Off-diagonal entries of a symmetric positive semi-definite matrix: Prove $a_{ij}^2 leq a_{ii}a_{jj}$ for $i...












0















If $A=(a_{ij})$ is positive semi-definite, prove that $a_{ij}^2 leq a_{ii}a_{jj}$ for $i neq j$.




I've only been provided with the following definition:




A symmetric matrix A of order $n$ is semi-definite if for all non-zero $x in mathbb{R}^n, x^tAx geq 0$.




I have tried substituting a vector $x$ with non-zero entries (1 in this case) in the $i$ and $j$ position, and $0$ otherwise.



The resulting equation I get is
begin{equation}
a_{ij}+a_{ji}+a_{ii}+a_{jj} geq 0
implies 2a_{ij}+a_{ii}+a_{jj} geq 0
end{equation}
as $A$ is symmetric.



How do I manipulate this into the required form? ($a_{ij}^2 leq a_{ii}a_{jj}$)






linear-algebra







share|cite|improve this question






















  • Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
    – Casper Thalen
    Nov 29 '18 at 13:52
















0















If $A=(a_{ij})$ is positive semi-definite, prove that $a_{ij}^2 leq a_{ii}a_{jj}$ for $i neq j$.




I've only been provided with the following definition:




A symmetric matrix A of order $n$ is semi-definite if for all non-zero $x in mathbb{R}^n, x^tAx geq 0$.




I have tried substituting a vector $x$ with non-zero entries (1 in this case) in the $i$ and $j$ position, and $0$ otherwise.



The resulting equation I get is
begin{equation}
a_{ij}+a_{ji}+a_{ii}+a_{jj} geq 0
implies 2a_{ij}+a_{ii}+a_{jj} geq 0
end{equation}
as $A$ is symmetric.



How do I manipulate this into the required form? ($a_{ij}^2 leq a_{ii}a_{jj}$)






linear-algebra







share|cite|improve this question






















  • Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
    – Casper Thalen
    Nov 29 '18 at 13:52














0












0








0


1






If $A=(a_{ij})$ is positive semi-definite, prove that $a_{ij}^2 leq a_{ii}a_{jj}$ for $i neq j$.




I've only been provided with the following definition:




A symmetric matrix A of order $n$ is semi-definite if for all non-zero $x in mathbb{R}^n, x^tAx geq 0$.




I have tried substituting a vector $x$ with non-zero entries (1 in this case) in the $i$ and $j$ position, and $0$ otherwise.



The resulting equation I get is
begin{equation}
a_{ij}+a_{ji}+a_{ii}+a_{jj} geq 0
implies 2a_{ij}+a_{ii}+a_{jj} geq 0
end{equation}
as $A$ is symmetric.



How do I manipulate this into the required form? ($a_{ij}^2 leq a_{ii}a_{jj}$)






linear-algebra







share|cite|improve this question














If $A=(a_{ij})$ is positive semi-definite, prove that $a_{ij}^2 leq a_{ii}a_{jj}$ for $i neq j$.




I've only been provided with the following definition:




A symmetric matrix A of order $n$ is semi-definite if for all non-zero $x in mathbb{R}^n, x^tAx geq 0$.




I have tried substituting a vector $x$ with non-zero entries (1 in this case) in the $i$ and $j$ position, and $0$ otherwise.



The resulting equation I get is
begin{equation}
a_{ij}+a_{ji}+a_{ii}+a_{jj} geq 0
implies 2a_{ij}+a_{ii}+a_{jj} geq 0
end{equation}
as $A$ is symmetric.



How do I manipulate this into the required form? ($a_{ij}^2 leq a_{ii}a_{jj}$)






linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Nov 29 '18 at 13:47









JanJan

453




453












  • Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
    – Casper Thalen
    Nov 29 '18 at 13:52


















  • Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
    – Casper Thalen
    Nov 29 '18 at 13:52
















Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
– Casper Thalen
Nov 29 '18 at 13:52




Although you were right in gathering insight by working out the general case, I'm afraid you worked out the equation for $n=2$ without naming $i,j$ as $1$ and $2$. I recommend trying to work out $x^tAx$ again!
– Casper Thalen
Nov 29 '18 at 13:52










3 Answers
3






active

oldest

votes


















4














Take $x=e_i+lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}geq 0$ for any $lambdainmathbb{R}.$ This implies the quadratic form $a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}leq 0.$ Hence the result.






share|cite|improve this answer





























    2














    The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
    $|a_{ij}|leq {sqrt {a_{ii}a_{jj}}}$
    From here, obviously, follows what you want to prove.






    share|cite|improve this answer





























      1














      You're on the right track, but just plugging in $1$ isn't enough in this case. Let $e_1,dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position. Setting $x = x_i e_i + x_j e_j$, we compute
      $$
      x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = pmatrix{x_i & x_j}pmatrix{a_{ii} & a_{ij}\a_{ij} & a_{jj}}pmatrix{x_i\x_j}
      $$
      Now, it suffices to note that if $(begin{smallmatrix}a_{ii} & a_{ij}\a_{ij} & a_{jj}end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

