Mixing
Rank of Matrix Determination
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
$X=(I+ab^T)A(I+ba^T)$;
$A$ is symmetric and positive definite matrix of $n times n$. $I$ is the Identity matrix of $n times n$.
$a$ and $b$ are vectors of $n times 1$.
$a.b neq -1$ and $a.b neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $leq 3$
But I am not getting how it can be exactly of rank $2$..
linear-algebra
edited 48 mins ago
Ahmad Bazzi
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
 |Â
show 2 more comments
up vote
3
down vote
favorite
$X=(I+ab^T)A(I+ba^T)$;
$A$ is symmetric and positive definite matrix of $n times n$. $I$ is the Identity matrix of $n times n$.
$a$ and $b$ are vectors of $n times 1$.
$a.b neq -1$ and $a.b neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $leq 3$
But I am not getting how it can be exactly of rank $2$..
linear-algebra
edited 48 mins ago
Ahmad Bazzi
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
1
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
1
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago
 |Â
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$X=(I+ab^T)A(I+ba^T)$;
$A$ is symmetric and positive definite matrix of $n times n$. $I$ is the Identity matrix of $n times n$.
$a$ and $b$ are vectors of $n times 1$.
$a.b neq -1$ and $a.b neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $leq 3$
But I am not getting how it can be exactly of rank $2$..
linear-algebra
edited 48 mins ago
Ahmad Bazzi
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$X=(I+ab^T)A(I+ba^T)$;
$A$ is symmetric and positive definite matrix of $n times n$. $I$ is the Identity matrix of $n times n$.
$a$ and $b$ are vectors of $n times 1$.
$a.b neq -1$ and $a.b neq 0$
$a$ is not parallel to $Ab$
How do we show that $X-A$ is a rank $2$ matrix ?
Efforts:
$$X-A= ab^T A + Aba^T+ ab^T A ba^T$$
Hence each of the terms are having rank $1$. so the sum of all the terms can have rank $leq 3$
But I am not getting how it can be exactly of rank $2$..
linear-algebra
linear-algebra
edited 48 mins ago
Ahmad Bazzi
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 48 mins ago
Ahmad Bazzi
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 48 mins ago
Ahmad Bazzi
6,2761624
edited 48 mins ago
Ahmad Bazzi
6,2761624
edited 48 mins ago
Ahmad Bazzi
6,2761624
6,2761624
asked 58 mins ago
Debasish Jana
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 58 mins ago
Debasish Jana
214
asked 58 mins ago
Debasish Jana
214
214
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Debasish Jana is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
1
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
1
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago
 |Â
show 2 more comments
How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
1
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
1
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago
How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
1
1
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
1
1
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $operatornamerank(PQ) leq minoperatornamerank(P),operatornamerank(Q)$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.
answered 36 mins ago
Omnomnomnom
122k784170
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
add a comment |Â
up vote
3
down vote
$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
beginequation
P = Z(Z^TZ)^-1Z^T
endequation
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
beginequation
Z =
beginbmatrix
a & Ab
endbmatrix
endequation
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
beginequation
Pa = a tag1
endequation
and
beginequation
PAb = Ab tag2
endequation
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
beginequation
(I-P).(X-A) = 0
endequation
where $I-P$ is of rank $n-2$.
answered 34 mins ago
Ahmad Bazzi
6,2761624
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $operatornamerank(PQ) leq minoperatornamerank(P),operatornamerank(Q)$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.
answered 36 mins ago
Omnomnomnom
122k784170
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
add a comment |Â
up vote
3
down vote
accepted
We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $operatornamerank(PQ) leq minoperatornamerank(P),operatornamerank(Q)$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.
answered 36 mins ago
Omnomnomnom
122k784170
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $operatornamerank(PQ) leq minoperatornamerank(P),operatornamerank(Q)$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.
answered 36 mins ago
Omnomnomnom
122k784170
We can write $X - A$ as
$$
X - A = ab^T[A + Aba^T] + Aba^T = ab^TA[I + ba^T] + Aba^T
$$
Noting that $operatornamerank(PQ) leq minoperatornamerank(P),operatornamerank(Q)$, we can see that each term has rank at most $1$, which means that $X - A$ has rank at most $2$.
It now remains to be shown that the rank is not $1$ or $0$.
First, we must show that the first matrix in the sum is non-zero. That is, we wish to show that the product
$$
b^T A[I + ba^T]
$$
is not the zero matrix. To that end, we note that
$$
(b^T A[I + ba^T])b = b^T A[b + b(a^Tb)] = (1 + a^Tb)(b^TAb)b neq 0
$$
From there, it suffices to note that the first term has the span of $a$ as its column space, while the second term has the span of $Ab$ as its column space. Thus, the two column spaces are distinct and one-dimensional. It follows that the sum of the two non-zero rank $1$ matrices must have rank $2$.
answered 36 mins ago
Omnomnomnom
122k784170
edited 27 mins ago
answered 36 mins ago
Omnomnomnom
122k784170
answered 36 mins ago
Omnomnomnom
122k784170
answered 36 mins ago
Omnomnomnom
122k784170
122k784170
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
add a comment |Â
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
now that's a really nice argument ..
