Determinant of an n x n matrix
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
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
add a comment |
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
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08
add a comment |
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
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
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
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
edited Dec 27 '18 at 13:44
StubbornAtom
5,34411138
5,34411138
asked Dec 27 '18 at 13:34
Paul Vinur
416
asked Dec 27 '18 at 13:34
Paul Vinur
416
asked Dec 27 '18 at 13:34
Paul Vinur
416
416
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08
add a comment |
1
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08
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
Dec 27 '18 at 13:37
It is not clear what you need/want since you even didn't show us the matrix! Try to clarify your question.
– Sigur
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08
add a comment |
2 Answers
2
active
oldest
votes
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
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 leftSubtract 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 simplifies 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
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
add a comment |
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.
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
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 leftSubtract 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 simplifies 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
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
add a comment |
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
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 leftSubtract 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 simplifies 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
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
add a comment |
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
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 leftSubtract 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 simplifies 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
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
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
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 leftSubtract 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 simplifies 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
$$det[M_n] = det[(I_n - eta_n)M_n(I_n - eta_n)]
= (n-1)^{n-1}(lambda-1) + n^{n-2}lambda$$
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
edited Dec 30 '18 at 9:54
Martin Sleziak
44.7k7115270
44.7k7115270
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
answered Dec 27 '18 at 16:20
achille hui
95.5k5130256
95.5k5130256
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
add a comment |
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:07
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
Dec 27 '18 at 17:12
@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
Dec 27 '18 at 17:12
add a comment |
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.
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
add a comment |
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.
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
add a comment |
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.
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
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.
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
edited Dec 27 '18 at 19:39
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
answered Dec 27 '18 at 14:22
Hans Engler
10.1k11836
10.1k11836
add a comment |
add a comment |
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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
Dec 27 '18 at 13:37
Do you have a particular matrix in mind, or just an arbitrary $n times n$ matrix?
– Clive Newstead
Dec 27 '18 at 13:37
I am sorry, the link to the picture was not included. I added it now.
– Paul Vinur
Dec 27 '18 at 13:41
Have you tried expanding over the first column or row for example or calculate the determinant for small values of $n$?
– Test123
Dec 27 '18 at 14:08