Calculations of elements in matrix
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How do I calculate number of non zero elements of a strictly upper or lower triangular matrix of order (n x n)?
linear-algebra
asked Dec 23 '18 at 11:00
user628653user628653
13
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add a comment |
$begingroup$
How do I calculate number of non zero elements of a strictly upper or lower triangular matrix of order (n x n)?
linear-algebra
asked Dec 23 '18 at 11:00
user628653user628653
13
$endgroup$
$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
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– Damien
Dec 23 '18 at 11:03
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Do you have any more assumptions about what the matrix could be?
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– Aniruddh Agarwal
Dec 23 '18 at 11:12
add a comment |
$begingroup$
How do I calculate number of non zero elements of a strictly upper or lower triangular matrix of order (n x n)?
linear-algebra
asked Dec 23 '18 at 11:00
user628653user628653
13
$endgroup$
How do I calculate number of non zero elements of a strictly upper or lower triangular matrix of order (n x n)?
linear-algebra
linear-algebra
asked Dec 23 '18 at 11:00
user628653user628653
13
asked Dec 23 '18 at 11:00
user628653user628653
13
asked Dec 23 '18 at 11:00
user628653user628653
13
asked Dec 23 '18 at 11:00
user628653user628653
13
asked Dec 23 '18 at 11:00
user628653user628653
13
13
$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
$endgroup$
– Damien
Dec 23 '18 at 11:03
$begingroup$
Do you have any more assumptions about what the matrix could be?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 11:12
add a comment |
$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
$endgroup$
– Damien
Dec 23 '18 at 11:03
$begingroup$
Do you have any more assumptions about what the matrix could be?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 11:12
$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
$endgroup$
– Damien
Dec 23 '18 at 11:03
$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
$endgroup$
– Damien
Dec 23 '18 at 11:03
$begingroup$
Do you have any more assumptions about what the matrix could be?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 11:12
$begingroup$
Do you have any more assumptions about what the matrix could be?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 11:12
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For a strictly upper triangular matrix,
$$U=begin{bmatrix}0&a_{12}&ldots&0\0&0&ldots&0\vdots&vdots&ddots&vdots\0&0&ldots&0end{bmatrix}$$
we have $n$ zeroes on the diagonal, $n-1$ zeroes on the sub-diagonal and so on till we have only $1$ zero at the bottom left corner. The number of zero entries is at-least
$$
n + (n - 1) + ldots + 1
= frac{n (n + 1)}{2},
$$
at-least, because the entries above the diagonal can also be zero. The number of non-zero entries is at-most
$$
n^2 - frac{n (n + 1)}{2} = frac{n(n-1)}{2}.
$$
It is easy to see that the case of strictly lower triangular matrices is similar.
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a strictly upper triangular matrix,
$$U=begin{bmatrix}0&a_{12}&ldots&0\0&0&ldots&0\vdots&vdots&ddots&vdots\0&0&ldots&0end{bmatrix}$$
we have $n$ zeroes on the diagonal, $n-1$ zeroes on the sub-diagonal and so on till we have only $1$ zero at the bottom left corner. The number of zero entries is at-least
$$
n + (n - 1) + ldots + 1
= frac{n (n + 1)}{2},
$$
at-least, because the entries above the diagonal can also be zero. The number of non-zero entries is at-most
$$
n^2 - frac{n (n + 1)}{2} = frac{n(n-1)}{2}.
$$
It is easy to see that the case of strictly lower triangular matrices is similar.
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
$endgroup$
add a comment |
$begingroup$
For a strictly upper triangular matrix,
$$U=begin{bmatrix}0&a_{12}&ldots&0\0&0&ldots&0\vdots&vdots&ddots&vdots\0&0&ldots&0end{bmatrix}$$
we have $n$ zeroes on the diagonal, $n-1$ zeroes on the sub-diagonal and so on till we have only $1$ zero at the bottom left corner. The number of zero entries is at-least
$$
n + (n - 1) + ldots + 1
= frac{n (n + 1)}{2},
$$
at-least, because the entries above the diagonal can also be zero. The number of non-zero entries is at-most
$$
n^2 - frac{n (n + 1)}{2} = frac{n(n-1)}{2}.
$$
It is easy to see that the case of strictly lower triangular matrices is similar.
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
$endgroup$
add a comment |
$begingroup$
For a strictly upper triangular matrix,
$$U=begin{bmatrix}0&a_{12}&ldots&0\0&0&ldots&0\vdots&vdots&ddots&vdots\0&0&ldots&0end{bmatrix}$$
we have $n$ zeroes on the diagonal, $n-1$ zeroes on the sub-diagonal and so on till we have only $1$ zero at the bottom left corner. The number of zero entries is at-least
$$
n + (n - 1) + ldots + 1
= frac{n (n + 1)}{2},
$$
at-least, because the entries above the diagonal can also be zero. The number of non-zero entries is at-most
$$
n^2 - frac{n (n + 1)}{2} = frac{n(n-1)}{2}.
$$
It is easy to see that the case of strictly lower triangular matrices is similar.
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
$endgroup$
For a strictly upper triangular matrix,
$$U=begin{bmatrix}0&a_{12}&ldots&0\0&0&ldots&0\vdots&vdots&ddots&vdots\0&0&ldots&0end{bmatrix}$$
we have $n$ zeroes on the diagonal, $n-1$ zeroes on the sub-diagonal and so on till we have only $1$ zero at the bottom left corner. The number of zero entries is at-least
$$
n + (n - 1) + ldots + 1
= frac{n (n + 1)}{2},
$$
at-least, because the entries above the diagonal can also be zero. The number of non-zero entries is at-most
$$
n^2 - frac{n (n + 1)}{2} = frac{n(n-1)}{2}.
$$
It is easy to see that the case of strictly lower triangular matrices is similar.
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
edited Dec 23 '18 at 12:05
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
answered Dec 23 '18 at 11:20
Shubham JohriShubham Johri
5,475818
5,475818
add a comment |
add a comment |
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$begingroup$
Maybe simply $2S+n=n^2$ corresponding to upper + lower + diagonal = total matrix
$endgroup$
– Damien
Dec 23 '18 at 11:03
$begingroup$
Do you have any more assumptions about what the matrix could be?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 11:12