Is it possible to get the original eigenvector after scaling a matrix?

Question

Let ${\mathbf{X}}\in\mathbb{R}^{n\times 1}$ and ${\mathbf{Y}}\in\mathbb{R}^{n\times 1}$ and let $\mathbf{A}\in\mathbb{R}^{n\times 2}$ be defined as \begin{equation} \mathbf{A} = \left[{\mathbf{X}}|{\mathbf{Y}}\right] \end{equation} If $\lambda$ be the smallest eigenvalues of $\mathbf{A}^T\mathbf{A}$ and its corresponding eigenvector $\mathbf{u}$ then \begin{equation} \mathbf{A}^T\mathbf{A}\mathbf{u} = \lambda\mathbf{u} \end{equation} Let us scale the matrix $\mathbf{A}$ in the following way \begin{equation} \begin{split} \mathbf{B} = \left[{\mathbf{X}}|{\mathbf{Y}}\right]\Sigma^{-1/2}\\ \Sigma^{-1/2} = \left[ \begin{array}{cc} 1/\sqrt{\sigma_1} & 0 \\ 0 & 1/\sqrt{\sigma_2} \end{array} \right] \end{split} \end{equation} where $\sigma_1$ and $\sigma_2$ are scalar. The corresponding eigenvalue problem for $\mathbf{B}^T\mathbf{B}$ can be written as \begin{equation} \begin{split} \mathbf{B}^T\mathbf{B}\mathbf{u}_1 &= \lambda_1\mathbf{u}_1\\ \Sigma^{-1/2}\mathbf{A}^T\mathbf{A}\Sigma^{-1/2}\mathbf{u}_1 &= \lambda_1\mathbf{u}_1\\ \mathbf{A}^T\mathbf{A}\Sigma^{-1/2}\mathbf{u}_1 &= \lambda_1\Sigma^{1/2}\mathbf{u}_1 \end{split} \end{equation} where, $\lambda_1$ is the smallest eigenvalue of $\mathbf{B}^T\mathbf{B}$.

My question is: Is $\lambda = \lambda_1$? and,

is it possible to get back $\mathbf{u}$ from $\mathbf{u}_1$?

Answer 1

Of course $\lambda\not=\lambda_1$.

By the power method, you can obtain the greatest eigenvalue $\lambda_2=tr(B^TB)-\lambda_1 $ of $B^TB$ and an associated eigenvector $v_2$. Let $P=[v_1,v_2]\in M_2$; then you know $B^TB=Pdiag(\lambda_1,\lambda_2)P^{-1}$; finally you must remove the scaling: $A^TA=\Sigma^{1/2}Pdiag(\lambda_1,\lambda_2)P^{-1}\Sigma^{1/2}$.

Answer 2

Counterexamples:
- with $A=I_2$ the identity matrix and $\Sigma^{-\frac{1}{2}}=\operatorname{diag}\left(\frac{1}{2},2\right)$ we get $A^{\top}A=I_2$ and $B^{\top}B=\operatorname{diag}\left(\frac{1}{4},4\right)$. Thus, both matrices have different eigenvalues.
- with $A=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ also all eigenvectors are different.