Determinant of the $ntimes n$ matrix whose $jk$-th entry is $1/min{j,k}$












0














So on my exam I got a True/False question that asked the following:




For any $ninmathbb N$, compute the determinant of the matrix



begin{bmatrix}
1&1&1&cdots&1
\1&1/2&1/2&cdots&1/2
\1&1/2&1/3&cdots&1/3
\vdots&vdots&vdots&ddots&vdots
\1&1/2&1/3&cdots&1/n
end{bmatrix}



This is the $ntimes n$ matrix whose $jk$-th entry is $1/min{j,k}$.




I recognize there has to be some formula that makes calculating the determinant easier, but I'm confused as to what it is.






linear-algebra matrices determinant examples-counterexamples







share|cite|improve this question




















  • 1




    One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
    – Servaes
    Nov 29 at 13:53












  • I don't actually, would you care to explain how that works.
    – M.Maric
    Nov 29 at 13:57
















0














So on my exam I got a True/False question that asked the following:




For any $ninmathbb N$, compute the determinant of the matrix



begin{bmatrix}
1&1&1&cdots&1
\1&1/2&1/2&cdots&1/2
\1&1/2&1/3&cdots&1/3
\vdots&vdots&vdots&ddots&vdots
\1&1/2&1/3&cdots&1/n
end{bmatrix}



This is the $ntimes n$ matrix whose $jk$-th entry is $1/min{j,k}$.




I recognize there has to be some formula that makes calculating the determinant easier, but I'm confused as to what it is.






linear-algebra matrices determinant examples-counterexamples







share|cite|improve this question




















  • 1




    One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
    – Servaes
    Nov 29 at 13:53












  • I don't actually, would you care to explain how that works.
    – M.Maric
    Nov 29 at 13:57














0












0








0


1





So on my exam I got a True/False question that asked the following:




For any $ninmathbb N$, compute the determinant of the matrix



begin{bmatrix}
1&1&1&cdots&1
\1&1/2&1/2&cdots&1/2
\1&1/2&1/3&cdots&1/3
\vdots&vdots&vdots&ddots&vdots
\1&1/2&1/3&cdots&1/n
end{bmatrix}



This is the $ntimes n$ matrix whose $jk$-th entry is $1/min{j,k}$.




I recognize there has to be some formula that makes calculating the determinant easier, but I'm confused as to what it is.






linear-algebra matrices determinant examples-counterexamples







share|cite|improve this question















So on my exam I got a True/False question that asked the following:




For any $ninmathbb N$, compute the determinant of the matrix



begin{bmatrix}
1&1&1&cdots&1
\1&1/2&1/2&cdots&1/2
\1&1/2&1/3&cdots&1/3
\vdots&vdots&vdots&ddots&vdots
\1&1/2&1/3&cdots&1/n
end{bmatrix}



This is the $ntimes n$ matrix whose $jk$-th entry is $1/min{j,k}$.




I recognize there has to be some formula that makes calculating the determinant easier, but I'm confused as to what it is.






linear-algebra matrices determinant examples-counterexamples




linear-algebra matrices determinant examples-counterexamples






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 15:26









StubbornAtom

5,19511138




5,19511138










asked Nov 29 at 13:47









M.Maric

294




294








  • 1




    One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
    – Servaes
    Nov 29 at 13:53












  • I don't actually, would you care to explain how that works.
    – M.Maric
    Nov 29 at 13:57














  • 1




    One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
    – Servaes
    Nov 29 at 13:53












  • I don't actually, would you care to explain how that works.
    – M.Maric
    Nov 29 at 13:57








1




1




One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
– Servaes
Nov 29 at 13:53






One formula would by $det=(-1)^{n+1}frac{n}{(n!)^2}$, but then you should just happen to know that one...
– Servaes
Nov 29 at 13:53














I don't actually, would you care to explain how that works.
– M.Maric
Nov 29 at 13:57




I don't actually, would you care to explain how that works.
– M.Maric
Nov 29 at 13:57










2 Answers
2






active

oldest

votes


















0














For any given $ninBbb{N}$ denote the matrix by $A_n$. For $ngeq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $frac{1}{n}$) an working right to left yields
$$det A_n=frac{1}{n}det A_{n-1}-frac{1}{n-1}det A_{n-1}=-frac{1}{n(n-1)}det A_{n-1},$$
because the subdeterminants at the coefficients $frac{1}{1}$ through $frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $det A_1=1$ so
$$det A_n=(-1)^{n+1}frac{1}{n!(n-1)!}=(-1)^{n+1}frac{n}{(n!)^2}.$$






share|cite|improve this answer



















  • 1




    It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
    – Olivier Moschetta
    Nov 29 at 14:13










