Intuition/Simple Proofs required about $kernel$, $rank$, $co-rank$

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This question might be so elementary , but I love to see some geometric/algebraic approach for these facts concerning transpose matrix. I know some basic thing about DUAL space but those are not satisfying in the way I know.



$1. : rank (A)=rank(A^t)$



$2. : ker (A) = co-image(A^t)$



$3. : co-ker(A)= image(A^t)$



P.S. One may re-state the third fact as



$$Ax=0 , x^ty=0 implies exists z: A^tz=y$$



Is there any simple way just using basic algebra to prove the fact?



Thanks community




linear-algebra matrices soft-question intuition matrix-rank




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  • $Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
    – Peter Melech
    Aug 17 at 8:49














up vote
2
down vote

favorite
2












This question might be so elementary , but I love to see some geometric/algebraic approach for these facts concerning transpose matrix. I know some basic thing about DUAL space but those are not satisfying in the way I know.



$1. : rank (A)=rank(A^t)$



$2. : ker (A) = co-image(A^t)$



$3. : co-ker(A)= image(A^t)$



P.S. One may re-state the third fact as



$$Ax=0 , x^ty=0 implies exists z: A^tz=y$$



Is there any simple way just using basic algebra to prove the fact?



Thanks community




linear-algebra matrices soft-question intuition matrix-rank




share|cite|improve this question






















  • $Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
    – Peter Melech
    Aug 17 at 8:49












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2





This question might be so elementary , but I love to see some geometric/algebraic approach for these facts concerning transpose matrix. I know some basic thing about DUAL space but those are not satisfying in the way I know.



$1. : rank (A)=rank(A^t)$



$2. : ker (A) = co-image(A^t)$



$3. : co-ker(A)= image(A^t)$



P.S. One may re-state the third fact as



$$Ax=0 , x^ty=0 implies exists z: A^tz=y$$



Is there any simple way just using basic algebra to prove the fact?



Thanks community




linear-algebra matrices soft-question intuition matrix-rank




share|cite|improve this question














This question might be so elementary , but I love to see some geometric/algebraic approach for these facts concerning transpose matrix. I know some basic thing about DUAL space but those are not satisfying in the way I know.



$1. : rank (A)=rank(A^t)$



$2. : ker (A) = co-image(A^t)$



$3. : co-ker(A)= image(A^t)$



P.S. One may re-state the third fact as



$$Ax=0 , x^ty=0 implies exists z: A^tz=y$$



Is there any simple way just using basic algebra to prove the fact?



Thanks community





linear-algebra matrices soft-question intuition matrix-rank





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 16:14

























asked Aug 17 at 8:12









Leila

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  • $Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
    – Peter Melech
    Aug 17 at 8:49
















  • $Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
    – Peter Melech
    Aug 17 at 8:49















$Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
– Peter Melech
Aug 17 at 8:49




$Ax=0$ iff $x^tA^t=0$ and thus $ker(A)=im(A^t)^perp$, what is $co$?
– Peter Melech
Aug 17 at 8:49















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