Question

Do elementary row operations change eigenvalues?

Answer 1

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)])`

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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.