Determinants and uniqueness of a solution












2












$begingroup$


I have a question here which says:



"If det$(A)neq0,$then $Ax=x$ has a unique solution."



The $x$ on the right side isn't a typo. It's not supposed to be $b.$



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I'm not really sure how to go about "proving" this.



I do notice that $Ax-x=0$ and thus $x(A-I)=0$. However, I can't really see where to go from there.






linear-algebra determinant







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$endgroup$












  • $begingroup$
    Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
    $endgroup$
    – jgon
    Nov 30 '18 at 3:24
















2












$begingroup$


I have a question here which says:



"If det$(A)neq0,$then $Ax=x$ has a unique solution."



The $x$ on the right side isn't a typo. It's not supposed to be $b.$



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I'm not really sure how to go about "proving" this.



I do notice that $Ax-x=0$ and thus $x(A-I)=0$. However, I can't really see where to go from there.






linear-algebra determinant







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
    $endgroup$
    – jgon
    Nov 30 '18 at 3:24














2












2








2


1



$begingroup$


I have a question here which says:



"If det$(A)neq0,$then $Ax=x$ has a unique solution."



The $x$ on the right side isn't a typo. It's not supposed to be $b.$



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I'm not really sure how to go about "proving" this.



I do notice that $Ax-x=0$ and thus $x(A-I)=0$. However, I can't really see where to go from there.






linear-algebra determinant







share|cite|improve this question









$endgroup$




I have a question here which says:



"If det$(A)neq0,$then $Ax=x$ has a unique solution."



The $x$ on the right side isn't a typo. It's not supposed to be $b.$



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I'm not really sure how to go about "proving" this.



I do notice that $Ax-x=0$ and thus $x(A-I)=0$. However, I can't really see where to go from there.






linear-algebra determinant




linear-algebra determinant






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 3:20









Future Math personFuture Math person

972717




972717












  • $begingroup$
    Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
    $endgroup$
    – jgon
    Nov 30 '18 at 3:24


















  • $begingroup$
    Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
    $endgroup$
    – jgon
    Nov 30 '18 at 3:24
















$begingroup$
Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
$endgroup$
– jgon
Nov 30 '18 at 3:24




$begingroup$
Well, is there an invertible matrix $A$ where $A-1$ has multiple vectors in the kernel?
$endgroup$
– jgon
Nov 30 '18 at 3:24










1 Answer
1






active

oldest

votes


















3












$begingroup$

It's clearly false. Take



$A = I; tag 1$



then



$det(A) = 1 ne 0, tag 2$



but



$forall x, ; Ax = Ix = x; tag 3$



the solution to



$Ax = x tag 4$



is certainly not unique!



This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $det(A) ne 0$ does not preclude $det (A - I) = 0$; if



$det(A - I) ne 0, tag 5$



then



$(A - I)x = 0 tag 6$



has the unique solution $0$; but when $det(A - I)$ vanishes, in general we will have



$ker (A - I) ne {0}; tag 7$



indeed, $dim(ker(A - I))$ may be as great as $text{size}(A)$, as the above example indicates.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I wonder how I missed that... Thanks. Wow. That felt so obvious.
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:26






  • 1




    $begingroup$
    ha ha..........
    $endgroup$
    – Randall
    Nov 30 '18 at 3:26











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

It's clearly false. Take



$A = I; tag 1$



then



$det(A) = 1 ne 0, tag 2$



but



$forall x, ; Ax = Ix = x; tag 3$



the solution to



$Ax = x tag 4$



is certainly not unique!



This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $det(A) ne 0$ does not preclude $det (A - I) = 0$; if



$det(A - I) ne 0, tag 5$



then



$(A - I)x = 0 tag 6$



has the unique solution $0$; but when $det(A - I)$ vanishes, in general we will have



$ker (A - I) ne {0}; tag 7$



indeed, $dim(ker(A - I))$ may be as great as $text{size}(A)$, as the above example indicates.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I wonder how I missed that... Thanks. Wow. That felt so obvious.
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:26






  • 1




    $begingroup$
    ha ha..........
    $endgroup$
    – Randall
    Nov 30 '18 at 3:26
















3












$begingroup$

It's clearly false. Take



$A = I; tag 1$



then



$det(A) = 1 ne 0, tag 2$



but



$forall x, ; Ax = Ix = x; tag 3$



the solution to



$Ax = x tag 4$



is certainly not unique!



This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $det(A) ne 0$ does not preclude $det (A - I) = 0$; if



$det(A - I) ne 0, tag 5$



then



$(A - I)x = 0 tag 6$



has the unique solution $0$; but when $det(A - I)$ vanishes, in general we will have



$ker (A - I) ne {0}; tag 7$



indeed, $dim(ker(A - I))$ may be as great as $text{size}(A)$, as the above example indicates.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    I wonder how I missed that... Thanks. Wow. That felt so obvious.
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:26






  • 1




    $begingroup$
    ha ha..........
    $endgroup$
    – Randall
    Nov 30 '18 at 3:26














3












3








3





$begingroup$

It's clearly false. Take



$A = I; tag 1$



then



$det(A) = 1 ne 0, tag 2$



but



$forall x, ; Ax = Ix = x; tag 3$



the solution to



$Ax = x tag 4$



is certainly not unique!



This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $det(A) ne 0$ does not preclude $det (A - I) = 0$; if



$det(A - I) ne 0, tag 5$



then



$(A - I)x = 0 tag 6$



has the unique solution $0$; but when $det(A - I)$ vanishes, in general we will have



$ker (A - I) ne {0}; tag 7$



indeed, $dim(ker(A - I))$ may be as great as $text{size}(A)$, as the above example indicates.






share|cite|improve this answer











$endgroup$



It's clearly false. Take



$A = I; tag 1$



then



$det(A) = 1 ne 0, tag 2$



but



$forall x, ; Ax = Ix = x; tag 3$



the solution to



$Ax = x tag 4$



is certainly not unique!



This example shows that the proposed hypothetical is not true, but why? I think the most concise reason I can give is that $det(A) ne 0$ does not preclude $det (A - I) = 0$; if



$det(A - I) ne 0, tag 5$



then



$(A - I)x = 0 tag 6$



has the unique solution $0$; but when $det(A - I)$ vanishes, in general we will have



$ker (A - I) ne {0}; tag 7$



indeed, $dim(ker(A - I))$ may be as great as $text{size}(A)$, as the above example indicates.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 30 '18 at 3:33

























answered Nov 30 '18 at 3:25









Robert LewisRobert Lewis

44.6k22964




44.6k22964








  • 1




    $begingroup$
    I wonder how I missed that... Thanks. Wow. That felt so obvious.
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:26






  • 1




    $begingroup$
    ha ha..........
    $endgroup$
    – Randall
    Nov 30 '18 at 3:26














  • 1




    $begingroup$
    I wonder how I missed that... Thanks. Wow. That felt so obvious.
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:26






  • 1




    $begingroup$
    ha ha..........
    $endgroup$
    – Randall
    Nov 30 '18 at 3:26








1




1




$begingroup$
I wonder how I missed that... Thanks. Wow. That felt so obvious.
$endgroup$
– Future Math person
Nov 30 '18 at 3:26




$begingroup$
I wonder how I missed that... Thanks. Wow. That felt so obvious.
$endgroup$
– Future Math person
Nov 30 '18 at 3:26




1




1




$begingroup$
ha ha..........
$endgroup$
– Randall
Nov 30 '18 at 3:26




$begingroup$
ha ha..........
$endgroup$
– Randall
Nov 30 '18 at 3:26


















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