3

Problem

Let $A$, $B \in M_n(\mathbb{C})$. Suppose the eigenvalues of $A$, $B$ are all non-negative real numbers, and that null($A$) $=$ null($A^2$) and null($B$) $=$ null($B^2$). If $A^2 = B^2$, prove that $A=B$.

user26857
  • 52,094
Star Chou
  • 1,351
  • 1
    Is $\operatorname{null}$ the nullity (a number) or the kernel (a subspace)? – Kenny Lau Dec 31 '17 at 14:55
  • @KennyLau in this case the number is sufficient, so I suspect that is what they mean. – Ben Grossmann Dec 31 '17 at 16:23
  • 1
    The nullity condition simply means that $A$ and $B$ have not any nontrivial nilpotent Jordan block in their Jordan forms (i.e. if zero is an eigenvalue, it is semisimple). It is here to ensure that, subject to this condition, the square root of a zero matrix is unique. The rest are pretty standard: just find a square root $X$ of $A^2$ that has a nonnegative spectrum, satisfies the nullity condition, and is a polynomial in $A^2$; then show that both $A,B$ are equal to $X$. – user1551 Jan 01 '18 at 05:40
  • construct $S \in GL_n(\mathbb C)$ having its first vectors as a basis for $\ker A^2$ and the rest as a basis for $\text{im } A^2$, then $S^{-1}AS =\begin{bmatrix} \mathbf 0 & \mathbf 0 \ \mathbf 0 & A' \end{bmatrix}$ and $S^{-1}BS =\begin{bmatrix} \mathbf 0 & \mathbf 0 \ \mathbf 0 & B' \end{bmatrix}$ where $A' = B'$ follows from the duplicate link – user8675309 Oct 23 '23 at 19:22

0 Answers0