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Problems #11,pg.272 and #7,pg.279 are graded.
#11,pg.272: Find a basis for the eigenspace corresponding to each listed eigenvalue: A=[1−3−45];λ=−1,7.
Solution: Recall that given any eigenvalue λ, the eigenspace is Nul(A−λI). First let us find the eigenspace associated with the eigenvalue λ=−1. This means we need to find a basis for
Nul(A−(−1)I)=Nul([1−(−1)−3−45−(−1)])=Nul([2−3−46]).
Recall that the basis of a nullspace can be easy found from a reduced echelon form. So we row reduce [2−3−46] and find
[2−3−46]∼[1−3200].
Hence we conclude that the solution vector →x to the homogeneous equation [2−3−46]→x=→0 is
→x=[x1x2]=[32x2x2]=x2[321].
Hence we conclude that
Nul([2−3−46])=span{[321]}.
Since the set {[321]} is linearly independent, we conclude that
B={[321]}
is a basis for the eigenspace associated with λ=−1.
#7,pg.279: Find the characteristic polynomial and the real eigenvalues of the matrix [53−44].
Solution: Recall that the characteristic polynomial is the polynomial in variable λ given by det(A−λI). The roots of this polynomial are the eigenvalues, i.e. we must find λ that solves det(A−λI)=0. So compute
A−λI=[5−λ3−44−λ],
so
det(A−λI)=(5−λ)(4−λ)−(3)(−4)=x2−9x+32.
We can compute the roots of this polynomial via the quadratic formula:
x=9±√(−9)2−4(1)(32)2=92±√81−1282=92±√472i.
Hence there are no real eigenvalues of this matrix.
