# If $A \in \text{End}(\bigwedge^k \mathbb{R}^d)$ is a complex power, is it a real power up to a sign?

Let $$1 be an integer. Let $$A \in \text{End}(\bigwedge^k \mathbb{R}^d)$$, and suppose that $$A=\bigwedge^k B$$ for some complex $$B \in \text{End}(\mathbb{C}^d)$$.

Does there exist $$M \in \text{End}(\mathbb{R}^d)$$ such that $$A=\bigwedge^k M$$ or $$A=-\bigwedge^k M$$?

More formally, I mean that we have an element $$A \in \text{End}(\bigwedge^k \mathbb{C}^d)$$, such that $$A(\bigwedge^k \mathbb{R}^d) \subseteq \bigwedge^k\mathbb{R}^d$$ (so in this sense $$A$$ is real), and $$A$$ has a complex "root". The question is whether $$A$$ must be a real power up to a sign.

The minus option can occur: Take $$A = -\operatorname{Id}_{\bigwedge^2\mathbb{C}^3}$$; then $$A|_{\bigwedge^2\mathbb{R}^3}=-\operatorname{Id}_{\bigwedge^2\mathbb{R}^3}$$, and $$A=\bigwedge^2 (i\operatorname{Id}_{\mathbb{C}^3})$$. $$A$$ does not admit a "real source".

Here is a possible approach for the invertible case $$A \in \text{GL}$$:

(In that case, if a real source exist, then it is unique up to a sign).

Since $$A=\bigwedge^k B$$, and $$A(\bigwedge^k \mathbb{R}^d) \subseteq \bigwedge^k\mathbb{R}^d$$, $$Bv_1 \wedge \ldots \wedge Bv_k \in \bigwedge^k\mathbb{R}^d$$ for every $$v_1,\ldots,v_k \in \mathbb{R}^d$$.

In other words, $$Bv_1 \wedge \ldots \wedge Bv_k$$ is decomposable in $$\bigwedge^k\mathbb{C}^d$$, and belongs to $$\bigwedge^k\mathbb{R}^d$$.

If this implies that it is also decomposable in $$\bigwedge^k\mathbb{R}^d$$, then $$B$$ is a complex matrix which maps real $$k$$-dimensional subspaces (over $$\mathbb{C}$$) to real subspaces (over $$\mathbb{C}$$).

We need to be careful here:

$$Bv_1 \wedge \ldots \wedge Bv_k$$ is decomposable in $$\bigwedge^k\mathbb{R}^d$$, means that there exist $$w_1,\ldots,w_k \in \mathbb{R}^d$$ such that $$Bv_1 \wedge \ldots \wedge Bv_k =w_1 \wedge \ldots \wedge w_k.$$

Now, thinking on the last equality as an equality of elements in $$\bigwedge^k\mathbb{C}^d$$, we deduce that $$\text{span}_{\mathbb{C}}(Bv_1,\ldots,Bv_k)=\text{span}_{\mathbb{C}}(w_1,\ldots,w_k)$$.

The stronger statement $$\text{span}_{\mathbb{R}}(Bv_1,\ldots,Bv_k)=\text{span}_{\mathbb{R}}(w_1,\ldots,w_k)$$ is false in general! Indeed, take $$B=i\operatorname{Id}_{\mathbb{C}^3}$$ from the example above.

Maybe this fact forces $$B$$ to be a (complex) scalar multiple of a real matrix.

However, I am not sure that that every "real" element which is "complex-decomposable" is also "real-decomposable".