Mixing
Rank of real/imaginary part of a complex matrix of rank 1
$begingroup$
$newcommand{rank}{operatorname{rank}}$
Given a complex matrix $M in mathbb{C}^{p times q}$ of rank $1$.
It appears that $rank Re ( M ) leq 2$ and $rank Im ( M ) leq 2$. If this is true, how to actually prove that?
linear-algebra matrices complex-numbers
edited Mar 13 '18 at 12:36
Henrik
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
$endgroup$
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$
Given a complex matrix $M in mathbb{C}^{p times q}$ of rank $1$.
It appears that $rank Re ( M ) leq 2$ and $rank Im ( M ) leq 2$. If this is true, how to actually prove that?
linear-algebra matrices complex-numbers
edited Mar 13 '18 at 12:36
Henrik
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
$endgroup$
add a comment |
$begingroup$
$newcommand{rank}{operatorname{rank}}$
Given a complex matrix $M in mathbb{C}^{p times q}$ of rank $1$.
It appears that $rank Re ( M ) leq 2$ and $rank Im ( M ) leq 2$. If this is true, how to actually prove that?
linear-algebra matrices complex-numbers
edited Mar 13 '18 at 12:36
Henrik
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
$endgroup$
$newcommand{rank}{operatorname{rank}}$
Given a complex matrix $M in mathbb{C}^{p times q}$ of rank $1$.
It appears that $rank Re ( M ) leq 2$ and $rank Im ( M ) leq 2$. If this is true, how to actually prove that?
linear-algebra matrices complex-numbers
linear-algebra matrices complex-numbers
edited Mar 13 '18 at 12:36
Henrik
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
edited Mar 13 '18 at 12:36
Henrik
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
edited Mar 13 '18 at 12:36
Henrik
6,04592030
edited Mar 13 '18 at 12:36
Henrik
6,04592030
edited Mar 13 '18 at 12:36
Henrik
6,04592030
6,04592030
asked Mar 8 '18 at 9:43
pumanpuman
21
asked Mar 8 '18 at 9:43
pumanpuman
21
asked Mar 8 '18 at 9:43
pumanpuman
21
21
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A:= Re ( M )$ and $B:= Im ( M )$, hence $M=A+iB$. We define$newcommand{rank}{operatorname{rank}}$
$overline{M}:= A-iB$. Then $rank(overline{M})=1$. Thus
$rank(2A)= rank(M+overline{M}) le 1+1=2$. Hence $rank(A) le 2$.
Similar: $rank(B) le 2$
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
Hint : Prove that $overline{M}$ also has rank $1$, and then use the fact that $Re ( M )=frac{M+overline{M}}{2}$ and $Im( M )=frac{M-overline{M}}{2i}$.
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
Let $A:= Re ( M )$ and $B:= Im ( M )$, hence $M=A+iB$. We define$newcommand{rank}{operatorname{rank}}$
$overline{M}:= A-iB$. Then $rank(overline{M})=1$. Thus
$rank(2A)= rank(M+overline{M}) le 1+1=2$. Hence $rank(A) le 2$.
Similar: $rank(B) le 2$
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
Let $A:= Re ( M )$ and $B:= Im ( M )$, hence $M=A+iB$. We define$newcommand{rank}{operatorname{rank}}$
$overline{M}:= A-iB$. Then $rank(overline{M})=1$. Thus
$rank(2A)= rank(M+overline{M}) le 1+1=2$. Hence $rank(A) le 2$.
Similar: $rank(B) le 2$
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
$endgroup$
add a comment |
$begingroup$
Let $A:= Re ( M )$ and $B:= Im ( M )$, hence $M=A+iB$. We define$newcommand{rank}{operatorname{rank}}$
$overline{M}:= A-iB$. Then $rank(overline{M})=1$. Thus
$rank(2A)= rank(M+overline{M}) le 1+1=2$. Hence $rank(A) le 2$.
Similar: $rank(B) le 2$
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
$endgroup$
Let $A:= Re ( M )$ and $B:= Im ( M )$, hence $M=A+iB$. We define$newcommand{rank}{operatorname{rank}}$
$overline{M}:= A-iB$. Then $rank(overline{M})=1$. Thus
$rank(2A)= rank(M+overline{M}) le 1+1=2$. Hence $rank(A) le 2$.
Similar: $rank(B) le 2$
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
edited Jan 30 at 17:47
Martin Sleziak
44.9k10122277
44.9k10122277
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
answered Mar 8 '18 at 10:16
FredFred
48.6k11849
48.6k11849
add a comment |
add a comment |
$begingroup$
Hint : Prove that $overline{M}$ also has rank $1$, and then use the fact that $Re ( M )=frac{M+overline{M}}{2}$ and $Im( M )=frac{M-overline{M}}{2i}$.
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
add a comment |
$begingroup$
Hint : Prove that $overline{M}$ also has rank $1$, and then use the fact that $Re ( M )=frac{M+overline{M}}{2}$ and $Im( M )=frac{M-overline{M}}{2i}$.
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
add a comment |
$begingroup$
Hint : Prove that $overline{M}$ also has rank $1$, and then use the fact that $Re ( M )=frac{M+overline{M}}{2}$ and $Im( M )=frac{M-overline{M}}{2i}$.
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
Hint : Prove that $overline{M}$ also has rank $1$, and then use the fact that $Re ( M )=frac{M+overline{M}}{2}$ and $Im( M )=frac{M-overline{M}}{2i}$.
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
answered Mar 8 '18 at 10:12
Arnaud D.Arnaud D.
16.2k52445
16.2k52445
add a comment |
add a comment |
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