– Ahmad Bazzi
28 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
Yours is pretty nifty as well, Ahmad.
– Omnomnomnom
22 mins ago
add a comment |Â
up vote
3
down vote
$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
beginequation
P = Z(Z^TZ)^-1Z^T
endequation
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
beginequation
Z =
beginbmatrix
a & Ab
endbmatrix
endequation
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
beginequation
Pa = a tag1
endequation
and
beginequation
PAb = Ab tag2
endequation
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
beginequation
(I-P).(X-A) = 0
endequation
where $I-P$ is of rank $n-2$.
answered 34 mins ago
Ahmad Bazzi
6,2761624
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
add a comment |Â
up vote
3
down vote
$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
beginequation
P = Z(Z^TZ)^-1Z^T
endequation
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
beginequation
Z =
beginbmatrix
a & Ab
endbmatrix
endequation
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
beginequation
Pa = a tag1
endequation
and
beginequation
PAb = Ab tag2
endequation
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
beginequation
(I-P).(X-A) = 0
endequation
where $I-P$ is of rank $n-2$.
answered 34 mins ago
Ahmad Bazzi
6,2761624
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
beginequation
P = Z(Z^TZ)^-1Z^T
endequation
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
beginequation
Z =
beginbmatrix
a & Ab
endbmatrix
endequation
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
beginequation
Pa = a tag1
endequation
and
beginequation
PAb = Ab tag2
endequation
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
beginequation
(I-P).(X-A) = 0
endequation
where $I-P$ is of rank $n-2$.
answered 34 mins ago
Ahmad Bazzi
6,2761624
$(X-A)$ is spanned by two vectors only
We can solve this using projector matrices, i.e. matrices of the form
beginequation
P = Z(Z^TZ)^-1Z^T
endequation
The number of columns of $X$ will determine the rank of the matrix $X -A $. If we stack in $Z$,
beginequation
Z =
beginbmatrix
a & Ab
endbmatrix
endequation
We are sure that $Z$ is full column rank because $a$ is not parallel to $Ab$, Hence it is easy to see that
beginequation
Pa = a tag1
endequation
and
beginequation
PAb = Ab tag2
endequation
Hence
$$P(X-A) = P(ab^T A + Aba^T+ ab^T A ba^T)$$
which is
$$P(X-A) = Pab^T A + PAba^T+ Pab^T A ba^T$$
Using equations $(1,2)$, we get
$$P(X-A) = ab^T A + Aba^T+ ab^T A ba^T = X-A$$
Hence $X-A$ is spanned by two vectors, i.e. rank $2$.
Therefore $P$ spans a two dimensional space, which is the span of the column space of $X -A$. On the other hand, we can also show that
beginequation
(I-P).(X-A) = 0
endequation
where $I-P$ is of rank $n-2$.
answered 34 mins ago
Ahmad Bazzi
6,2761624
edited 29 mins ago
answered 34 mins ago
Ahmad Bazzi
6,2761624
answered 34 mins ago
Ahmad Bazzi
6,2761624
answered 34 mins ago
Ahmad Bazzi
6,2761624
6,2761624
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
add a comment |Â
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
Perhaps I'm missing something, but I don't see how you've ruled out the possibility that $X - A$ has rank $1$ or $0$.
– Omnomnomnom
15 mins ago
add a comment |Â
Debasish Jana is a new contributor. Be nice, and check out our Code of Conduct.
Debasish Jana is a new contributor. Be nice, and check out our Code of Conduct.
Debasish Jana is a new contributor. Be nice, and check out our Code of Conduct.
Debasish Jana is a new contributor. Be nice, and check out our Code of Conduct.
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How to show that X−A is a rank 2 matrix??
– Debasish Jana
58 mins ago
Welcome to MSE. It is more likely that you will get responses when you share your efforts.
– Ahmad Bazzi
56 mins ago
$X-A= ab^T A + Aba^T+ ab^T A ba^T$ ... Hence each of the terms are having rank 1. so the sum of all the terms can have rank $leq 3$.. But I am not getting how it can be exactly of rank 2..
– Debasish Jana
50 mins ago
1
Also $+1$ for showing your efforts :)
– Ahmad Bazzi
47 mins ago
1
@StammeringMathematician for instance, note that the column space of $ab^T$ is the span of $a$
– Omnomnomnom
35 mins ago