Given A, B matrices, prove BA is singular

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Given $A_5x7$, $B_7x5$, prove that $BA$ is singular.



This is what I did:



Matrix $A$ is $5x7$. That means, there are 7 parameters with 5 equations. There are at least 2 free variables, so the solution for $Aunderlinex=underline0$ is non-trivial.
BA means each $i$ column of B multiplies each $i$ column of A, where $1≤i≤7$. The column with the free variables also multiplied by the corresponding B column. This leads to free variables multiplied by a scalar, so the outcome solution is still non-trivial.



I'm afraid that is not how the proof goes...




linear-algebra matrices




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    There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
    – copper.hat
    Aug 13 at 6:32






  • 2




    I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
    – Lord Shark the Unknown
    Aug 13 at 6:34














up vote
0
down vote

favorite












Given $A_5x7$, $B_7x5$, prove that $BA$ is singular.



This is what I did:



Matrix $A$ is $5x7$. That means, there are 7 parameters with 5 equations. There are at least 2 free variables, so the solution for $Aunderlinex=underline0$ is non-trivial.
BA means each $i$ column of B multiplies each $i$ column of A, where $1≤i≤7$. The column with the free variables also multiplied by the corresponding B column. This leads to free variables multiplied by a scalar, so the outcome solution is still non-trivial.



I'm afraid that is not how the proof goes...




linear-algebra matrices




share|cite|improve this question
















  • 3




    There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
    – copper.hat
    Aug 13 at 6:32






  • 2




    I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
    – Lord Shark the Unknown
    Aug 13 at 6:34












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given $A_5x7$, $B_7x5$, prove that $BA$ is singular.



This is what I did:



Matrix $A$ is $5x7$. That means, there are 7 parameters with 5 equations. There are at least 2 free variables, so the solution for $Aunderlinex=underline0$ is non-trivial.
BA means each $i$ column of B multiplies each $i$ column of A, where $1≤i≤7$. The column with the free variables also multiplied by the corresponding B column. This leads to free variables multiplied by a scalar, so the outcome solution is still non-trivial.



I'm afraid that is not how the proof goes...




linear-algebra matrices




share|cite|improve this question












Given $A_5x7$, $B_7x5$, prove that $BA$ is singular.



This is what I did:



Matrix $A$ is $5x7$. That means, there are 7 parameters with 5 equations. There are at least 2 free variables, so the solution for $Aunderlinex=underline0$ is non-trivial.
BA means each $i$ column of B multiplies each $i$ column of A, where $1≤i≤7$. The column with the free variables also multiplied by the corresponding B column. This leads to free variables multiplied by a scalar, so the outcome solution is still non-trivial.



I'm afraid that is not how the proof goes...





linear-algebra matrices





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 13 at 6:30









Michael Sazonov

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  • 3




    There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
    – copper.hat
    Aug 13 at 6:32






  • 2




    I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
    – Lord Shark the Unknown
    Aug 13 at 6:34












  • 3




    There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
    – copper.hat
    Aug 13 at 6:32






  • 2




    I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
    – Lord Shark the Unknown
    Aug 13 at 6:34







3




3




There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
– copper.hat
Aug 13 at 6:32




There is some $x ne 0$ such that $Ax=0$. Then $BA x= 0$ and hence it is singular.
– copper.hat
Aug 13 at 6:32




2




2




I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
– Lord Shark the Unknown
Aug 13 at 6:34




I'd say your solution isn't well written but the idea is sound: as $5<7$ there is a nonzero vector $x$ with $Ax=0$. Then $BAx=0$, but if $BA$ were nonsingular then $x=0$, a contradiction. We conclude that $BA$ is singular.
– Lord Shark the Unknown
Aug 13 at 6:34










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Hint:



$A$ is $5 times 7$ implies $Ax=0$ has a non zero solution.






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    Hint:



    $A$ is $5 times 7$ implies $Ax=0$ has a non zero solution.






    share|cite|improve this answer
























      up vote
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      Hint:



      $A$ is $5 times 7$ implies $Ax=0$ has a non zero solution.






      share|cite|improve this answer






















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        Hint:



        $A$ is $5 times 7$ implies $Ax=0$ has a non zero solution.






        share|cite|improve this answer












        Hint:



        $A$ is $5 times 7$ implies $Ax=0$ has a non zero solution.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Aug 13 at 6:33









        Chinnapparaj R

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