Different eigenvectors obtained with Mathematica

Question

Suppose that we have a matrix $$ \begin{pmatrix} a&c\\c&b \end{pmatrix}. $$ The eigenvalues of this matrix are given by $\lambda_1=(a+b-D)/2$ and $\lambda_1=(a+b+D)/2$, where $D=\sqrt{(a-b)^2+4c^2}$. The eigenvectors are given by $$ \tag{1} \begin{pmatrix}1\\(\lambda_1-a)/c\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}1\\(\lambda_2-a)/c\end{pmatrix} $$ provided that $c\ne0$. However, when I use Mathematica to calculate the eigenvectors of this matrix, I get $$ \tag{2} \begin{pmatrix}-(\lambda_2-a)/c\\1\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}-(\lambda_1-a)/c\\1\end{pmatrix}. $$ As far as I know, the eigenvectors are defined up to a multiplication by a non-zero constant. But multiplication by a constant does not transform eigenvectors $(1)$ into eigenvectors $(2)$. Or am I wrong?

Could anyone explain why I get different eigenvectors when I use Mathematica? Or are these eigenvectors the same?

Any help is much appreciated!

Answer 1

The characteristic polynomial of your matrix $A$ is

$$ p_A(x) = x^2 - (a + b)x + (ab - c^2) = (x - \lambda_1)(x - \lambda_2) = x^2 - (\lambda_1 + \lambda_2)x + \lambda_1 \lambda_2. $$

Assume that $c \neq 0$. If you multiply your first eigenvector by $-\frac{\lambda_2 - a}{c}$, you get

$$ -\frac{\lambda_2 - a}{c} \begin{pmatrix} 1 \\ \frac{\lambda_1 - a}{c} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ - \frac{(\lambda_1 - a)(\lambda_2 - a)}{c^2} \end{pmatrix}. $$

Now,

$$ (\lambda_1 - a)(\lambda_2 - a) = \lambda_1 \lambda_2 - a (\lambda_1 + \lambda_2) + a^2 = (ab - c^2) - a(a + b) + a^2 = -c^2$$

so indeed we get

$$ -\frac{\lambda_2 - a}{c} \begin{pmatrix} 1 \\ \frac{\lambda_1 - a}{c} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ - \frac{(\lambda_1 - a)(\lambda_2 - a)}{c^2} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ 1 \end{pmatrix} $$

so your eigenvector and Mathematica's eigenvector are equal up to a multiplication by a non-zero constant.

Answer 2

Actually the pairs of vectors do match up to scalar multiples. Your first vector corresponds to Mathematica's second and vice versa.

To see this: Take your first vector and multiply by the first component of Mathematica's second vector. Note that when a is subtracted from each eigenvalue and the differences are multiplied together, that product is -c^2. Do the same with your second vector and Mathematica's first.