Prove that $langlemathbf{A}, mathbf{C}rangle leq delta$ equals with $|mathbf{A}|_*leqdelta$












5












$begingroup$


Given an arbitrary matrix $mathbf{A}in R^{ntimes n}$ and the basis matrix set $mathbb{S}={mathbf{C}in R^{ntimes n}: mathbf{C}^Tmathbf{C}=mathbf{I}_n}$. Then how to prove:



1:If we have $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$ then it holds that $|mathbf{A}|_*leqdelta$.



2:Also, if we have $|mathbf{A}|_*leqdelta$, then we obtain $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$.



In the question, $|mathbf{A}|_*$ is the nuclear norm, and $langlemathbf{A}, mathbf{C}rangle = operatorname{Tr} (mathbf{A}^Tmathbf{C})$.






calculus linear-algebra matrices







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  • $begingroup$
    @OmG, nuclear norm, see the last line of the question
    $endgroup$
    – pointguard0
    Jan 21 at 9:17


















5












$begingroup$


Given an arbitrary matrix $mathbf{A}in R^{ntimes n}$ and the basis matrix set $mathbb{S}={mathbf{C}in R^{ntimes n}: mathbf{C}^Tmathbf{C}=mathbf{I}_n}$. Then how to prove:



1:If we have $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$ then it holds that $|mathbf{A}|_*leqdelta$.



2:Also, if we have $|mathbf{A}|_*leqdelta$, then we obtain $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$.



In the question, $|mathbf{A}|_*$ is the nuclear norm, and $langlemathbf{A}, mathbf{C}rangle = operatorname{Tr} (mathbf{A}^Tmathbf{C})$.






calculus linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    @OmG, nuclear norm, see the last line of the question
    $endgroup$
    – pointguard0
    Jan 21 at 9:17
















5












5








5


1



$begingroup$


Given an arbitrary matrix $mathbf{A}in R^{ntimes n}$ and the basis matrix set $mathbb{S}={mathbf{C}in R^{ntimes n}: mathbf{C}^Tmathbf{C}=mathbf{I}_n}$. Then how to prove:



1:If we have $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$ then it holds that $|mathbf{A}|_*leqdelta$.



2:Also, if we have $|mathbf{A}|_*leqdelta$, then we obtain $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$.



In the question, $|mathbf{A}|_*$ is the nuclear norm, and $langlemathbf{A}, mathbf{C}rangle = operatorname{Tr} (mathbf{A}^Tmathbf{C})$.






calculus linear-algebra matrices







share|cite|improve this question











$endgroup$




Given an arbitrary matrix $mathbf{A}in R^{ntimes n}$ and the basis matrix set $mathbb{S}={mathbf{C}in R^{ntimes n}: mathbf{C}^Tmathbf{C}=mathbf{I}_n}$. Then how to prove:



1:If we have $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$ then it holds that $|mathbf{A}|_*leqdelta$.



2:Also, if we have $|mathbf{A}|_*leqdelta$, then we obtain $langlemathbf{A}, mathbf{C}rangle leq delta$ for all $mathbf{C}subsetmathbb{S}$.



In the question, $|mathbf{A}|_*$ is the nuclear norm, and $langlemathbf{A}, mathbf{C}rangle = operatorname{Tr} (mathbf{A}^Tmathbf{C})$.






calculus linear-algebra matrices




calculus linear-algebra matrices






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edited Jan 22 at 12:17







olivia

















asked Jan 21 at 9:06









oliviaolivia

791616




791616












  • $begingroup$
    @OmG, nuclear norm, see the last line of the question
    $endgroup$
    – pointguard0
    Jan 21 at 9:17




















  • $begingroup$
    @OmG, nuclear norm, see the last line of the question
    $endgroup$
    – pointguard0
    Jan 21 at 9:17


















$begingroup$
@OmG, nuclear norm, see the last line of the question
$endgroup$
– pointguard0
Jan 21 at 9:17






$begingroup$
@OmG, nuclear norm, see the last line of the question
$endgroup$
– pointguard0
Jan 21 at 9:17












1 Answer
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$begingroup$

We can use the polar decomposition
$$
A=Psqrt{A^*A}
$$ where $sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $text{tr}(AB)=text{tr}(BA)$ and $text{tr}(A)=text{tr}(A^T)$ hold for any $ntimes n$ real matrices $A,B$.