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        active

        oldest

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        4














        Take $x=e_i+lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}geq 0$ for any $lambdainmathbb{R}.$ This implies the quadratic form $a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}leq 0.$ Hence the result.






        share|cite|improve this answer


























          4














          Take $x=e_i+lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}geq 0$ for any $lambdainmathbb{R}.$ This implies the quadratic form $a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}leq 0.$ Hence the result.






          share|cite|improve this answer
























            4












            4








            4






            Take $x=e_i+lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}geq 0$ for any $lambdainmathbb{R}.$ This implies the quadratic form $a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}leq 0.$ Hence the result.






            share|cite|improve this answer












            Take $x=e_i+lambda e_j$ where $e_i$ is the vector with $1$ at $i$-th entry and $0$ for the other co-ordinates. Then $x'Ax=a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}geq 0$ for any $lambdainmathbb{R}.$ This implies the quadratic form $a_{ii}+2lambda a_{ij}+lambda^2 a_{jj}$ can have atmost one real root. So, we must have $4a_{ij}^2-4a_{ii}a_{jj}leq 0.$ Hence the result.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 29 '18 at 13:55









            John_WickJohn_Wick

            1,476111




            1,476111























                2














                The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
                $|a_{ij}|leq {sqrt {a_{ii}a_{jj}}}$
                From here, obviously, follows what you want to prove.






                share|cite|improve this answer


























                  2














                  The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
                  $|a_{ij}|leq {sqrt {a_{ii}a_{jj}}}$
                  From here, obviously, follows what you want to prove.






                  share|cite|improve this answer
























                    2












                    2








                    2






                    The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
                    $|a_{ij}|leq {sqrt {a_{ii}a_{jj}}}$
                    From here, obviously, follows what you want to prove.






                    share|cite|improve this answer












                    The diagonal entries $a_{ii}$ are real and non-negative. As a consequence the trace, $tr(A) ge 0$. Now, since every principal sub-matrix (in particular, 2-by-2) is positive definite,
                    $|a_{ij}|leq {sqrt {a_{ii}a_{jj}}}$
                    From here, obviously, follows what you want to prove.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 29 '18 at 13:57









                    FrobeniusFrobenius

                    613




                    613























                        1














                        You're on the right track, but just plugging in $1$ isn't enough in this case. Let $e_1,dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position. Setting $x = x_i e_i + x_j e_j$, we compute
                        $$
                        x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = pmatrix{x_i & x_j}pmatrix{a_{ii} & a_{ij}\a_{ij} & a_{jj}}pmatrix{x_i\x_j}
                        $$
                        Now, it suffices to note that if $(begin{smallmatrix}a_{ii} & a_{ij}\a_{ij} & a_{jj}end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.






                        share|cite|improve this answer


























                          1














                          You're on the right track, but just plugging in $1$ isn't enough in this case. Let $e_1,dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position. Setting $x = x_i e_i + x_j e_j$, we compute
                          $$
                          x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = pmatrix{x_i & x_j}pmatrix{a_{ii} & a_{ij}\a_{ij} & a_{jj}}pmatrix{x_i\x_j}
                          $$
                          Now, it suffices to note that if $(begin{smallmatrix}a_{ii} & a_{ij}\a_{ij} & a_{jj}end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.






                          share|cite|improve this answer
























                            1












                            1








                            1






                            You're on the right track, but just plugging in $1$ isn't enough in this case. Let $e_1,dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position. Setting $x = x_i e_i + x_j e_j$, we compute
                            $$
                            x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = pmatrix{x_i & x_j}pmatrix{a_{ii} & a_{ij}\a_{ij} & a_{jj}}pmatrix{x_i\x_j}
                            $$
                            Now, it suffices to note that if $(begin{smallmatrix}a_{ii} & a_{ij}\a_{ij} & a_{jj}end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.






                            share|cite|improve this answer












                            You're on the right track, but just plugging in $1$ isn't enough in this case. Let $e_1,dots,e_n$ denote the standard basis vectors, so that $x_i e_i + x_j e_j$ is a vector with $x_i$ in the $i$th position and $x_j$ in the $j$th position. Setting $x = x_i e_i + x_j e_j$, we compute
                            $$
                            x^TAx = a_{ii} x_{i}^2 + a_{jj} x_j^2 + 2 a_ia_j x_ix_j = pmatrix{x_i & x_j}pmatrix{a_{ii} & a_{ij}\a_{ij} & a_{jj}}pmatrix{x_i\x_j}
                            $$
                            Now, it suffices to note that if $(begin{smallmatrix}a_{ii} & a_{ij}\a_{ij} & a_{jj}end{smallmatrix})$ is positive semidefinite, then its determinant $a_{ii}a_{jj} - a_{ij}^2$ must be positive.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 29 '18 at 13:58









                            OmnomnomnomOmnomnomnom

                            127k788176




                            127k788176






























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