Do elementary row operations change eigenvalues?

1 Answer
Truong-Son N.
May 2, 2016

Yes. For a given matrix #hatA#, elementary row operations do NOT retain the eigenvalues of #hatA#.

For instance, take the following matrix:

#color(green)(hatA = [(2,2),(0,1)])#

The eigenvalues are determined by solving

#\mathbf(hatAvecv = lambdavecv),#

such that #|lambdaI - hatA| = 0#. Then, the eigenvectors #vecv# form a basis acquired from solving #[lambdaI - hatA]vecv = vec0# for #vecv#.

#|lambdaI - hatA| = 0#

#= |[(lambda,0),(0,lambda)] - [(2,2),(0,1)]|#

#= |(lambda - 2, -2),(0,lambda - 1)|#

#= (lambda-2)(lambda-1) - 0#

From this we acquire the characteristic equation:

#=> color(green)((lambda - 2)(lambda - 1) = 0),#

And we get the eigenvalues

# => color(blue)(lambda = 1, 2),#

whose eigenvectors are...

#\mathbf(lambda = 1)#:

#[(lambda - 2, - 2),(0,lambda - 1)][(v_1),(v_2)] = [(0),(0)]#

#= [(-1,-2),(0,0)][(v_1),(v_2)] = [(0),(0)]#

#-> v_1 = -2v_2 -> color(blue)(vecv = v_1[(1),(-2)])#

#\mathbf(lambda = 2)#:

#[(lambda - 2, - 2),(0,lambda - 1)][(v_1),(v_2)] = [(0),(0)]#

#= [(0,-2),(0,1)][(v_1),(v_2)] = [(0),(0)]#

#-> v_2 = 0, v_1 = "anything"#, :. let #v_1 = 9000#. Then, #color(blue)(vecv = v_1[(9000),(0)])#

Of course, had you row-reduced #hatA#, you would have gotten:

#hatA = [(2,2),(0,1)]#

#stackrel(1/2R_1; -R_2+R_1" ")(->)[(1,0),(0,1)],#

where the notation #cR_i + R_j# implies that #c# times row #i# is added to row #j# and the result is stored into row #j#.

That would give you the characteristic equation #|lambdaI - hatA| = (lambda - 1)^2 = 0#, and thus give you one eigenvalue #lambda = 1# of multiplicity #2#.

However, without row-reduction, we had gotten two distinct eigenvalues: #lambda = 1,2#. Thus, the eigenvalues were not retained as a result of elementary row operations.

Related questions
See all questions in Elementary Row Operations
Impact of this question
43851 views around the world
You can reuse this answer
Creative Commons License