Rank of an NxN matrix

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This question very similar to this one.

I have this problem I'm working on. I think I have it figured out but I just want to make sure I am not misunderstanding how rank works.

The question goes like this,

Consider an nXn matrix where the elements go from 1 to $ n^2 $ as you read across and then down. For instance a 3X3 matrix where n = 3 looks like this:

$ left( beginarray_
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 endarray right)$

What is the rank of the nXn matrix, as a function of n, for n>=2?

Drawing out the matrices for n=2 to n=5 I can spot a pretty obvious pattern in the matrix construction and the whole thing can be written in terms of n:

$ left( beginarray_
1 & 2 & 3 & ... & n \
n+1 & n+2 & n+3 & ... & 2n \
2n+1 & 2n+2 & 2n+3 & ... & 3n \
3n+1 & 3n+2 & 3n+3 & ... & 4n \
... &... &... & ... & ... \
n^2-n+1 & n^2-n+2 & n^2-n+3 & ... & n^2 endarray right)$

From this I can spot that as soon as you get past the 3rd row, you can sum the prior rows to cancel out that row.

For instance take row 4 from above for the first column.
If I take $ row 3 + row 2 - row 1 $ I get this value

$ (2n+1) + (n+1) - (1) = 3n+1 $

Which is the next rows value in row 4.

And that can be carried out on the rest of the columns to produce the same result.

So for any arbitrary row k beyond row 3, I can make that row all zeros by doing a row reduction of:

R_k -> R_k - R_k-1 - R_k-2 + R_k-3

Thus the maximum rank this matrix can have is 3 correct?

Is this a reasonable way to solve this problem or there a more direct approach I should be taking?

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

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up vote
2
down vote

favorite

This question very similar to this one.

I have this problem I'm working on. I think I have it figured out but I just want to make sure I am not misunderstanding how rank works.

The question goes like this,

Consider an nXn matrix where the elements go from 1 to $ n^2 $ as you read across and then down. For instance a 3X3 matrix where n = 3 looks like this:

$ left( beginarray_
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 endarray right)$

What is the rank of the nXn matrix, as a function of n, for n>=2?

Drawing out the matrices for n=2 to n=5 I can spot a pretty obvious pattern in the matrix construction and the whole thing can be written in terms of n:

From this I can spot that as soon as you get past the 3rd row, you can sum the prior rows to cancel out that row.

For instance take row 4 from above for the first column.
If I take $ row 3 + row 2 - row 1 $ I get this value

$ (2n+1) + (n+1) - (1) = 3n+1 $

Which is the next rows value in row 4.

And that can be carried out on the rest of the columns to produce the same result.

So for any arbitrary row k beyond row 3, I can make that row all zeros by doing a row reduction of:

R_k -> R_k - R_k-1 - R_k-2 + R_k-3

Thus the maximum rank this matrix can have is 3 correct?

Is this a reasonable way to solve this problem or there a more direct approach I should be taking?

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

add a commentÂ |Â

up vote
2
down vote

favorite

This question very similar to this one.

I have this problem I'm working on. I think I have it figured out but I just want to make sure I am not misunderstanding how rank works.

The question goes like this,

Consider an nXn matrix where the elements go from 1 to $ n^2 $ as you read across and then down. For instance a 3X3 matrix where n = 3 looks like this:

$ left( beginarray_
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 endarray right)$

What is the rank of the nXn matrix, as a function of n, for n>=2?

Drawing out the matrices for n=2 to n=5 I can spot a pretty obvious pattern in the matrix construction and the whole thing can be written in terms of n:

From this I can spot that as soon as you get past the 3rd row, you can sum the prior rows to cancel out that row.

For instance take row 4 from above for the first column.
If I take $ row 3 + row 2 - row 1 $ I get this value

$ (2n+1) + (n+1) - (1) = 3n+1 $

Which is the next rows value in row 4.

And that can be carried out on the rest of the columns to produce the same result.

So for any arbitrary row k beyond row 3, I can make that row all zeros by doing a row reduction of:

R_k -> R_k - R_k-1 - R_k-2 + R_k-3

Thus the maximum rank this matrix can have is 3 correct?

Is this a reasonable way to solve this problem or there a more direct approach I should be taking?

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

This question very similar to this one.

I have this problem I'm working on. I think I have it figured out but I just want to make sure I am not misunderstanding how rank works.

The question goes like this,

Consider an nXn matrix where the elements go from 1 to $ n^2 $ as you read across and then down. For instance a 3X3 matrix where n = 3 looks like this:

$ left( beginarray_
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 endarray right)$

What is the rank of the nXn matrix, as a function of n, for n>=2?

Drawing out the matrices for n=2 to n=5 I can spot a pretty obvious pattern in the matrix construction and the whole thing can be written in terms of n:

From this I can spot that as soon as you get past the 3rd row, you can sum the prior rows to cancel out that row.

