analytical proof for invertible matrix
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How can I prove the matrix following type to be invertible, (the dimensions of my actual matrix are larger)
$beginbmatrix1 & 0 & 0 \ -a_21 & 1 & -a_23&\-a_31&-a_32&1 endbmatrix$
Row sum of last $2$ rows is zero, (diagonally dominant matrix).
Since the dimensions of my actual matrix are large I cannot use determined for proof.
linear-algebra
 |Â
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up vote
-1
down vote
favorite
How can I prove the matrix following type to be invertible, (the dimensions of my actual matrix are larger)
$beginbmatrix1 & 0 & 0 \ -a_21 & 1 & -a_23&\-a_31&-a_32&1 endbmatrix$
Row sum of last $2$ rows is zero, (diagonally dominant matrix).
Since the dimensions of my actual matrix are large I cannot use determined for proof.
linear-algebra
see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36
 |Â
show 14 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
How can I prove the matrix following type to be invertible, (the dimensions of my actual matrix are larger)
$beginbmatrix1 & 0 & 0 \ -a_21 & 1 & -a_23&\-a_31&-a_32&1 endbmatrix$
Row sum of last $2$ rows is zero, (diagonally dominant matrix).
Since the dimensions of my actual matrix are large I cannot use determined for proof.
linear-algebra
How can I prove the matrix following type to be invertible, (the dimensions of my actual matrix are larger)
$beginbmatrix1 & 0 & 0 \ -a_21 & 1 & -a_23&\-a_31&-a_32&1 endbmatrix$
Row sum of last $2$ rows is zero, (diagonally dominant matrix).
Since the dimensions of my actual matrix are large I cannot use determined for proof.
linear-algebra
linear-algebra
edited Aug 30 at 8:18
Arthur
101k795176
101k795176
asked Aug 30 at 8:16
faisal
62
62
see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36
 |Â
show 14 more comments
see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36
see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36
 |Â
show 14 more comments
1 Answer
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since the first row contains all zeros except the first element, i can ignore the first row and first column, without affecting the determinant, and the remaining matrix has ($n-1,n-1$) dimensions. then i add and subtract the first column, $c_(n-1,1)$
$A_(n-1,n-1)+diag(c_(n-1,1))-diag(c_(n-1,1))$
since elements of $c_(n-1,1)$ are negative $A_(n-1,n-1)+diag(c_(n-1,1))$ forms a modfied laplacian matrix, $L$, with row sum equal to zero. (see the defination of laplacian matrix)
$L-diag(c_(n-1,1))$ is positive definite, since the elements of $c_(n-1,1))$ are negative, (proof available in literature).
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
since the first row contains all zeros except the first element, i can ignore the first row and first column, without affecting the determinant, and the remaining matrix has ($n-1,n-1$) dimensions. then i add and subtract the first column, $c_(n-1,1)$
$A_(n-1,n-1)+diag(c_(n-1,1))-diag(c_(n-1,1))$
since elements of $c_(n-1,1)$ are negative $A_(n-1,n-1)+diag(c_(n-1,1))$ forms a modfied laplacian matrix, $L$, with row sum equal to zero. (see the defination of laplacian matrix)
$L-diag(c_(n-1,1))$ is positive definite, since the elements of $c_(n-1,1))$ are negative, (proof available in literature).
add a comment |Â
up vote
0
down vote
since the first row contains all zeros except the first element, i can ignore the first row and first column, without affecting the determinant, and the remaining matrix has ($n-1,n-1$) dimensions. then i add and subtract the first column, $c_(n-1,1)$
$A_(n-1,n-1)+diag(c_(n-1,1))-diag(c_(n-1,1))$
since elements of $c_(n-1,1)$ are negative $A_(n-1,n-1)+diag(c_(n-1,1))$ forms a modfied laplacian matrix, $L$, with row sum equal to zero. (see the defination of laplacian matrix)
$L-diag(c_(n-1,1))$ is positive definite, since the elements of $c_(n-1,1))$ are negative, (proof available in literature).
add a comment |Â
up vote
0
down vote
up vote
0
down vote
since the first row contains all zeros except the first element, i can ignore the first row and first column, without affecting the determinant, and the remaining matrix has ($n-1,n-1$) dimensions. then i add and subtract the first column, $c_(n-1,1)$
$A_(n-1,n-1)+diag(c_(n-1,1))-diag(c_(n-1,1))$
since elements of $c_(n-1,1)$ are negative $A_(n-1,n-1)+diag(c_(n-1,1))$ forms a modfied laplacian matrix, $L$, with row sum equal to zero. (see the defination of laplacian matrix)
$L-diag(c_(n-1,1))$ is positive definite, since the elements of $c_(n-1,1))$ are negative, (proof available in literature).
since the first row contains all zeros except the first element, i can ignore the first row and first column, without affecting the determinant, and the remaining matrix has ($n-1,n-1$) dimensions. then i add and subtract the first column, $c_(n-1,1)$
$A_(n-1,n-1)+diag(c_(n-1,1))-diag(c_(n-1,1))$
since elements of $c_(n-1,1)$ are negative $A_(n-1,n-1)+diag(c_(n-1,1))$ forms a modfied laplacian matrix, $L$, with row sum equal to zero. (see the defination of laplacian matrix)
$L-diag(c_(n-1,1))$ is positive definite, since the elements of $c_(n-1,1))$ are negative, (proof available in literature).
edited Sep 7 at 7:48
answered Aug 31 at 12:05
faisal
62
62
add a comment |Â
add a comment |Â
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see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
â Dr. Sonnhard Graubner
Aug 30 at 8:20
It is not clear what the general form of the matrix is in higher dimensions.
â Alexey
Aug 30 at 8:25
You mention that your matrix's "dimension" is larger, which means to me that it has more rows and columns...and thus there can be huge differences. The determinant of the matrix you wrote is $;1-a_23a_32;$ , so it is invertible iff $;a_23a_32neq1;$ ...With other matrices you can also check the determinant.
â DonAntonio
Aug 30 at 8:26
If the sum of the last two rows is zero, as you say, then the matrix is not invertible...
â 5xum
Aug 30 at 8:30
@5xum, not the sum of the last two rows, but the row sum of each of the last two rows.
â Alexey
Aug 30 at 8:36