1. We can take $B=P$ and it follows
$$
|A|_*=text{tr}(sqrt{A^*A})=text{tr}(B^TPsqrt{A^*A})=text{tr}(Psqrt{A^*A}B^T)=text{tr}(AB^T)=langle A,Brangle le delta.
$$



2. We have $$begin{eqnarray}langle A,Brangle &=&text{tr}(AB^T)\&=&text{tr}(Psqrt{A^*A}B^T)\&=&text{tr}(B^TPsqrt{A^*A})=text{tr}(tilde{B}sqrt{A^*A})end{eqnarray}$$ where $tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)$ is positive semi-definite, we have
$$
text{tr}left((I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)right)ge 0.
$$ This leads to
$$
text{tr}left(tilde{B}sqrt{A^*A}right)+text{tr}left(sqrt{A^*A}tilde{B}^Tright)letext{tr}left(sqrt{A^*A}right)+text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright).
$$ Since $$text{tr}left(sqrt{A^*A}tilde{B}^Tright)=text{tr}left((tilde{B}^T)^Tsqrt{A^*A}^Tright)=text{tr}left(tilde{B}sqrt{A^*A}right)$$ and $$text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright)=text{tr}left(tilde{B}^Ttilde{B}sqrt{A^*A}right)=text{tr}left(sqrt{A^*A}right),$$
it follows$$
langle A,Brangle=text{tr}left(sqrt{A^*A}tilde{B}^Tright)le text{tr}left(sqrt{A^*A}right)=|A|_*le delta.
$$






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    4












    $begingroup$

    We can use the polar decomposition
    $$
    A=Psqrt{A^*A}
    $$     where $sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $text{tr}(AB)=text{tr}(BA)$ and $text{tr}(A)=text{tr}(A^T)$ hold for any $ntimes n$ real matrices $A,B$.



    1. We can take $B=P$ and it follows
    $$
    |A|_*=text{tr}(sqrt{A^*A})=text{tr}(B^TPsqrt{A^*A})=text{tr}(Psqrt{A^*A}B^T)=text{tr}(AB^T)=langle A,Brangle le delta.
    $$



    2. We have $$begin{eqnarray}langle A,Brangle &=&text{tr}(AB^T)\&=&text{tr}(Psqrt{A^*A}B^T)\&=&text{tr}(B^TPsqrt{A^*A})=text{tr}(tilde{B}sqrt{A^*A})end{eqnarray}$$ where $tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)$ is positive semi-definite, we have
    $$
    text{tr}left((I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)right)ge 0.
    $$     This leads to
    $$
    text{tr}left(tilde{B}sqrt{A^*A}right)+text{tr}left(sqrt{A^*A}tilde{B}^Tright)letext{tr}left(sqrt{A^*A}right)+text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright).
    $$     Since $$text{tr}left(sqrt{A^*A}tilde{B}^Tright)=text{tr}left((tilde{B}^T)^Tsqrt{A^*A}^Tright)=text{tr}left(tilde{B}sqrt{A^*A}right)$$ and $$text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright)=text{tr}left(tilde{B}^Ttilde{B}sqrt{A^*A}right)=text{tr}left(sqrt{A^*A}right),$$
    it follows$$
    langle A,Brangle=text{tr}left(sqrt{A^*A}tilde{B}^Tright)le text{tr}left(sqrt{A^*A}right)=|A|_*le delta.
    $$






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      We can use the polar decomposition
      $$
      A=Psqrt{A^*A}
      $$       where $sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $text{tr}(AB)=text{tr}(BA)$ and $text{tr}(A)=text{tr}(A^T)$ hold for any $ntimes n$ real matrices $A,B$.



      1. We can take $B=P$ and it follows
      $$
      |A|_*=text{tr}(sqrt{A^*A})=text{tr}(B^TPsqrt{A^*A})=text{tr}(Psqrt{A^*A}B^T)=text{tr}(AB^T)=langle A,Brangle le delta.
      $$



      2. We have $$begin{eqnarray}langle A,Brangle &=&text{tr}(AB^T)\&=&text{tr}(Psqrt{A^*A}B^T)\&=&text{tr}(B^TPsqrt{A^*A})=text{tr}(tilde{B}sqrt{A^*A})end{eqnarray}$$ where $tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)$ is positive semi-definite, we have
      $$
      text{tr}left((I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)right)ge 0.
      $$       This leads to
      $$
      text{tr}left(tilde{B}sqrt{A^*A}right)+text{tr}left(sqrt{A^*A}tilde{B}^Tright)letext{tr}left(sqrt{A^*A}right)+text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright).
      $$       Since $$text{tr}left(sqrt{A^*A}tilde{B}^Tright)=text{tr}left((tilde{B}^T)^Tsqrt{A^*A}^Tright)=text{tr}left(tilde{B}sqrt{A^*A}right)$$ and $$text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright)=text{tr}left(tilde{B}^Ttilde{B}sqrt{A^*A}right)=text{tr}left(sqrt{A^*A}right),$$
      it follows$$
      langle A,Brangle=text{tr}left(sqrt{A^*A}tilde{B}^Tright)le text{tr}left(sqrt{A^*A}right)=|A|_*le delta.
      $$






