A, B and C are matrices. Prove that $A^{T}AB = A^{T}AC$ iff AB = AC












1












$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







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  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59
















1












$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







share|cite|improve this question









$endgroup$












  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59














1












1








1





$begingroup$


Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices







share|cite|improve this question









$endgroup$




Let $A in mathbb{R}^{m times n}$, $B in mathbb{R}^{n times p}$, and $C in mathbb{R}^{n times p}$. Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$.



What I have done so far: The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$, then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$, I would have at least one column of $B$ and $C$, say $b_i$ and $c_i$, such that $Ab_i neq Ac_i$, but $A^{T}Ab_i = A^{T}Ac_i$. This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.






linear-algebra matrices




linear-algebra matrices






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share|cite|improve this question











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asked Jan 2 at 15:55









BoboBobo

1226




1226












  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59


















  • $begingroup$
    I'd try to show that $A^TA$ and $A$ have the same rank.
    $endgroup$
    – Lord Shark the Unknown
    Jan 2 at 15:59
















$begingroup$
I'd try to show that $A^TA$ and $A$ have the same rank.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 15:59




$begingroup$
I'd try to show that $A^TA$ and $A$ have the same rank.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 15:59










1 Answer
1






active

oldest

votes


















4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09














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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09


















4












$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09
















4












4








4





$begingroup$

Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$






share|cite|improve this answer









$endgroup$



Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 2 at 16:00









A.Γ.A.Γ.

22.9k32656




22.9k32656












  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09




















  • $begingroup$
    so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
    $endgroup$
    – Bobo
    Jan 2 at 16:06










  • $begingroup$
    @Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
    $endgroup$
    – A.Γ.
    Jan 2 at 16:08












  • $begingroup$
    <v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
    $endgroup$
    – Joel Pereira
    Jan 2 at 16:08










  • $begingroup$
    @Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
    $endgroup$
    – A.Γ.
    Jan 2 at 16:09


















$begingroup$
so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
$endgroup$
– Bobo
Jan 2 at 16:06




$begingroup$
so are you saying that since $langle (AB-AC)x,(AB-AC)xrangle = 0$ for all $x$ that $AB=AC$?
$endgroup$
– Bobo
Jan 2 at 16:06












$begingroup$
@Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
$endgroup$
– A.Γ.
Jan 2 at 16:08






$begingroup$
@Bobo One can do it via that as $langle (AB-AC)x,(AB-AC)xrangle =|(AB-AC)x|^2$. If it is zero then $(AB-AC)x=0$ for all $x$.
$endgroup$
– A.Γ.
Jan 2 at 16:08














$begingroup$
<v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
$endgroup$
– Joel Pereira
Jan 2 at 16:08




$begingroup$
<v,v> = 0 iff v =0. Thus every x is in the Null Space of AB-AC. Thus AB-AC = 0.
$endgroup$
– Joel Pereira
Jan 2 at 16:08












$begingroup$
@Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
$endgroup$
– A.Γ.
Jan 2 at 16:09






$begingroup$
@Bobo Another way is to take the trace $$operatorname{trace}((AB-AC)^T(AB-AC))=|AB-AC|_{text{frobenius}}^2=0.$$
$endgroup$
– A.Γ.
Jan 2 at 16:09




















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