A, B and C are matrices. Prove that $A^{T}AB = A^{T}AC$ iff AB = AC
$begingroup$
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
asked Jan 2 at 15:55
BoboBobo
1226
$endgroup$
add a comment |
$begingroup$
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
asked Jan 2 at 15:55
BoboBobo
1226
$endgroup$
$begingroup$
I'd try to show that $A^TA$ and $A$ have the same rank.
$endgroup$
– Lord Shark the Unknown
Jan 2 at 15:59
add a comment |
$begingroup$
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
asked Jan 2 at 15:55
BoboBobo
1226
$endgroup$
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
asked Jan 2 at 15:55
BoboBobo
1226
asked Jan 2 at 15:55
BoboBobo
1226
asked Jan 2 at 15:55
BoboBobo
1226
asked Jan 2 at 15:55
BoboBobo
1226
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
oldest
votes
$begingroup$
Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
$endgroup$
Hint:
$$
A^T(AB-AC)=0qquadRightarrow qquad\
(B-C)^TA^T(AB-AC)=(AB-AC)^T(AB-AC)=0.
$$
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
answered Jan 2 at 16:00
A.Γ.A.Γ.
22.9k32656
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
add a comment |
$begingroup$
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
add a comment |
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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