Mixing
Advice on showing this matrix equality
$begingroup$
Let $A$ be an $n times m$ matrix where the first collumn of $A$ is a $1$ vector
So that $A_{i,1}=1 $for $i = 1,2,dots,n$.
Define $G$ as a generalised inverse of the matrix $A^T A$, so that
$$A^T A G A^T A = A^T A$$,
Further define $B$ as the $ntimes n$ matrix of $1$'s. So that $B_{i,j}=1$ for all $i,j$.
My problem is to show that
$$AGA^TB = B$$
I already proved by properties of $G$ that
$$A G A^T A = A$$
And so, at least to me, I think I need to use this fact combined with the fact the first collumn of $A$ is the $1$ vector to show
$$AGA^T B = B$$
I was wondering if it would be right to partition $A$ as
$$A = [1 space C]$$
for some matrix $C$. Then, assuming this is a valid use of partitioning (??), this would give
$$AGA^T [1 space C] = [1 space C]$$
And hence
$$AGA^T 1 = 1$$
$$A G A^T 11^T = 11^T$$
$$AGA^T B = B$$
But I feel like is not valid. Any advice?
linear-algebra matrices
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $A$ be an $n times m$ matrix where the first collumn of $A$ is a $1$ vector
So that $A_{i,1}=1 $for $i = 1,2,dots,n$.
Define $G$ as a generalised inverse of the matrix $A^T A$, so that
$$A^T A G A^T A = A^T A$$,
Further define $B$ as the $ntimes n$ matrix of $1$'s. So that $B_{i,j}=1$ for all $i,j$.
My problem is to show that
$$AGA^TB = B$$
I already proved by properties of $G$ that
$$A G A^T A = A$$
And so, at least to me, I think I need to use this fact combined with the fact the first collumn of $A$ is the $1$ vector to show
$$AGA^T B = B$$
I was wondering if it would be right to partition $A$ as
$$A = [1 space C]$$
for some matrix $C$. Then, assuming this is a valid use of partitioning (??), this would give
$$AGA^T [1 space C] = [1 space C]$$
And hence
$$AGA^T 1 = 1$$
$$A G A^T 11^T = 11^T$$
$$AGA^T B = B$$
But I feel like is not valid. Any advice?
linear-algebra matrices
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
$endgroup$
$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47
add a comment |
$begingroup$
Let $A$ be an $n times m$ matrix where the first collumn of $A$ is a $1$ vector
So that $A_{i,1}=1 $for $i = 1,2,dots,n$.
Define $G$ as a generalised inverse of the matrix $A^T A$, so that
$$A^T A G A^T A = A^T A$$,
Further define $B$ as the $ntimes n$ matrix of $1$'s. So that $B_{i,j}=1$ for all $i,j$.
My problem is to show that
$$AGA^TB = B$$
I already proved by properties of $G$ that
$$A G A^T A = A$$
And so, at least to me, I think I need to use this fact combined with the fact the first collumn of $A$ is the $1$ vector to show
$$AGA^T B = B$$
I was wondering if it would be right to partition $A$ as
$$A = [1 space C]$$
for some matrix $C$. Then, assuming this is a valid use of partitioning (??), this would give
$$AGA^T [1 space C] = [1 space C]$$
And hence
$$AGA^T 1 = 1$$
$$A G A^T 11^T = 11^T$$
$$AGA^T B = B$$
But I feel like is not valid. Any advice?
linear-algebra matrices
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
$endgroup$
Let $A$ be an $n times m$ matrix where the first collumn of $A$ is a $1$ vector
So that $A_{i,1}=1 $for $i = 1,2,dots,n$.
Define $G$ as a generalised inverse of the matrix $A^T A$, so that
$$A^T A G A^T A = A^T A$$,
Further define $B$ as the $ntimes n$ matrix of $1$'s. So that $B_{i,j}=1$ for all $i,j$.
My problem is to show that
$$AGA^TB = B$$
I already proved by properties of $G$ that
$$A G A^T A = A$$
And so, at least to me, I think I need to use this fact combined with the fact the first collumn of $A$ is the $1$ vector to show
$$AGA^T B = B$$
I was wondering if it would be right to partition $A$ as
$$A = [1 space C]$$
for some matrix $C$. Then, assuming this is a valid use of partitioning (??), this would give
$$AGA^T [1 space C] = [1 space C]$$
And hence
$$AGA^T 1 = 1$$
$$A G A^T 11^T = 11^T$$
$$AGA^T B = B$$
But I feel like is not valid. Any advice?
linear-algebra matrices
linear-algebra matrices
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
edited Jan 14 at 13:46
Xiaomi
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
asked Jan 14 at 13:40
XiaomiXiaomi
1,057115
1,057115
$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47
add a comment |
$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47
$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
When computing a general matrix product $Z = XY$, any column of $Z$ depends only on the left factor $X$ and the corresponding column of the right factor $Y$.
Now apply this to the product $AGA^T cdot A = A$.
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When computing a general matrix product $Z = XY$, any column of $Z$ depends only on the left factor $X$ and the corresponding column of the right factor $Y$.
Now apply this to the product $AGA^T cdot A = A$.
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
add a comment |
$begingroup$
When computing a general matrix product $Z = XY$, any column of $Z$ depends only on the left factor $X$ and the corresponding column of the right factor $Y$.
Now apply this to the product $AGA^T cdot A = A$.
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
$endgroup$
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
add a comment |
$begingroup$
When computing a general matrix product $Z = XY$, any column of $Z$ depends only on the left factor $X$ and the corresponding column of the right factor $Y$.
Now apply this to the product $AGA^T cdot A = A$.
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
$endgroup$
When computing a general matrix product $Z = XY$, any column of $Z$ depends only on the left factor $X$ and the corresponding column of the right factor $Y$.
Now apply this to the product $AGA^T cdot A = A$.
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
answered Jan 14 at 13:58
Hans EnglerHans Engler
10.5k11836
10.5k11836
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
add a comment |
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
$begingroup$
Thanks a lot, very simple but clear explanation
$endgroup$
– Xiaomi
Jan 15 at 0:59
add a comment |
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$begingroup$
What is $J$? Is $J=B$?
$endgroup$
– Song
Jan 14 at 13:43
$begingroup$
Sorry, I accidently changed notation midway through. Corrected now
$endgroup$
– Xiaomi
Jan 14 at 13:47