# Eigenvalues and Eigenvectors or Pseudoinverse Matrix

In Gilbert Strang's "Linear Algebra and Learning from Data" he asks the question

Why do $$A$$ and $$A^+$$ have the same rank? If $$A$$ is square, do $$A$$ and $$A^+$$ have the same eigenvectors? What are the eigenvalues of $$A^+$$?

I've managed to answer the first two questions:

1. Using the SVD one can see that the $$r$$ positive singular values of $$A$$ lead to $$r$$ positive singular values for $$A^+$$ via inversion.
2. $$A=\begin{bmatrix}1 & 2 \\ 1 & 2 \end{bmatrix}$$ has eigenvectors $$[1,1]^T$$ and $$[2,-1]^T$$. $$A^+=\begin{bmatrix} 1/10 & 1/10 \\ 1/5 & 1/5 \end{bmatrix}$$ doesn't have any of these as eigenvectors.

Regarding the eigenvalues of $$A^+$$: I tried playing around with some particular matrices, but I couldn't see a simple relationship with the eigenvalues of $$A$$. I suspect there is no such relationship between the eigenvalues. Is that the case?

Thanks.

• Nov 14, 2019 at 10:25

Let v be an eigenvector of A and suppose its corresponding eigenvalue $$\lambda$$ is not zero. Because $$\lambda$$ is different from zero, we have that v is in the row space of A ($$Row(A) = Null(A)^{\perp}$$). Page 125 of states "$$A^{+}Ax=x$$ exactly when x is in the row space". Hence $$A^{+}Av=v$$. Thus $$A^{+}Av=A^{+}\lambda v$$ implies $$\frac{1}{\lambda}v = A^{+}v$$. Which means $$\frac{1}{\lambda}$$ is an eigenvalue of $$A^+$$. Note that $$A^+v$$ makes sense because we are refering to a square matrix A (in order to talk about eigenvalues).
On the other hand, if we let $$\lambda = 0$$ (recall v cannot equal the zero vector) we then have $$A^{+}Av=A^{+}\lambda v= \lambda A^{+}v = 0$$. Which implies $$\lambda$$ is also an eigenvalue of $$A^{+}$$.
Thus the eigenvalues of $$A^{+}$$ are the reciprocal of those of A, or zero when it is the case. However i don't know how to show that these are the only eigenvalues. I suppose it is thanks to the definition of pseudoinverse.