Mixing
If $A in text{End}(bigwedge^k mathbb{R}^d)$ is a complex power, is it a real power up to a sign?
$begingroup$
Let $1<k<d$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$ (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 $A = -operatorname{Id}_{bigwedge^2mathbb{C}^3}$; then $A|_{bigwedge^2mathbb{R}^3}=-operatorname{Id}_{bigwedge^2mathbb{R}^3}$, and $A=bigwedge^2 (ioperatorname{Id}_{mathbb{C}^3})$. $A$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$, $Bv_1 wedge ldots wedge Bv_k in bigwedge^kmathbb{R}^d$ 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^kmathbb{C}^d$, and belongs to $bigwedge^kmathbb{R}^d$.
If this implies that it is also decomposable in $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{C}^d$, 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 $B=ioperatorname{Id}_{mathbb{C}^3}$ 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".
complex-numbers complex-geometry multilinear-algebra exterior-algebra grassmannian
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
$endgroup$
add a comment |
$begingroup$
Let $1<k<d$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$ (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 $A = -operatorname{Id}_{bigwedge^2mathbb{C}^3}$; then $A|_{bigwedge^2mathbb{R}^3}=-operatorname{Id}_{bigwedge^2mathbb{R}^3}$, and $A=bigwedge^2 (ioperatorname{Id}_{mathbb{C}^3})$. $A$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$, $Bv_1 wedge ldots wedge Bv_k in bigwedge^kmathbb{R}^d$ 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^kmathbb{C}^d$, and belongs to $bigwedge^kmathbb{R}^d$.
If this implies that it is also decomposable in $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{C}^d$, 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 $B=ioperatorname{Id}_{mathbb{C}^3}$ 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".
complex-numbers complex-geometry multilinear-algebra exterior-algebra grassmannian
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
$endgroup$
add a comment |
$begingroup$
Let $1<k<d$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$ (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 $A = -operatorname{Id}_{bigwedge^2mathbb{C}^3}$; then $A|_{bigwedge^2mathbb{R}^3}=-operatorname{Id}_{bigwedge^2mathbb{R}^3}$, and $A=bigwedge^2 (ioperatorname{Id}_{mathbb{C}^3})$. $A$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$, $Bv_1 wedge ldots wedge Bv_k in bigwedge^kmathbb{R}^d$ 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^kmathbb{C}^d$, and belongs to $bigwedge^kmathbb{R}^d$.
If this implies that it is also decomposable in $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{C}^d$, 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 $B=ioperatorname{Id}_{mathbb{C}^3}$ 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".
complex-numbers complex-geometry multilinear-algebra exterior-algebra grassmannian
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
$endgroup$
Let $1<k<d$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$ (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 $A = -operatorname{Id}_{bigwedge^2mathbb{C}^3}$; then $A|_{bigwedge^2mathbb{R}^3}=-operatorname{Id}_{bigwedge^2mathbb{R}^3}$, and $A=bigwedge^2 (ioperatorname{Id}_{mathbb{C}^3})$. $A$ 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 $A(bigwedge^k mathbb{R}^d) subseteq bigwedge^kmathbb{R}^d$, $Bv_1 wedge ldots wedge Bv_k in bigwedge^kmathbb{R}^d$ 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^kmathbb{C}^d$, and belongs to $bigwedge^kmathbb{R}^d$.
If this implies that it is also decomposable in $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{R}^d$, 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 $bigwedge^kmathbb{C}^d$, 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 $B=ioperatorname{Id}_{mathbb{C}^3}$ 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".
complex-numbers complex-geometry multilinear-algebra exterior-algebra grassmannian
complex-numbers complex-geometry multilinear-algebra exterior-algebra grassmannian
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
edited Jan 29 at 10:36
Asaf Shachar
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
asked Jan 29 at 8:54
Asaf ShacharAsaf Shachar
5,77031145
5,77031145
add a comment |
add a comment |
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