$Ain mathbbR^mtimes n, mge n$ and $operatornamerank(A) = n$. Prove that $A^tA$ is nonsingular

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Let $Ain mathbbR^mtimes n, mge n$ and $operatornamerank(A) = n$. Prove
that $A^tA$ is nonsingular




I found this answer https://math.stackexchange.com/a/1840814/166180



which gives some hints. It says that $A$ and $A^tA$ have the same null space, that is, if $x$ is such that $A^tAx = 0$, then



$$A^tAx = 0 implies Ax = 0$$



To say that $A^tA$ is nonsingular (invertible) is the same as saying that the null space of $A^tA$ is $0$, which is the same, by above, as saying that the null space of $A$ is $0$.



So by hypothesis on the exercise we have that $rank(A)=n$ and $mge n$. What should daying that the rank of $A$ is $n$ mean? I think that since the trace is less than $m$, the system $Ax=0$ has multiple solutions (just a guess, I don't know how to formalize this), so the null space of $A$ wouldn't be $0$.






linear-algebra systems-of-equations







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  • you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
    – dezdichado
    Sep 5 at 4:41










  • Trace is only defined for square matrices.
    – xbh
    Sep 5 at 4:48










  • Thanks, I corrected
    – Paprika
    Sep 5 at 4:57










  • By the rank-nullity theorem the null space of $A$ is $0$.
    – amsmath
    Sep 5 at 7:14














up vote
2
down vote

favorite













Let $Ain mathbbR^mtimes n, mge n$ and $operatornamerank(A) = n$. Prove
that $A^tA$ is nonsingular




I found this answer https://math.stackexchange.com/a/1840814/166180



which gives some hints. It says that $A$ and $A^tA$ have the same null space, that is, if $x$ is such that $A^tAx = 0$, then



$$A^tAx = 0 implies Ax = 0$$



To say that $A^tA$ is nonsingular (invertible) is the same as saying that the null space of $A^tA$ is $0$, which is the same, by above, as saying that the null space of $A$ is $0$.



So by hypothesis on the exercise we have that $rank(A)=n$ and $mge n$. What should daying that the rank of $A$ is $n$ mean? I think that since the trace is less than $m$, the system $Ax=0$ has multiple solutions (just a guess, I don't know how to formalize this), so the null space of $A$ wouldn't be $0$.






linear-algebra systems-of-equations







share|cite|improve this question























  • you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
    – dezdichado
    Sep 5 at 4:41










  • Trace is only defined for square matrices.
    – xbh
    Sep 5 at 4:48










  • Thanks, I corrected
    – Paprika
    Sep 5 at 4:57










  • By the rank-nullity theorem the null space of $A$ is $0$.
    – amsmath
    Sep 5 at 7:14












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Let $Ain mathbbR^mtimes n, mge n$ and $operatornamerank(A) = n$. Prove
that $A^tA$ is nonsingular




I found this answer https://math.stackexchange.com/a/1840814/166180



which gives some hints. It says that $A$ and $A^tA$ have the same null space, that is, if $x$ is such that $A^tAx = 0$, then



$$A^tAx = 0 implies Ax = 0$$



To say that $A^tA$ is nonsingular (invertible) is the same as saying that the null space of $A^tA$ is $0$, which is the same, by above, as saying that the null space of $A$ is $0$.



So by hypothesis on the exercise we have that $rank(A)=n$ and $mge n$. What should daying that the rank of $A$ is $n$ mean? I think that since the trace is less than $m$, the system $Ax=0$ has multiple solutions (just a guess, I don't know how to formalize this), so the null space of $A$ wouldn't be $0$.






linear-algebra systems-of-equations







share|cite|improve this question
















Let $Ain mathbbR^mtimes n, mge n$ and $operatornamerank(A) = n$. Prove
that $A^tA$ is nonsingular




I found this answer https://math.stackexchange.com/a/1840814/166180



which gives some hints. It says that $A$ and $A^tA$ have the same null space, that is, if $x$ is such that $A^tAx = 0$, then



$$A^tAx = 0 implies Ax = 0$$



To say that $A^tA$ is nonsingular (invertible) is the same as saying that the null space of $A^tA$ is $0$, which is the same, by above, as saying that the null space of $A$ is $0$.



So by hypothesis on the exercise we have that $rank(A)=n$ and $mge n$. What should daying that the rank of $A$ is $n$ mean? I think that since the trace is less than $m$, the system $Ax=0$ has multiple solutions (just a guess, I don't know how to formalize this), so the null space of $A$ wouldn't be $0$.






linear-algebra systems-of-equations




linear-algebra systems-of-equations






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edited Sep 5 at 9:37









Bernard

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asked Sep 5 at 4:35









Paprika

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  • you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
    – dezdichado
    Sep 5 at 4:41










  • Trace is only defined for square matrices.
    – xbh
    Sep 5 at 4:48










  • Thanks, I corrected
    – Paprika
    Sep 5 at 4:57










  • By the rank-nullity theorem the null space of $A$ is $0$.
    – amsmath
    Sep 5 at 7:14
















  • you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
    – dezdichado
    Sep 5 at 4:41










  • Trace is only defined for square matrices.
    – xbh
    Sep 5 at 4:48










  • Thanks, I corrected
    – Paprika
    Sep 5 at 4:57










  • By the rank-nullity theorem the null space of $A$ is $0$.
    – amsmath
    Sep 5 at 7:14















you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
– dezdichado
Sep 5 at 4:41




you are confusing trace and ranks of a matrix for the statement as you put it is completely false.
– dezdichado
Sep 5 at 4:41












Trace is only defined for square matrices.
– xbh
Sep 5 at 4:48




Trace is only defined for square matrices.
– xbh
Sep 5 at 4:48












Thanks, I corrected
– Paprika
Sep 5 at 4:57




Thanks, I corrected
– Paprika
Sep 5 at 4:57












By the rank-nullity theorem the null space of $A$ is $0$.
– amsmath
Sep 5 at 7:14




By the rank-nullity theorem the null space of $A$ is $0$.
– amsmath
Sep 5 at 7:14










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Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.






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    up vote
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    Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.






    share|cite|improve this answer
























      up vote
      2
      down vote



      accepted










      Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.






      share|cite|improve this answer






















        up vote
        2
        down vote



        accepted







        up vote
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        accepted






        Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.






        share|cite|improve this answer












        Hint: $Ax = 0$ is equivalent to $||Ax|| = 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 5 at 5:37









        Seewoo Lee

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