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Prove AN*=0 where AN=0 and NN*=N*N
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Prove AN*=0
where
AN=0
and
NN*=N*N
(N is normal)
Where N and A is complex matrices
I thought about proving that the columns space of N is the same as the columns space of N* and therefore they both forfill the equation Ax=0
But didn't saw how to
linear-algebra
asked Aug 1 at 10:39
roydouek
12
add a comment |Â
up vote
0
down vote
favorite
Prove AN*=0
where
AN=0
and
NN*=N*N
(N is normal)
Where N and A is complex matrices
I thought about proving that the columns space of N is the same as the columns space of N* and therefore they both forfill the equation Ax=0
But didn't saw how to
linear-algebra
asked Aug 1 at 10:39
roydouek
12
Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove AN*=0
where
AN=0
and
NN*=N*N
(N is normal)
Where N and A is complex matrices
I thought about proving that the columns space of N is the same as the columns space of N* and therefore they both forfill the equation Ax=0
But didn't saw how to
linear-algebra
asked Aug 1 at 10:39
roydouek
12
Prove AN*=0
where
AN=0
and
NN*=N*N
(N is normal)
Where N and A is complex matrices
I thought about proving that the columns space of N is the same as the columns space of N* and therefore they both forfill the equation Ax=0
But didn't saw how to
linear-algebra
asked Aug 1 at 10:39
roydouek
12
edited Aug 1 at 10:45
asked Aug 1 at 10:39
roydouek
12
asked Aug 1 at 10:39
roydouek
12
asked Aug 1 at 10:39
roydouek
12
12
Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31
add a comment |Â
Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31
Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31
add a comment |Â
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Consider two cases: either $N$ is invertible or it isn't. If it is, then $AN=textbf0iff ANN^*=textbf0N^*$.
– Git Gud
Aug 1 at 10:49
If you're over an inner product space, then remember that $$AN^*=0 iff langle AN^*, AN^* rangle = 0$$
– Rab
Aug 1 at 10:51
I didn't quite found the way to prove it yet with your comments, could you go to more details?
– roydouek
Aug 1 at 14:31