a.
A=⎣⎡321141123⎦⎤
A−λI=⎣⎡3−λ2114−λ1123−λ⎦⎤
detA−λI=∣∣3−λ2114−λ1123−λ∣∣
=(3−λ)2(4−λ)+2+2−(4−λ)−2(3−λ)
−2(3−λ)=36−24λ+4λ2−9λ+6λ2−λ3
+4−4+λ−12+4λ=−λ3+10λ2−28λ+24=0
−λ2(λ−2)+8λ(λ−2)−12(λ−2)=0
−(λ−2)2(λ−6)=0 λ1=6,λ2=2,λ3=2
These are the eigenvalues.
b. Find the eigenvectors.
λ=6
⎣⎡3−λ2114−λ1123−λ⎦⎤=⎣⎡−3211−2112−3⎦⎤ R1=R1/(−3)
⎣⎡121−1/3−21−1/32−3⎦⎤R2=R2−2R1
⎣⎡101−1/3−4/31−1/38/3−3⎦⎤ R3=R3−R1
⎣⎡100−1/3−4/34/3−1/38/3−8/3⎦⎤ R2=−3R2/4
⎣⎡100−1/314/3−1/3−2−8/3⎦⎤ R1=R1+R2/3
⎣⎡100014/3−1−2−8/3⎦⎤ R3=R3−4R2/3
⎣⎡100010−1−20⎦⎤
⎣⎡100010−1−20⎦⎤⎣⎡x1x2x3⎦⎤=⎣⎡000⎦⎤If we take x3=t, then x1=t,x2=2t.
The eigenvector is ⎣⎡121⎦⎤
λ=2
⎣⎡3−λ2114−λ1123−λ⎦⎤=⎣⎡121121121⎦⎤ R2=R2−2R1
⎣⎡101101101⎦⎤ R3=R3−R1
⎣⎡100100100⎦⎤
⎣⎡100100100⎦⎤⎣⎡x1x2x3⎦⎤=⎣⎡000⎦⎤If we take x2=t,x3=s, then x1=−t−s.
The eigenvectors are ⎣⎡−110⎦⎤,⎣⎡−101⎦⎤.
P=⎣⎡121−110−101⎦⎤,D=⎣⎡600020002⎦⎤
PDP−1=A
The matrices A=⎣⎡321141123⎦⎤ and D=⎣⎡600020002⎦⎤ are similar.
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