Mixing
Why $det_mathcal B_n(e_1,…,e_p,d_p+1,d_p+2+c_p+2,…,c_n+d_n)=0$?
Clash Royale CLAN TAG#URR8PPP
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The aim of the exercise is to prove that $$detbeginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$$
where $Bin mathbb R^ptimes p$, $Cin mathbb R^qtimes q$, $n=p+q$ and $M:=beginpmatrixB&D\0&Cendpmatrixinmathbb R^ntimes n$.
Let $mathcal B_n(e_1,...,e_n)$ the canonical basis of $mathbb R^n$, $E=Spane_1,...,e_p$ and $E'=Spane_p+1,...,e_n$. Therefore $mathbb R^n=Eoplus E'$.
Set $M=(m_j)_1leq jleq n$ where $m_j$ are the column of $M$ and are defined as $m_j=b_j+0$ if $1leq jleq p$ and $m_j=d_j+c_j$ if $p+1leq jleq n$. Therefore
$$det(M)=det_mathcal B_n(m_1,...,m_n)=det_mathcal B_n(b_1,...,b_p,c_p+1+d_p+1,...,c_n+d_n).$$
Let $$alpha :E^pto mathbb R$$
defined as $$alpha (x_1,...,x_p)=det_mathcal B_n(x_1,...,x_p,c_p+1+d_p+1,...,c_n+d_n).$$
Recall that $e_1,...,e_p$ is a basis of $E$. Therefore, by linearity :
$$alpha (e_1,...,e_p)=det_mathcal B_n(e_1,...,e_p,c_p+1,...,c_n+d_n)+underbracedet_mathcal B_n(e_1,...,e_p,d_p+1,...,c_n+d_n)_=0,$$
but I don't understand why the last term is $0$.
linear-algebra determinant
asked Aug 2 at 16:41
Peter
326112
add a comment |Â
up vote
2
down vote
favorite
The aim of the exercise is to prove that $$detbeginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$$
where $Bin mathbb R^ptimes p$, $Cin mathbb R^qtimes q$, $n=p+q$ and $M:=beginpmatrixB&D\0&Cendpmatrixinmathbb R^ntimes n$.
Let $mathcal B_n(e_1,...,e_n)$ the canonical basis of $mathbb R^n$, $E=Spane_1,...,e_p$ and $E'=Spane_p+1,...,e_n$. Therefore $mathbb R^n=Eoplus E'$.
Set $M=(m_j)_1leq jleq n$ where $m_j$ are the column of $M$ and are defined as $m_j=b_j+0$ if $1leq jleq p$ and $m_j=d_j+c_j$ if $p+1leq jleq n$. Therefore
$$det(M)=det_mathcal B_n(m_1,...,m_n)=det_mathcal B_n(b_1,...,b_p,c_p+1+d_p+1,...,c_n+d_n).$$
Let $$alpha :E^pto mathbb R$$
defined as $$alpha (x_1,...,x_p)=det_mathcal B_n(x_1,...,x_p,c_p+1+d_p+1,...,c_n+d_n).$$
Recall that $e_1,...,e_p$ is a basis of $E$. Therefore, by linearity :
$$alpha (e_1,...,e_p)=det_mathcal B_n(e_1,...,e_p,c_p+1,...,c_n+d_n)+underbracedet_mathcal B_n(e_1,...,e_p,d_p+1,...,c_n+d_n)_=0,$$
but I don't understand why the last term is $0$.
linear-algebra determinant
asked Aug 2 at 16:41
Peter
326112
1
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The aim of the exercise is to prove that $$detbeginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$$
where $Bin mathbb R^ptimes p$, $Cin mathbb R^qtimes q$, $n=p+q$ and $M:=beginpmatrixB&D\0&Cendpmatrixinmathbb R^ntimes n$.
Let $mathcal B_n(e_1,...,e_n)$ the canonical basis of $mathbb R^n$, $E=Spane_1,...,e_p$ and $E'=Spane_p+1,...,e_n$. Therefore $mathbb R^n=Eoplus E'$.
