
Es . A =/ 11,1 B =/ d,o| row equivalent,
but net similar.
Theorem If A and B are similar,
then they have the
← same eigenvalues ; and the eigenspaces for A haveD= PAF '
ptBp=Athe same dimensions as the eigenspaces for B
.
Let x be an egenoalne of A.
Then the eyen space
for A ± X B Nul ( A- XI ).
I .to .↳FFBEII
's
i. BP , = PHD =xP±
I - Py Is an eigenvector of B y e. value 7.
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