  • @OlivierMoschetta Corrected, thanks for noticing!
    – Servaes
    Nov 29 at 14:14



















1














I'll take $n=4$ for simplicity of exposition. We develop against the last line:
$$-begin{vmatrix}
1 & 1 & 1 \
1/2 & 1/2 & 1/2 \
1/2 & 1/3 & 1/3
end{vmatrix}
+dfrac{1}{2}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/3 & 1/3
end{vmatrix}
-dfrac{1}{3}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}
+dfrac{1}{4}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}$$
The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that
$$D_4=-dfrac{1}{12}D_3$$
if $D_n$ is the determinant of your $ntimes n$ matrix.
You can generalize these calculations to prove that
$$D_{n+1}=-dfrac{1}{n(n+1)}D_n$$
Together with $D_1=1$ this is enough to find a general formula for $D_n$.






share|cite|improve this answer





















  • Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
    – M.Maric
    Nov 29 at 16:10











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














For any given $ninBbb{N}$ denote the matrix by $A_n$. For $ngeq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $frac{1}{n}$) an working right to left yields
$$det A_n=frac{1}{n}det A_{n-1}-frac{1}{n-1}det A_{n-1}=-frac{1}{n(n-1)}det A_{n-1},$$
because the subdeterminants at the coefficients $frac{1}{1}$ through $frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $det A_1=1$ so
$$det A_n=(-1)^{n+1}frac{1}{n!(n-1)!}=(-1)^{n+1}frac{n}{(n!)^2}.$$






share|cite|improve this answer



















  • 1




    It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
    – Olivier Moschetta
    Nov 29 at 14:13










  • @OlivierMoschetta Corrected, thanks for noticing!
    – Servaes
    Nov 29 at 14:14
















0














For any given $ninBbb{N}$ denote the matrix by $A_n$. For $ngeq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $frac{1}{n}$) an working right to left yields
$$det A_n=frac{1}{n}det A_{n-1}-frac{1}{n-1}det A_{n-1}=-frac{1}{n(n-1)}det A_{n-1},$$
because the subdeterminants at the coefficients $frac{1}{1}$ through $frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $det A_1=1$ so
$$det A_n=(-1)^{n+1}frac{1}{n!(n-1)!}=(-1)^{n+1}frac{n}{(n!)^2}.$$






share|cite|improve this answer



















  • 1




    It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
    – Olivier Moschetta
    Nov 29 at 14:13










  • @OlivierMoschetta Corrected, thanks for noticing!
    – Servaes
    Nov 29 at 14:14














0












0








0






For any given $ninBbb{N}$ denote the matrix by $A_n$. For $ngeq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $frac{1}{n}$) an working right to left yields
$$det A_n=frac{1}{n}det A_{n-1}-frac{1}{n-1}det A_{n-1}=-frac{1}{n(n-1)}det A_{n-1},$$
because the subdeterminants at the coefficients $frac{1}{1}$ through $frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $det A_1=1$ so
$$det A_n=(-1)^{n+1}frac{1}{n!(n-1)!}=(-1)^{n+1}frac{n}{(n!)^2}.$$






share|cite|improve this answer














For any given $ninBbb{N}$ denote the matrix by $A_n$. For $ngeq 2$, applying Laplace expansion along the last row and starting at the bottom right (the entry $frac{1}{n}$) an working right to left yields
$$det A_n=frac{1}{n}det A_{n-1}-frac{1}{n-1}det A_{n-1}=-frac{1}{n(n-1)}det A_{n-1},$$
because the subdeterminants at the coefficients $frac{1}{1}$ through $frac{1}{n-2}$ are all zero; removing the $k$-th column makes the rows $k-1$, $k$ and $k+1$ linearly dependent. Of course $det A_1=1$ so
$$det A_n=(-1)^{n+1}frac{1}{n!(n-1)!}=(-1)^{n+1}frac{n}{(n!)^2}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 29 at 14:14

























answered Nov 29 at 14:06









Servaes

22.3k33793




22.3k33793








  • 1




    It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
    – Olivier Moschetta
    Nov 29 at 14:13










  • @OlivierMoschetta Corrected, thanks for noticing!
    – Servaes
    Nov 29 at 14:14