Set $M=(m_j)_1leq jleq n$ where $m_j$ are the column of $M$ and are defined as $m_j=b_j+0$ if $1leq jleq p$ and $m_j=d_j+c_j$ if $p+1leq jleq n$. Therefore
$$det(M)=det_mathcal B_n(m_1,...,m_n)=det_mathcal B_n(b_1,...,b_p,c_p+1+d_p+1,...,c_n+d_n).$$
Let $$alpha :E^pto mathbb R$$
defined as $$alpha (x_1,...,x_p)=det_mathcal B_n(x_1,...,x_p,c_p+1+d_p+1,...,c_n+d_n).$$
Recall that $e_1,...,e_p$ is a basis of $E$. Therefore, by linearity :
$$alpha (e_1,...,e_p)=det_mathcal B_n(e_1,...,e_p,c_p+1,...,c_n+d_n)+underbracedet_mathcal B_n(e_1,...,e_p,d_p+1,...,c_n+d_n)_=0,$$
but I don't understand why the last term is $0$.
linear-algebra determinant
asked Aug 2 at 16:41
Peter
326112
The aim of the exercise is to prove that $$detbeginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$$
where $Bin mathbb R^ptimes p$, $Cin mathbb R^qtimes q$, $n=p+q$ and $M:=beginpmatrixB&D\0&Cendpmatrixinmathbb R^ntimes n$.
Let $mathcal B_n(e_1,...,e_n)$ the canonical basis of $mathbb R^n$, $E=Spane_1,...,e_p$ and $E'=Spane_p+1,...,e_n$. Therefore $mathbb R^n=Eoplus E'$.
Set $M=(m_j)_1leq jleq n$ where $m_j$ are the column of $M$ and are defined as $m_j=b_j+0$ if $1leq jleq p$ and $m_j=d_j+c_j$ if $p+1leq jleq n$. Therefore
$$det(M)=det_mathcal B_n(m_1,...,m_n)=det_mathcal B_n(b_1,...,b_p,c_p+1+d_p+1,...,c_n+d_n).$$
Let $$alpha :E^pto mathbb R$$
defined as $$alpha (x_1,...,x_p)=det_mathcal B_n(x_1,...,x_p,c_p+1+d_p+1,...,c_n+d_n).$$
Recall that $e_1,...,e_p$ is a basis of $E$. Therefore, by linearity :
$$alpha (e_1,...,e_p)=det_mathcal B_n(e_1,...,e_p,c_p+1,...,c_n+d_n)+underbracedet_mathcal B_n(e_1,...,e_p,d_p+1,...,c_n+d_n)_=0,$$
but I don't understand why the last term is $0$.
linear-algebra determinant
asked Aug 2 at 16:41
Peter
326112
asked Aug 2 at 16:41
Peter
326112
asked Aug 2 at 16:41
Peter
326112
asked Aug 2 at 16:41
Peter
326112
326112
1
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23
add a comment |Â
1
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23
1
1
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Here is another way:
Note that if $C$ is invertible we have
$beginbmatrix B & D \ 0 & Cendbmatrix = beginbmatrix I & DC^-1 \ 0 & Iendbmatrix beginbmatrix B & 0 \ 0 & Cendbmatrix$ and so we see that
$det beginbmatrix B & D \ 0 & Cendbmatrix = = det B det C$.
Since this is true for all invertible $C$, $det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.
answered Aug 2 at 17:02
copper.hat
122k557155
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
add a comment |Â
up vote
0
down vote
As Mike Hawk mentioned $d_p+1$ is a linear combination of $(e_i)_1le ile p$ since it is an element of $mathbb R^p$ which has as base $(e_i)_1le ile p$
answered Aug 2 at 17:06
Ahmed Lazhar
618
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Here is another way:
Note that if $C$ is invertible we have
$beginbmatrix B & D \ 0 & Cendbmatrix = beginbmatrix I & DC^-1 \ 0 & Iendbmatrix beginbmatrix B & 0 \ 0 & Cendbmatrix$ and so we see that
$det beginbmatrix B & D \ 0 & Cendbmatrix = = det B det C$.
Since this is true for all invertible $C$, $det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.
answered Aug 2 at 17:02
copper.hat
122k557155
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
add a comment |Â
up vote
2
down vote
Here is another way:
Note that if $C$ is invertible we have
$beginbmatrix B & D \ 0 & Cendbmatrix = beginbmatrix I & DC^-1 \ 0 & Iendbmatrix beginbmatrix B & 0 \ 0 & Cendbmatrix$ and so we see that
$det beginbmatrix B & D \ 0 & Cendbmatrix = = det B det C$.
Since this is true for all invertible $C$, $det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.
answered Aug 2 at 17:02
copper.hat
122k557155
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is another way:
Note that if $C$ is invertible we have
$beginbmatrix B & D \ 0 & Cendbmatrix = beginbmatrix I & DC^-1 \ 0 & Iendbmatrix beginbmatrix B & 0 \ 0 & Cendbmatrix$ and so we see that
$det beginbmatrix B & D \ 0 & Cendbmatrix = = det B det C$.
