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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018638%2fdeterminant-of-the-n-times-n-matrix-whose-jk-th-entry-is-1-min-j-k%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3018638%2fdeterminant-of-the-n-times-n-matrix-whose-jk-th-entry-is-1-min-j-k%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc
Read more

Fiat S.p.A.

Image
Cet article court présente un sujet plus développé dans : Fiat et Fiat Chrysler Automobiles. Fiat S.p.A. Fiat S.p.A. ( Fabbrica Italiana Automobili Torino , acronyme FIAT) est un groupe, notamment connu comme société-mère du constructeur automobile italien, Fiat. Portail des entreprises Portail de Turin Portail de l’automobile This page is only for reference, If you need detailed information, please check here
Read more

Type 'String' is not a subtype of type 'int' of 'index'

Image
up vote 0 down vote favorite My App can authenticate successfully when debugging against the Android emulator, but if I try to authenticate using debugging against the physical device (with same OS version), an error message appears after more than one minute waiting: Exception has occurred. _TypeError (type 'String' is not a subtype of type 'int' of 'index') The error message points to the following code: if (responseJson[AuthUtils.authTokenKey] != null) { AuthUtils.insertDetails( userNameController.text, passwordController.text, responseJson); ... } else { ... }; And in the DEBUG CONSOLE I get the following: I/flutter ( 9531): Auth: SocketException: OS Error: Connection timed out, errno = 110, address = example.com, port = 3897
Read more