Multiplicity of the zero eigenvalue when a symmetric matrix has $m$ identical rows

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Let $A$ be a real symmetric matrix of order $n$.
Show that if $A$ has $m$ identical rows then $A$ has a zero eigenvalue of multiplicity at least $m-1$.




My try:



Since $A$ has $m$ identical rows by elementary row operations we have $m-1$ rows of $A$ equal to $0$.



Hence $A$ has geometric multiplicity of $0$ to be equal to atleast $m-1$.
Since $A$ is symmetric so the algebraic multiplicity of $0$ equals its geometric multiplicity and so it is atleast $m-1$.



But how to show that it is equal to exactly $m-1$?



Will you please help?




linear-algebra matrices eigenvalues-eigenvectors




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    I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
    – user1551
    Aug 26 at 9:17











  • @user1551; thanks but will u say what to assume so that the problem is true
    – PureMathematics
    Aug 26 at 9:18










  • @user1551;i guess here the problem is itself wrong
    – PureMathematics
    Aug 26 at 9:18






  • 1




    For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
    – user1551
    Aug 26 at 9:24














up vote
4
down vote

favorite













Let $A$ be a real symmetric matrix of order $n$.
Show that if $A$ has $m$ identical rows then $A$ has a zero eigenvalue of multiplicity at least $m-1$.




My try:



Since $A$ has $m$ identical rows by elementary row operations we have $m-1$ rows of $A$ equal to $0$.



Hence $A$ has geometric multiplicity of $0$ to be equal to atleast $m-1$.
Since $A$ is symmetric so the algebraic multiplicity of $0$ equals its geometric multiplicity and so it is atleast $m-1$.



But how to show that it is equal to exactly $m-1$?



Will you please help?




linear-algebra matrices eigenvalues-eigenvectors




share|cite|improve this question


















  • 1




    I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
    – user1551
    Aug 26 at 9:17











  • @user1551; thanks but will u say what to assume so that the problem is true
    – PureMathematics
    Aug 26 at 9:18










  • @user1551;i guess here the problem is itself wrong
    – PureMathematics
    Aug 26 at 9:18






  • 1




    For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
    – user1551
    Aug 26 at 9:24












up vote
4
down vote

favorite









up vote
4
down vote

favorite












Let $A$ be a real symmetric matrix of order $n$.
Show that if $A$ has $m$ identical rows then $A$ has a zero eigenvalue of multiplicity at least $m-1$.




My try:



Since $A$ has $m$ identical rows by elementary row operations we have $m-1$ rows of $A$ equal to $0$.



Hence $A$ has geometric multiplicity of $0$ to be equal to atleast $m-1$.
Since $A$ is symmetric so the algebraic multiplicity of $0$ equals its geometric multiplicity and so it is atleast $m-1$.



But how to show that it is equal to exactly $m-1$?



Will you please help?




linear-algebra matrices eigenvalues-eigenvectors




share|cite|improve this question















Let $A$ be a real symmetric matrix of order $n$.
Show that if $A$ has $m$ identical rows then $A$ has a zero eigenvalue of multiplicity at least $m-1$.




My try:



Since $A$ has $m$ identical rows by elementary row operations we have $m-1$ rows of $A$ equal to $0$.



Hence $A$ has geometric multiplicity of $0$ to be equal to atleast $m-1$.
Since $A$ is symmetric so the algebraic multiplicity of $0$ equals its geometric multiplicity and so it is atleast $m-1$.



But how to show that it is equal to exactly $m-1$?



Will you please help?





linear-algebra matrices eigenvalues-eigenvectors





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 26 at 9:13









user1551

67.2k565123




67.2k565123










asked Aug 26 at 8:51









PureMathematics

976




976







  • 1




    I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
    – user1551
    Aug 26 at 9:17











  • @user1551; thanks but will u say what to assume so that the problem is true
    – PureMathematics
    Aug 26 at 9:18










  • @user1551;i guess here the problem is itself wrong
    – PureMathematics
    Aug 26 at 9:18






  • 1




    For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
    – user1551
    Aug 26 at 9:24












  • 1




    I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
    – user1551
    Aug 26 at 9:17











  • @user1551; thanks but will u say what to assume so that the problem is true
    – PureMathematics
    Aug 26 at 9:18










  • @user1551;i guess here the problem is itself wrong
    – PureMathematics
    Aug 26 at 9:18






  • 1




    For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
    – user1551
    Aug 26 at 9:24







1




1




I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
– user1551
Aug 26 at 9:17





I've edited the problem statement, because the original one is false (counterexample: $A=pmatrix1&1&0\ 1&1&0\ 0&0&0$ has $m=2$ identical rows but the multiplicity of the zero eigenvalue is $m$).
– user1551
Aug 26 at 9:17













@user1551; thanks but will u say what to assume so that the problem is true
– PureMathematics
Aug 26 at 9:18




@user1551; thanks but will u say what to assume so that the problem is true
– PureMathematics
Aug 26 at 9:18












@user1551;i guess here the problem is itself wrong
– PureMathematics
Aug 26 at 9:18




@user1551;i guess here the problem is itself wrong
– PureMathematics
Aug 26 at 9:18




1




1




For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
– user1551
Aug 26 at 9:24




For the original statement to hold, you need the other $n-m$ rows to be linearly independent of the $m$ identical rows.
– user1551
Aug 26 at 9:24















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