Since this is true for all invertible $C$, $det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.
answered Aug 2 at 17:02
copper.hat
122k557155
Here is another way:
Note that if $C$ is invertible we have
$beginbmatrix B & D \ 0 & Cendbmatrix = beginbmatrix I & DC^-1 \ 0 & Iendbmatrix beginbmatrix B & 0 \ 0 & Cendbmatrix$ and so we see that
$det beginbmatrix B & D \ 0 & Cendbmatrix = = det B det C$.
Since this is true for all invertible $C$, $det$ is continuous and the invertible matrices are dense we see that it is true for all $C$.
answered Aug 2 at 17:02
copper.hat
122k557155
answered Aug 2 at 17:02
copper.hat
122k557155
answered Aug 2 at 17:02
copper.hat
122k557155
answered Aug 2 at 17:02
copper.hat
122k557155
122k557155
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
add a comment |Â
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
This doesn't answer my question. By the why, your hint is equivalent to prove that $det beginpmatrixB&D\0&Cendpmatrix=det(B)det(C)$, so I don't really see in what it helps.
– Peter
Aug 2 at 17:15
Why the downvote?
– copper.hat
Aug 2 at 17:18
Why the downvote?
– copper.hat
Aug 2 at 17:18
1
1
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
@Peter: The computation of $det$ for the first matrix is straightforward since there is exactly one permutation that needs to be examined. The computation for the second is also straightforward for similar reasons.
– copper.hat
Aug 2 at 17:22
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
I did downvoted (please don't take it against you). I'm not criticizing your mathematical skills that are very huge. The reason for my downvote it's that proving your hint is almost equivalent to prove what I asked (even if it's more straightforward, it's almost the same)
– Peter
Aug 2 at 22:54
add a comment |Â
up vote
0
down vote
As Mike Hawk mentioned $d_p+1$ is a linear combination of $(e_i)_1le ile p$ since it is an element of $mathbb R^p$ which has as base $(e_i)_1le ile p$
answered Aug 2 at 17:06
Ahmed Lazhar
618
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
add a comment |Â
up vote
0
down vote
As Mike Hawk mentioned $d_p+1$ is a linear combination of $(e_i)_1le ile p$ since it is an element of $mathbb R^p$ which has as base $(e_i)_1le ile p$
answered Aug 2 at 17:06
Ahmed Lazhar
618
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As Mike Hawk mentioned $d_p+1$ is a linear combination of $(e_i)_1le ile p$ since it is an element of $mathbb R^p$ which has as base $(e_i)_1le ile p$
answered Aug 2 at 17:06
Ahmed Lazhar
618
As Mike Hawk mentioned $d_p+1$ is a linear combination of $(e_i)_1le ile p$ since it is an element of $mathbb R^p$ which has as base $(e_i)_1le ile p$
answered Aug 2 at 17:06
Ahmed Lazhar
618
answered Aug 2 at 17:06
Ahmed Lazhar
618
answered Aug 2 at 17:06
Ahmed Lazhar
618
answered Aug 2 at 17:06
Ahmed Lazhar
618
618
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
add a comment |Â
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
First, you just copy a comment (that is not very well see). Then, how do you know that $d_p+1$ is a linear combinaison of $(e_i)_1leq ileq p$ ?
– Peter
Aug 2 at 17:14
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
Yes I know, I did that since I am new and I cannot comment, so it was my only solution. As I told you $(e_i)_ile p$ is a base of $mathbb R^p$. And $d_p+1$ is in $mathbb R^p$
– Ahmed Lazhar
Aug 2 at 17:25
1
1
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
+1: This is a perfectly good answer, I do not understand the downvote.
– copper.hat
Aug 2 at 17:25
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
@copper.hat: In fact I did downvote, but I forgot that less 50 reputation can't comment. I tried to cancel my downvote, but it doesn't work...
– Peter
Aug 2 at 22:35
1
1
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
Sorry for the downvote, I forgot the fact that you can't comment. And thank you for your answer.
– Peter
Aug 2 at 22:36
add a comment |Â
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1
d_p+1 is a linear combination of e_i
– Mike Hawk
Aug 2 at 16:45
@MikeHawk: Thank you but how do you know that ?
– Peter
Aug 2 at 16:46
Petulant downvoting is not a great strategy. Certainly does not encourage one to help you.
– copper.hat
Aug 2 at 17:23