Proving false or true a matrix statement

up vote
2
down vote

favorite

I am suppose to figure out if this is false or true.

If $I$ is the $n$ x $n$ identity matrix, and $J$ is an $n$ x $n$ matrix consisting entirely of ones, then the matrix $A = I - fracJn+1$ is idempotent (ie $A^2$ = $A$)

I understand obviously what $I$ and $J$ are, my issue is with the $A = I - fracJn+1$

I searched my textbook and found no reference to it.

What does it mean?

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

add a commentÂ |Â

up vote
2
down vote

favorite

I am suppose to figure out if this is false or true.

If $I$ is the $n$ x $n$ identity matrix, and $J$ is an $n$ x $n$ matrix consisting entirely of ones, then the matrix $A = I - fracJn+1$ is idempotent (ie $A^2$ = $A$)

I understand obviously what $I$ and $J$ are, my issue is with the $A = I - fracJn+1$

I searched my textbook and found no reference to it.

What does it mean?

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

add a commentÂ |Â

up vote
2
down vote

favorite

I am suppose to figure out if this is false or true.

If $I$ is the $n$ x $n$ identity matrix, and $J$ is an $n$ x $n$ matrix consisting entirely of ones, then the matrix $A = I - fracJn+1$ is idempotent (ie $A^2$ = $A$)

I understand obviously what $I$ and $J$ are, my issue is with the $A = I - fracJn+1$

I searched my textbook and found no reference to it.

What does it mean?

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

I am suppose to figure out if this is false or true.

If $I$ is the $n$ x $n$ identity matrix, and $J$ is an $n$ x $n$ matrix consisting entirely of ones, then the matrix $A = I - fracJn+1$ is idempotent (ie $A^2$ = $A$)

I understand obviously what $I$ and $J$ are, my issue is with the $A = I - fracJn+1$

I searched my textbook and found no reference to it.

What does it mean?

linear-algebra

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

edited 2 hours ago

lisyarus

10.1k21433

edited 2 hours ago

lisyarus

10.1k21433

edited 2 hours ago

lisyarus

10.1k21433

asked 2 hours ago

Forextrader

112

New contributor

asked 2 hours ago

Forextrader

112

asked 2 hours ago

Forextrader

112

New contributor

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

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
3
down vote

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

answered 2 hours ago

lisyarus

10.1k21433

add a commentÂ |Â

up vote
3
down vote

The expression $A = I - fracJn+1$ is matrix addition, where $fracJn+1$ represents scalar multiplication of the matrix $J$ by $frac1n+1$. Since the matrix $J$ consists completely of $1$s, each entry of the scaled matrix $fracJn+1$ is $frac1n+1$. Here is the matrix addition:

$$ I - fracJn+1 =
left[beginarraycccc
1 & 0 & cdots & 0 \
0 & 1 & cdots & 0 \
vdots & vdots & ddots & vdots \
0 & 0 & cdots & 1 \
endarrayright]
-
left[beginarraycccc
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
vdots & vdots & ddots & vdots \
frac1n+1 & frac1n+1 & cdots & frac1n+1 \
endarrayright]$$

$$ implies A = left[beginarraycccc
fracnn+1 & frac-1n+1 & cdots & frac-1n+1 \
frac-1n+1 & fracnn+1 & cdots & frac-1n+1 \
vdots & vdots & ddots & vdots \
frac-1n+1 & frac-1n+1 & cdots & fracnn+1 \
endarrayright] $$

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

add a commentÂ |Â

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

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

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2963656%2fproving-false-or-true-a-matrix-statement%23new-answer', 'question_page');

);

Post as a guest

Name

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

answered 2 hours ago

lisyarus

10.1k21433

add a commentÂ |Â

up vote
3
down vote

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

answered 2 hours ago

lisyarus

10.1k21433

add a commentÂ |Â

up vote
3
down vote

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

answered 2 hours ago

lisyarus

10.1k21433

It is not hard to manually obtain $J^2=nJ$. Now it's just a matter of some matrix algebra:

$$A^2=(I-frac1n+1J)^2=I^2-frac1n+1IJ-frac1n+1JI+frac1(n+1)^2J^2 = \
= I - frac2n+1J + fracn(n+1)^2J = I + fracn-2(n+1)(n+1)^2J = \
= I + frac-n-1(n+1)^2J = I - frac1n+1J = A$$

answered 2 hours ago

lisyarus

10.1k21433

answered 2 hours ago

lisyarus

10.1k21433

answered 2 hours ago

lisyarus

10.1k21433

answered 2 hours ago

lisyarus

10.1k21433

add a commentÂ |Â

up vote
3
down vote

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

add a commentÂ |Â

up vote
3
down vote

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

add a commentÂ |Â

up vote
3
down vote

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

An interesting matrix. You could describe it by $A_ij=fracnn+1$ if $i=j$ and $A_ij=frac-1n+1$ if $ineq j$.

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

edited 1 hour ago

answered 2 hours ago

M. Nestor

5339

answered 2 hours ago

M. Nestor

5339

answered 2 hours ago

M. Nestor

5339

add a commentÂ |Â

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

draft saved

draft discarded

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

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky