Determinant of an n x n matrix












4














I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?



This is the matrix :



$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$






linear-algebra matrices determinant







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  • 1




    It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
    – Sigur
    7 hours ago












  • Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
    – Clive Newstead
    7 hours ago










  • I am sorry, the link to the picture was not included. I added it now.
    – Paul Vinur
    7 hours ago










  • Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
    – Test123
    6 hours ago
















4














I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?



This is the matrix :



$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$






linear-algebra matrices determinant







share|cite|improve this question









New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 1




    It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
    – Sigur
    7 hours ago












  • Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
    – Clive Newstead
    7 hours ago










  • I am sorry, the link to the picture was not included. I added it now.
    – Paul Vinur
    7 hours ago










  • Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
    – Test123
    6 hours ago














4












4








4







I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?



This is the matrix :



$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$






linear-algebra matrices determinant







share|cite|improve this question









New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I do not know what this kind of matrix is called, it does not really look Circulant, but I tried to do many row and columns operation in order to make it into an upper triangular matrix so the determinant would be the product of the diagonal elements but I couldn't find a way. Any thoughts?



This is the matrix :



$$begin{bmatrix}n&n-1&n-2&cdots&2&1\1&n&n-1&cdots&3&2\1&1&n&cdots&4&3\vdots&vdots&vdots&ddots&vdots&vdots\1&1&1&cdots&n&n-1\1&1&1&cdots&1&lambdaend{bmatrix}$$






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question









New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 7 hours ago









StubbornAtom

5,26711138




5,26711138






New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 7 hours ago









Paul Vinur

335




335




New contributor




Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Paul Vinur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
    – Sigur
    7 hours ago












  • Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
    – Clive Newstead
    7 hours ago










  • I am sorry, the link to the picture was not included. I added it now.
    – Paul Vinur
    7 hours ago










  • Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
    – Test123
    6 hours ago














  • 1




    It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
    – Sigur
    7 hours ago












  • Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
    – Clive Newstead
    7 hours ago










  • I am sorry, the link to the picture was not included. I added it now.
    – Paul Vinur
    7 hours ago










  • Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
    – Test123
    6 hours ago








1




1




It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
7 hours ago






It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
7 hours ago














Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
7 hours ago




Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
7 hours ago












I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
7 hours ago




I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
7 hours ago












Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
6 hours ago




Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
6 hours ago










2 Answers
2






active

oldest

votes


















4














Let $M_n$ be your matrix.



Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you




  1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.

    This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left


  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).

    This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.



After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$



From this, you can deduce



$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$






share|cite|improve this answer























  • This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
    – Paul Vinur
    3 hours ago








  • 2




    @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
    – achille hui
    3 hours ago












  • Thank you so much.
    – Paul Vinur
    3 hours ago



















4














This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.



Then $det(A) = det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.



This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.






share|cite|improve this answer























  • Thanks, that was a typo. It has been corrected.
    – Hans Engler
    1 hour ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














Let $M_n$ be your matrix.



Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you




  1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.

    This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left


  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).

    This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.



After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$



From this, you can deduce



$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$






share|cite|improve this answer























  • This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
    – Paul Vinur
    3 hours ago








  • 2




    @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
    – achille hui
    3 hours ago












  • Thank you so much.
    – Paul Vinur
    3 hours ago
















4














Let $M_n$ be your matrix.



Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you




  1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.

    This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left


  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).

    This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.



After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$



From this, you can deduce



$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$






share|cite|improve this answer























  • This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
    – Paul Vinur
    3 hours ago








  • 2




    @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
    – achille hui
    3 hours ago












  • Thank you so much.
    – Paul Vinur
    3 hours ago














4












4








4






Let $M_n$ be your matrix.



Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you




  1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.

    This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left


  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).

    This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.



After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$



From this, you can deduce



$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$






share|cite|improve this answer














Let $M_n$ be your matrix.



