proof of non singularity of matrix

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i have a matrix $A$ (similar to a Laplacian matrix) with following properties;



$A=beginbmatrix1 & 0 & 0\-0.3 & 1& -0.7\ -0.22& -0.78& 1endbmatrix$ (looks like this)



i. diagonal elements of $A$ are all $1$.



ii. off diagonal element of $A$ and all negative and sums to $-1$.



iii. (except the first row) All row sums of $A$ are equal to $0$ , (since off diagonal elements sum to $-1$).



iv. off diagonal elements in first row of are all zeros.



if the off diagonal elements in first row $A$ are non zero and sum to $-1$, then $A$ is normalized laplacian matrix with rank $=n-1$, and is singular.



In my case zero off diagonal elements in first row, makes it non-singular.



i have tested numerically that $A$ is non singular, but i don't know how to show it analytically.




linear-algebra




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  • Your matrix is diagonally dominant.
    – Bruce
    Aug 29 at 8:49














up vote
-1
down vote

favorite












i have a matrix $A$ (similar to a Laplacian matrix) with following properties;



$A=beginbmatrix1 & 0 & 0\-0.3 & 1& -0.7\ -0.22& -0.78& 1endbmatrix$ (looks like this)



i. diagonal elements of $A$ are all $1$.



ii. off diagonal element of $A$ and all negative and sums to $-1$.



iii. (except the first row) All row sums of $A$ are equal to $0$ , (since off diagonal elements sum to $-1$).



iv. off diagonal elements in first row of are all zeros.



if the off diagonal elements in first row $A$ are non zero and sum to $-1$, then $A$ is normalized laplacian matrix with rank $=n-1$, and is singular.



In my case zero off diagonal elements in first row, makes it non-singular.



i have tested numerically that $A$ is non singular, but i don't know how to show it analytically.




linear-algebra




share|cite|improve this question






















  • Your matrix is diagonally dominant.
    – Bruce
    Aug 29 at 8:49












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











i have a matrix $A$ (similar to a Laplacian matrix) with following properties;



$A=beginbmatrix1 & 0 & 0\-0.3 & 1& -0.7\ -0.22& -0.78& 1endbmatrix$ (looks like this)



i. diagonal elements of $A$ are all $1$.



ii. off diagonal element of $A$ and all negative and sums to $-1$.



iii. (except the first row) All row sums of $A$ are equal to $0$ , (since off diagonal elements sum to $-1$).



iv. off diagonal elements in first row of are all zeros.



if the off diagonal elements in first row $A$ are non zero and sum to $-1$, then $A$ is normalized laplacian matrix with rank $=n-1$, and is singular.



In my case zero off diagonal elements in first row, makes it non-singular.



i have tested numerically that $A$ is non singular, but i don't know how to show it analytically.




linear-algebra




share|cite|improve this question














i have a matrix $A$ (similar to a Laplacian matrix) with following properties;



$A=beginbmatrix1 & 0 & 0\-0.3 & 1& -0.7\ -0.22& -0.78& 1endbmatrix$ (looks like this)



i. diagonal elements of $A$ are all $1$.



ii. off diagonal element of $A$ and all negative and sums to $-1$.



iii. (except the first row) All row sums of $A$ are equal to $0$ , (since off diagonal elements sum to $-1$).



iv. off diagonal elements in first row of are all zeros.



if the off diagonal elements in first row $A$ are non zero and sum to $-1$, then $A$ is normalized laplacian matrix with rank $=n-1$, and is singular.



In my case zero off diagonal elements in first row, makes it non-singular.



i have tested numerically that $A$ is non singular, but i don't know how to show it analytically.





linear-algebra





share|cite|improve this question













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edited Aug 29 at 8:45

























asked Aug 29 at 8:36









faisal

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  • Your matrix is diagonally dominant.
    – Bruce
    Aug 29 at 8:49
















  • Your matrix is diagonally dominant.
    – Bruce
    Aug 29 at 8:49















Your matrix is diagonally dominant.
– Bruce
Aug 29 at 8:49




Your matrix is diagonally dominant.
– Bruce
Aug 29 at 8:49















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