The definition of orthogonal complement in the column space
$begingroup$
Denote $A^bot$ is the matrix satisfied $A'A^bot=0$ with the highest rank. Proof that:
(1)$I-(A')^-A'$ is a $A^bot$, here $A^-$ means pseudo inverse.
(2)$M(A^bot)=M(A)^bot$. $M(A)$ is the column space of matrix $A$.
In the first question, it is obvious that $A' (I-(A')^-A')=0$. But how to proof $I-(A')^-A'$ has the highest rank?
In the second question, the common way may be to proof $alphain M(A^bot)Longleftrightarrow alphain M(A)^bot$. Then I don't know how to go on since the definition of $A^bot$ is abstract to me.
linear-algebra orthogonality pseudoinverse
asked Dec 25 '18 at 11:17
breggbregg
235
$endgroup$
add a comment |
$begingroup$
Denote $A^bot$ is the matrix satisfied $A'A^bot=0$ with the highest rank. Proof that:
(1)$I-(A')^-A'$ is a $A^bot$, here $A^-$ means pseudo inverse.
(2)$M(A^bot)=M(A)^bot$. $M(A)$ is the column space of matrix $A$.
In the first question, it is obvious that $A' (I-(A')^-A')=0$. But how to proof $I-(A')^-A'$ has the highest rank?
In the second question, the common way may be to proof $alphain M(A^bot)Longleftrightarrow alphain M(A)^bot$. Then I don't know how to go on since the definition of $A^bot$ is abstract to me.
linear-algebra orthogonality pseudoinverse
asked Dec 25 '18 at 11:17
breggbregg
235
$endgroup$
add a comment |
$begingroup$
Denote $A^bot$ is the matrix satisfied $A'A^bot=0$ with the highest rank. Proof that:
(1)$I-(A')^-A'$ is a $A^bot$, here $A^-$ means pseudo inverse.
(2)$M(A^bot)=M(A)^bot$. $M(A)$ is the column space of matrix $A$.
In the first question, it is obvious that $A' (I-(A')^-A')=0$. But how to proof $I-(A')^-A'$ has the highest rank?
In the second question, the common way may be to proof $alphain M(A^bot)Longleftrightarrow alphain M(A)^bot$. Then I don't know how to go on since the definition of $A^bot$ is abstract to me.
linear-algebra orthogonality pseudoinverse
asked Dec 25 '18 at 11:17
breggbregg
235
$endgroup$
Denote $A^bot$ is the matrix satisfied $A'A^bot=0$ with the highest rank. Proof that:
(1)$I-(A')^-A'$ is a $A^bot$, here $A^-$ means pseudo inverse.
(2)$M(A^bot)=M(A)^bot$. $M(A)$ is the column space of matrix $A$.
In the first question, it is obvious that $A' (I-(A')^-A')=0$. But how to proof $I-(A')^-A'$ has the highest rank?
In the second question, the common way may be to proof $alphain M(A^bot)Longleftrightarrow alphain M(A)^bot$. Then I don't know how to go on since the definition of $A^bot$ is abstract to me.
linear-algebra orthogonality pseudoinverse
linear-algebra orthogonality pseudoinverse
asked Dec 25 '18 at 11:17
breggbregg
235
asked Dec 25 '18 at 11:17
breggbregg
235
edited Dec 26 '18 at 0:56
bregg
asked Dec 25 '18 at 11:17
breggbregg
235
asked Dec 25 '18 at 11:17
breggbregg
235
asked Dec 25 '18 at 11:17
breggbregg
235
235
add a comment |
add a comment |
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