For instance take row 4 from above for the first column.
If I take $ row 3 + row 2 - row 1 $ I get this value

$ (2n+1) + (n+1) - (1) = 3n+1 $

Which is the next rows value in row 4.

And that can be carried out on the rest of the columns to produce the same result.

So for any arbitrary row k beyond row 3, I can make that row all zeros by doing a row reduction of:

R_k -> R_k - R_k-1 - R_k-2 + R_k-3

Thus the maximum rank this matrix can have is 3 correct?

Is this a reasonable way to solve this problem or there a more direct approach I should be taking?

linear-algebra matrices matrix-rank

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

edited 6 hours ago

zahbaz

7,68521636

edited 6 hours ago

zahbaz

7,68521636

edited 6 hours ago

zahbaz

7,68521636

asked 6 hours ago

user2326106

1111

asked 6 hours ago

user2326106

1111

asked 6 hours ago

user2326106

1111

add a commentÂ |Â

2 Answers
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up vote
4
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Subtracting the first row from the others we obtain $n-1$ rows multiple of $(n,n,ldots,n)$ therefore $operatornamerank(A)=2$ for any $nge 2$.

answered 6 hours ago

gimusi

72.2k73888

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up vote
1
down vote

You approach is fine for this problem but sometimes it is very hard to see the pattern between rows or columns.

So a general way of finding the rank of the matrix is by reducing it to it's Row Echelon Form (note that row-reduced echelon form is not necessary). You just have to make all the elements zero below the pivot elements. So you have:

$$ left[ beginarray_
1 & 2 & 3 \
4 & 5 & 6 \
7 & 8 & 9 endarray right]xrightarrow[R_3=R_3-7R_1]R_2=R_2-4R_1 left[ beginarray_
1 & 2 & 3 \
0 &-3 &-6 \
0 &-6 &-12 endarray right] xrightarrowR_3=R_3-2R_2 left[ beginarray_
1 & 2 & 3 \
0 &-3 &-6 \
0 & 0 & 0 endarray right]$$

After reducing it you just have to count the number of non-zero rows and that's your rank. Here as you can see it's $2$. This can help you find the rank of any matrix.

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

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2 Answers
2

active

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2 Answers
2

active

oldest

votes

up vote
4
down vote

Subtracting the first row from the others we obtain $n-1$ rows multiple of $(n,n,ldots,n)$ therefore $operatornamerank(A)=2$ for any $nge 2$.

answered 6 hours ago

gimusi

72.2k73888

add a commentÂ |Â

up vote
4
down vote

Subtracting the first row from the others we obtain $n-1$ rows multiple of $(n,n,ldots,n)$ therefore $operatornamerank(A)=2$ for any $nge 2$.

answered 6 hours ago

gimusi

72.2k73888

add a commentÂ |Â

up vote
4
down vote

Subtracting the first row from the others we obtain $n-1$ rows multiple of $(n,n,ldots,n)$ therefore $operatornamerank(A)=2$ for any $nge 2$.

answered 6 hours ago

gimusi

72.2k73888

Subtracting the first row from the others we obtain $n-1$ rows multiple of $(n,n,ldots,n)$ therefore $operatornamerank(A)=2$ for any $nge 2$.

answered 6 hours ago

gimusi

72.2k73888

answered 6 hours ago

gimusi

72.2k73888

answered 6 hours ago

gimusi

72.2k73888

answered 6 hours ago

gimusi

72.2k73888

add a commentÂ |Â

up vote
1
down vote

You approach is fine for this problem but sometimes it is very hard to see the pattern between rows or columns.

After reducing it you just have to count the number of non-zero rows and that's your rank. Here as you can see it's $2$. This can help you find the rank of any matrix.

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

add a commentÂ |Â

up vote
1
down vote

You approach is fine for this problem but sometimes it is very hard to see the pattern between rows or columns.

After reducing it you just have to count the number of non-zero rows and that's your rank. Here as you can see it's $2$. This can help you find the rank of any matrix.

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

add a commentÂ |Â

up vote
1
down vote

You approach is fine for this problem but sometimes it is very hard to see the pattern between rows or columns.

After reducing it you just have to count the number of non-zero rows and that's your rank. Here as you can see it's $2$. This can help you find the rank of any matrix.

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

You approach is fine for this problem but sometimes it is very hard to see the pattern between rows or columns.

After reducing it you just have to count the number of non-zero rows and that's your rank. Here as you can see it's $2$. This can help you find the rank of any matrix.

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

edited 30 mins ago

answered 50 mins ago

paulplusx

875217

answered 50 mins ago

paulplusx

875217

answered 50 mins ago

paulplusx

875217

add a commentÂ |Â

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