Let $eta_n$ be the $ntimes n$ matrix with entry $1$ at the superdiagonal and $0$ 4 elsewhere. If you




  1. Subtract row $k+1$ from row $k$ for $k = 1,2,ldots,n-1$.

    This is equivalent to multiply $M_n$ by $I_n - eta_n$ from the left


  2. Subtract column $k-1$ from column $k$ for $k = n,n-1,ldots,2$ (notice the order of $k$).

    This is equivalent to multiply $(I_n-eta_n)M_n$ by $I_n - eta_n$ from the right.



After you do this, your matrix simplfies to
$$(I_n - eta_n) M_n (I_n - eta_n) =
begin{bmatrix}
n-1&-n&0&cdots&0&0&0\
0&n-1&-n&cdots&0&0&0\
0&0&n-1&ddots&0&0&0\
vdots&vdots&vdots&ddots&ddots&vdots&vdots\
0&0&0&cdots&n-1&-n&0\
0&0&0&cdots&0&n-1&-lambda\
1&0&0&cdots&0&0&lambda-1
end{bmatrix}$$



From this, you can deduce



$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 4 hours ago

























answered 4 hours ago









achille hui

95.3k5130256




95.3k5130256












  • This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
    – Paul Vinur
    3 hours ago








  • 2




    @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
    – achille hui
    3 hours ago












  • Thank you so much.
    – Paul Vinur
    3 hours ago


















  • This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
    – Paul Vinur
    3 hours ago








  • 2




    @PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
    – achille hui
    3 hours ago












  • Thank you so much.
    – Paul Vinur
    3 hours ago
















This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
3 hours ago






This is obviously the correct way since I checked the final answer. But the element "1" in the bottom left is still there and therefore not all elements (except diagonal and superdiagonal) are zeroes, and correct me if I am wrong, wouldn't that mean this is not a triangular matrix and we won't be able to find the det. by just multiplying the diagonal elements? And also I don't quite understand how we got (n^n-2 * λ)
– Paul Vinur
3 hours ago






2




2




@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
3 hours ago






@PaulVinur This is not a triangular matrix. Since the entries $a_{ij}$ vanishes unless $j = i text{ or } i+1 pmod n$. When you expand the determinant out completely, among all the $n!$ possible terms in the determinant, only two terms survive. i.e those of the form $prod_{i} a_{ii}$ and $prod_{} a_{i,i+1}$ ($i+1$ upto modulo $n$) survive. That's why the final expression is a sum of two terms.
– achille hui
3 hours ago














Thank you so much.
– Paul Vinur
3 hours ago




Thank you so much.
– Paul Vinur
3 hours ago











4














This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.



Then $det(A) = det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.



This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.






share|cite|improve this answer























  • Thanks, that was a typo. It has been corrected.
    – Hans Engler
    1 hour ago
















4














This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.



Then $det(A) = det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.



This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.






share|cite|improve this answer























  • Thanks, that was a typo. It has been corrected.
    – Hans Engler
    1 hour ago














4












4








4






This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.



Then $det(A) = det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.



This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.






share|cite|improve this answer














This is a rank one update of a triangular matrix. Let $A$ be the matrix in the post and let $B$ be the matrix with entries $b_{ij} = a_{ij} - 1$. Let $e$ be the column vector of 1's. Then $A = ee^T + B$.



Then $det(A) = det(B)(1+ e^TB^{-1}e)$. Since $B$ is triangular, $B^{-1}e$ is not hard to find.



This is defined if $lambda ne 1$. For $lambda = 1$, take the limit.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 1 hour ago

























answered 6 hours ago









Hans Engler

10.1k11836




10.1k11836












  • Thanks, that was a typo. It has been corrected.
    – Hans Engler
    1 hour ago


















  • Thanks, that was a typo. It has been corrected.
    – Hans Engler
    1 hour ago
















Thanks, that was a typo. It has been corrected.
– Hans Engler
1 hour ago




Thanks, that was a typo. It has been corrected.
– Hans Engler
1 hour ago










Paul Vinur is a new contributor. Be nice, and check out our Code of Conduct.










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