Mixing
Proving false or true a matrix statement
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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
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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?
linear-algebra
edited 2 hours ago
lisyarus
10.1k21433
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 comment |Â
up vote
2
down vote
favorite
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?
linear-algebra
edited 2 hours ago
lisyarus
10.1k21433
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.
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
linear-algebra
edited 2 hours ago
lisyarus
10.1k21433
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.
edited 2 hours ago
lisyarus
10.1k21433
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.
edited 2 hours ago
lisyarus
10.1k21433
edited 2 hours ago
lisyarus
10.1k21433
edited 2 hours ago
lisyarus
10.1k21433
10.1k21433
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.
asked 2 hours ago
Forextrader
112
asked 2 hours ago
Forextrader
112
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.
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.
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 comment |Â
add a comment |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 comment |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 $ineq j$.
answered 2 hours ago
M. Nestor
5339
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 comment |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 comment |Â
up vote
3
down vote
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
10.1k21433
add a comment |Â
add a comment |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 $ineq j$.
answered 2 hours ago
M. Nestor
5339
add a comment |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 $ineq j$.
answered 2 hours ago
M. Nestor
5339
add a comment |Â
up vote
3
down vote
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 $ineq j$.
answered 2 hours ago
M. Nestor
5339
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 $ineq j$.
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
5339
add a comment |Â
add a comment |Â
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