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.










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  • see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
    – Dr. Sonnhard Graubner
    Aug 30 at 8:20










  • It is not clear what the general form of the matrix is in higher dimensions.
    – 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.
    – 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...
    – 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.
    – Alexey
    Aug 30 at 8:36















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.










share|cite|improve this question























  • see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
    – Dr. Sonnhard Graubner
    Aug 30 at 8:20










  • It is not clear what the general form of the matrix is in higher dimensions.
    – 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.
    – 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...
    – 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.
    – Alexey
    Aug 30 at 8:36













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.










share|cite|improve this question















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






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edited Aug 30 at 8:18









Arthur

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asked Aug 30 at 8:16









faisal

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  • see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
    – Dr. Sonnhard Graubner
    Aug 30 at 8:20










  • It is not clear what the general form of the matrix is in higher dimensions.
    – 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.
    – 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...
    – 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.
    – Alexey
    Aug 30 at 8:36

















  • see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
    – Dr. Sonnhard Graubner
    Aug 30 at 8:20










  • It is not clear what the general form of the matrix is in higher dimensions.
    – 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.
    – 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...
    – 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.
    – Alexey
    Aug 30 at 8:36
















see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
– Dr. Sonnhard Graubner
Aug 30 at 8:20




see here mathworld.wolfram.com/InvertibleMatrixTheorem.html
– Dr. Sonnhard Graubner
Aug 30 at 8:20












It is not clear what the general form of the matrix is in higher dimensions.
– Alexey
Aug 30 at 8:25




It is not clear what the general form of the matrix is in higher dimensions.
– 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.
– 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.
– 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...
– 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...
– 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.
– 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.
– Alexey
Aug 30 at 8:36











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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).






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    1 Answer
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    active

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    1 Answer
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    up vote
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    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).






    share|cite|improve this answer


























      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).






      share|cite|improve this answer
























        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).






        share|cite|improve this answer














        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).







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Sep 7 at 7:48

























        answered Aug 31 at 12:05









        faisal

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