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        We can use the polar decomposition
        $$
        A=Psqrt{A^*A}
        $$         where $sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $text{tr}(AB)=text{tr}(BA)$ and $text{tr}(A)=text{tr}(A^T)$ hold for any $ntimes n$ real matrices $A,B$.



        1. We can take $B=P$ and it follows
        $$
        |A|_*=text{tr}(sqrt{A^*A})=text{tr}(B^TPsqrt{A^*A})=text{tr}(Psqrt{A^*A}B^T)=text{tr}(AB^T)=langle A,Brangle le delta.
        $$



        2. We have $$begin{eqnarray}langle A,Brangle &=&text{tr}(AB^T)\&=&text{tr}(Psqrt{A^*A}B^T)\&=&text{tr}(B^TPsqrt{A^*A})=text{tr}(tilde{B}sqrt{A^*A})end{eqnarray}$$ where $tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)$ is positive semi-definite, we have
        $$
        text{tr}left((I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)right)ge 0.
        $$         This leads to
        $$
        text{tr}left(tilde{B}sqrt{A^*A}right)+text{tr}left(sqrt{A^*A}tilde{B}^Tright)letext{tr}left(sqrt{A^*A}right)+text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright).
        $$         Since $$text{tr}left(sqrt{A^*A}tilde{B}^Tright)=text{tr}left((tilde{B}^T)^Tsqrt{A^*A}^Tright)=text{tr}left(tilde{B}sqrt{A^*A}right)$$ and $$text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright)=text{tr}left(tilde{B}^Ttilde{B}sqrt{A^*A}right)=text{tr}left(sqrt{A^*A}right),$$
        it follows$$
        langle A,Brangle=text{tr}left(sqrt{A^*A}tilde{B}^Tright)le text{tr}left(sqrt{A^*A}right)=|A|_*le delta.
        $$






        share|cite|improve this answer









        $endgroup$



        We can use the polar decomposition
        $$
        A=Psqrt{A^*A}
        $$         where $sqrt{A^*A}$ is positive semi-definite and $P$ is orthogonal, i.e. $PP^T=P^TP=I$. Note that $text{tr}(AB)=text{tr}(BA)$ and $text{tr}(A)=text{tr}(A^T)$ hold for any $ntimes n$ real matrices $A,B$.



        1. We can take $B=P$ and it follows
        $$
        |A|_*=text{tr}(sqrt{A^*A})=text{tr}(B^TPsqrt{A^*A})=text{tr}(Psqrt{A^*A}B^T)=text{tr}(AB^T)=langle A,Brangle le delta.
        $$



        2. We have $$begin{eqnarray}langle A,Brangle &=&text{tr}(AB^T)\&=&text{tr}(Psqrt{A^*A}B^T)\&=&text{tr}(B^TPsqrt{A^*A})=text{tr}(tilde{B}sqrt{A^*A})end{eqnarray}$$ where $tilde{B}=B^TP$ is an orthogonal matrix. Since $(I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)$ is positive semi-definite, we have
        $$
        text{tr}left((I-tilde{B})sqrt{A^*A}(I-tilde{B}^T)right)ge 0.
        $$         This leads to
        $$
        text{tr}left(tilde{B}sqrt{A^*A}right)+text{tr}left(sqrt{A^*A}tilde{B}^Tright)letext{tr}left(sqrt{A^*A}right)+text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright).
        $$         Since $$text{tr}left(sqrt{A^*A}tilde{B}^Tright)=text{tr}left((tilde{B}^T)^Tsqrt{A^*A}^Tright)=text{tr}left(tilde{B}sqrt{A^*A}right)$$ and $$text{tr}left(tilde{B}sqrt{A^*A}tilde{B}^Tright)=text{tr}left(tilde{B}^Ttilde{B}sqrt{A^*A}right)=text{tr}left(sqrt{A^*A}right),$$
        it follows$$
        langle A,Brangle=text{tr}left(sqrt{A^*A}tilde{B}^Tright)le text{tr}left(sqrt{A^*A}right)=|A|_*le delta.
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 21 at 9:54









        SongSong

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