Proof Verification: the orthogonal complement of the column space is the left nullspace
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$begingroup$
Can someone please check my proof and my definitions.
Let $A in mathbb{R}^{n times m}$ be my matrix.
The left null space of $A$ is written as,
$mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$
The orthogonal complement of the column space $mathcal{C}(A)$ is written as,
$mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$
We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $
First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$
Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.
Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$
Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.
I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
linear-algebra matrices proof-verification vector-spaces linear-transformations
$endgroup$
add a comment |
$begingroup$
Can someone please check my proof and my definitions.
Let $A in mathbb{R}^{n times m}$ be my matrix.
The left null space of $A$ is written as,
$mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$
The orthogonal complement of the column space $mathcal{C}(A)$ is written as,
$mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$
We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $
First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$
Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.
Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$
Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.
I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
linear-algebra matrices proof-verification vector-spaces linear-transformations
$endgroup$
add a comment |
$begingroup$
Can someone please check my proof and my definitions.
Let $A in mathbb{R}^{n times m}$ be my matrix.
The left null space of $A$ is written as,
$mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$
The orthogonal complement of the column space $mathcal{C}(A)$ is written as,
$mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$
We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $
First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$
Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.
Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$
Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.
I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
linear-algebra matrices proof-verification vector-spaces linear-transformations
$endgroup$
Can someone please check my proof and my definitions.
Let $A in mathbb{R}^{n times m}$ be my matrix.
The left null space of $A$ is written as,
$mathcal{N}(A^top) = {x in mathbb{R}^n| A^top x = 0}$
The orthogonal complement of the column space $mathcal{C}(A)$ is written as,
$mathcal{C}(A)^perp = {x in mathbb{R}^n | x^top y = 0, forall y in mathcal{C}(A)}$
We want to show that $mathcal{N}(A^top) = mathcal{C}(A)^perp $
First, we show, $mathcal{N}(A^top) subseteq mathcal{C}(A)^perp$
Let $x in mathcal{N}(A^top)$, then $A^top x = 0 implies x^top A = 0^top implies x^top Av= 0^top v, forall v in mathcal{C}(A) implies x^top y = 0 , y = Av$, $implies x in C(A)^perp$.
Next, we show, $mathcal{N}(A^top) supseteq mathcal{C}(A)^perp$
Let $x in C(A)^perp$, then $x^top y = 0$, forall $y in C(A)$. But $y = Av, forall v in mathbb{R}^n$. Hence, $x^top y = x^top Av = v^top A^top x.$ For all $v neq 0, A^top x = 0$, hence $x in mathcal{N}(A^top)$.
I'm pretty confident about the first proof. But the second proof is a bit more rough. Can someone please check for me.
linear-algebra matrices proof-verification vector-spaces linear-transformations
linear-algebra matrices proof-verification vector-spaces linear-transformations
edited Jan 25 at 16:56
Aåkon
asked Jan 25 at 16:13
AåkonAåkon
4,87131759
4,87131759
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$begingroup$
$y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.
So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.
$endgroup$
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$begingroup$
$y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.
So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.
$endgroup$
add a comment |
$begingroup$
$y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.
So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.
$endgroup$
add a comment |
$begingroup$
$y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.
So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.
$endgroup$
$y in C(A)$ means that there exists (at least one) $v$ of appropriate dimension such that $y = Av$.
So we can say: For $x in C(A)^{perp}$, then $x^T y = 0$ for every $y in C(A)$. For every $y in C(A)$, we can express $y = Av$ for some (nonzero) $v$. So we can always express $x^T y$ as $x^T Av$. So $$x^T y = x^T (A v) = (x^T A) v = (A^T x)^T v = 0^T v = 0$$ for $v neq 0$, so we must have $A^T x = 0$, i.e., $x in N(A^T)$.
edited Jan 25 at 17:11
answered Jan 25 at 16:51
jjjjjjjjjjjj
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1,204516
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