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.

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