Non determinant -class operators satisfing $A^2+I=0$

Question

Let $H$ be a complex Hilbert space and $A\in B(H)$ satisfies $A^2+I=0$. Does this implies that $A$ can be written in the form $A=\lambda I +T$ where $T$ is a trace classe operator and $\lambda$ is a complex scalar?

Answer 1

I think not: Define $A:l^2\to l^2$ by $$ Ax = i (x_1,-x_2,x_3,-x_4,\dots). $$ Then $A^2=-I$, but $A-\lambda I$ does not have finite trace for all $\lambda$.