  • 1




    It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
    – Olivier Moschetta
    Nov 29 at 14:13










  • @OlivierMoschetta Corrected, thanks for noticing!
    – Servaes
    Nov 29 at 14:14








1




1




It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
– Olivier Moschetta
Nov 29 at 14:13




It should read $det A_{n-1}$ instead of $det(A_n)$ on the right of the first formula.
– Olivier Moschetta
Nov 29 at 14:13












@OlivierMoschetta Corrected, thanks for noticing!
– Servaes
Nov 29 at 14:14




@OlivierMoschetta Corrected, thanks for noticing!
– Servaes
Nov 29 at 14:14











1














I'll take $n=4$ for simplicity of exposition. We develop against the last line:
$$-begin{vmatrix}
1 & 1 & 1 \
1/2 & 1/2 & 1/2 \
1/2 & 1/3 & 1/3
end{vmatrix}
+dfrac{1}{2}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/3 & 1/3
end{vmatrix}
-dfrac{1}{3}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}
+dfrac{1}{4}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}$$
The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that
$$D_4=-dfrac{1}{12}D_3$$
if $D_n$ is the determinant of your $ntimes n$ matrix.
You can generalize these calculations to prove that
$$D_{n+1}=-dfrac{1}{n(n+1)}D_n$$
Together with $D_1=1$ this is enough to find a general formula for $D_n$.






share|cite|improve this answer





















  • Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
    – M.Maric
    Nov 29 at 16:10
















1














I'll take $n=4$ for simplicity of exposition. We develop against the last line:
$$-begin{vmatrix}
1 & 1 & 1 \
1/2 & 1/2 & 1/2 \
1/2 & 1/3 & 1/3
end{vmatrix}
+dfrac{1}{2}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/3 & 1/3
end{vmatrix}
-dfrac{1}{3}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}
+dfrac{1}{4}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}$$
The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that
$$D_4=-dfrac{1}{12}D_3$$
if $D_n$ is the determinant of your $ntimes n$ matrix.
You can generalize these calculations to prove that
$$D_{n+1}=-dfrac{1}{n(n+1)}D_n$$
Together with $D_1=1$ this is enough to find a general formula for $D_n$.






share|cite|improve this answer





















  • Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
    – M.Maric
    Nov 29 at 16:10














1












1








1






I'll take $n=4$ for simplicity of exposition. We develop against the last line:
$$-begin{vmatrix}
1 & 1 & 1 \
1/2 & 1/2 & 1/2 \
1/2 & 1/3 & 1/3
end{vmatrix}
+dfrac{1}{2}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/3 & 1/3
end{vmatrix}
-dfrac{1}{3}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}
+dfrac{1}{4}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}$$
The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that
$$D_4=-dfrac{1}{12}D_3$$
if $D_n$ is the determinant of your $ntimes n$ matrix.
You can generalize these calculations to prove that
$$D_{n+1}=-dfrac{1}{n(n+1)}D_n$$
Together with $D_1=1$ this is enough to find a general formula for $D_n$.






share|cite|improve this answer












I'll take $n=4$ for simplicity of exposition. We develop against the last line:
$$-begin{vmatrix}
1 & 1 & 1 \
1/2 & 1/2 & 1/2 \
1/2 & 1/3 & 1/3
end{vmatrix}
+dfrac{1}{2}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/3 & 1/3
end{vmatrix}
-dfrac{1}{3}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}
+dfrac{1}{4}begin{vmatrix}
1 & 1 & 1 \
1 & 1/2 & 1/2 \
1 & 1/2 & 1/3
end{vmatrix}$$
The first determinant is $0$ (two proportional lines), the second is $0$ (two identical columns) the last two are the same and equal the original determinant in size $n-1=3$. This shows that
$$D_4=-dfrac{1}{12}D_3$$
if $D_n$ is the determinant of your $ntimes n$ matrix.
You can generalize these calculations to prove that
$$D_{n+1}=-dfrac{1}{n(n+1)}D_n$$
Together with $D_1=1$ this is enough to find a general formula for $D_n$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 at 14:06









Olivier Moschetta

2,7761411




2,7761411












  • Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
    – M.Maric
    Nov 29 at 16:10


















  • Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
    – M.Maric
    Nov 29 at 16:10
















Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
– M.Maric
Nov 29 at 16:10




Thanks, I figured out later that this was another way that we could do the problem, along with row reducing and making the matrix triangular and computing the determinant from there.
– M.Maric
Nov 29 at 16:10


















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