Can an upperbound constraint on the squared Frobenius norm of a matrix be expressed as a linear matrix...












1












$begingroup$


Is it possible to express an inequality constraint on the squared Frobenius norm of a matrix $X$:



$$|X|_F^2 = mathop{tr}( X^T X ) le t$$



as a linear matrix inequality?



I want to say that it's:



$$left[begin{array}{cc} I & X \ X^T & tIend{array}right] succcurlyeq 0$$



but I've lost confidence that this is correct since the Schur complement would be $tI - X^TX succcurlyeq 0 $ and I can't figure how the trace gets in there.






trace lmis matrix-norms







share|cite|improve this question









$endgroup$












  • $begingroup$
    Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
    $endgroup$
    – quarague
    Jan 9 at 9:14










  • $begingroup$
    I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 15:51










  • $begingroup$
    Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
    $endgroup$
    – quarague
    Jan 9 at 16:22






  • 1




    $begingroup$
    what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 18:38


















1












$begingroup$


Is it possible to express an inequality constraint on the squared Frobenius norm of a matrix $X$:



$$|X|_F^2 = mathop{tr}( X^T X ) le t$$



as a linear matrix inequality?



I want to say that it's:



$$left[begin{array}{cc} I & X \ X^T & tIend{array}right] succcurlyeq 0$$



but I've lost confidence that this is correct since the Schur complement would be $tI - X^TX succcurlyeq 0 $ and I can't figure how the trace gets in there.






trace lmis matrix-norms







share|cite|improve this question









$endgroup$












  • $begingroup$
    Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
    $endgroup$
    – quarague
    Jan 9 at 9:14










  • $begingroup$
    I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 15:51










  • $begingroup$
    Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
    $endgroup$
    – quarague
    Jan 9 at 16:22






  • 1




    $begingroup$
    what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 18:38
















1












1








1





$begingroup$


Is it possible to express an inequality constraint on the squared Frobenius norm of a matrix $X$:



$$|X|_F^2 = mathop{tr}( X^T X ) le t$$



as a linear matrix inequality?



I want to say that it's:



$$left[begin{array}{cc} I & X \ X^T & tIend{array}right] succcurlyeq 0$$



but I've lost confidence that this is correct since the Schur complement would be $tI - X^TX succcurlyeq 0 $ and I can't figure how the trace gets in there.






trace lmis matrix-norms







share|cite|improve this question









$endgroup$




Is it possible to express an inequality constraint on the squared Frobenius norm of a matrix $X$:



$$|X|_F^2 = mathop{tr}( X^T X ) le t$$



as a linear matrix inequality?



I want to say that it's:



$$left[begin{array}{cc} I & X \ X^T & tIend{array}right] succcurlyeq 0$$



but I've lost confidence that this is correct since the Schur complement would be $tI - X^TX succcurlyeq 0 $ and I can't figure how the trace gets in there.






trace lmis matrix-norms




trace lmis matrix-norms






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 5:14









Alec JacobsonAlec Jacobson

270111




270111












  • $begingroup$
    Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
    $endgroup$
    – quarague
    Jan 9 at 9:14










  • $begingroup$
    I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 15:51










  • $begingroup$
    Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
    $endgroup$
    – quarague
    Jan 9 at 16:22






  • 1




    $begingroup$
    what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 18:38




















  • $begingroup$
    Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
    $endgroup$
    – quarague
    Jan 9 at 9:14










  • $begingroup$
    I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 15:51










  • $begingroup$
    Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
    $endgroup$
    – quarague
    Jan 9 at 16:22






  • 1




    $begingroup$
    what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
    $endgroup$
    – Alec Jacobson
    Jan 9 at 18:38


















$begingroup$
Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
$endgroup$
– quarague
Jan 9 at 9:14




$begingroup$
Does your round greater equal sign stand for determinant? If so than what you wrote does not work for 2 by 2 matrices (unless I made a computation error somewhere).
$endgroup$
– quarague
Jan 9 at 9:14












$begingroup$
I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
$endgroup$
– Alec Jacobson
Jan 9 at 15:51




$begingroup$
I intended $M succcurlyeq 0$ to mean $M$ is positive semi-definite.
$endgroup$
– Alec Jacobson
Jan 9 at 15:51












$begingroup$
Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
$endgroup$
– quarague
Jan 9 at 16:22




$begingroup$
Positive semidefinite is equivalent to eigenvalues greater equal zero. I think that doesn't even work for 1 by 1 matrices.
$endgroup$
– quarague
Jan 9 at 16:22




1




1




$begingroup$
what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
$endgroup$
– Alec Jacobson
Jan 9 at 18:38






$begingroup$
what doesn't work for 1by1? The Schur complement in that case says $t - x^2 succcurlyeq 0$ or $c (t -x^2) c ge 0, forall c$ which simply means $(t -x^2)ge 0$ which in this case is the same as $mathop{tr}(x^2) = x^2 le t$.
$endgroup$
– Alec Jacobson
Jan 9 at 18:38












1 Answer
1






active

oldest

votes


















1












$begingroup$

There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^top X$. These can be calculated using



$$
Y_i = X,e_i
$$



with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^top Y_i$ is the $i$th diagonal term of $X^top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^top Y_i leq alpha_i$, namely



$$
begin{bmatrix}
I & Y_i \ Y_i^top & alpha_i
end{bmatrix} succeq 0,
$$



with $alpha_i in mathbb{R}$. Now a bound for $|X|_F^2$ can be found by summing all $alpha_i$, which should be smaller or equal to $t$



$$
sum alpha_i leq t,
$$



which is also a linear inequality.




By using an intermediate LMI for $X^top X$ you might also be able to write $X^top X preceq M$, with $M = M^top$, as



$$
begin{bmatrix}
I & X \ X^top & M
end{bmatrix} succeq 0.
$$



An upper bound for $|X|_F^2$ would then be $text{Tr}(M)$, so adding the linear inequality $text{Tr}(M) leq t$ would make this system of LMI's equivalent to your problem. To show that $X^top X preceq M$ also implies that $text{Tr}(X^top X) leq text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^top X preceq M$ is equivalent to $M - X^top X succeq 0$, thus $M - X^top X$ can only have non-negative eigenvalues and therefore $text{Tr}(M - X^top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $text{Tr}(X^top X) leq text{Tr}(M)$ is equivalent to $text{Tr}(M - X^top X) geq 0$ and in the previous sentence it was shown that it holds when $X^top X preceq M$, thus $text{Tr}(X^top X) leq text{Tr}(M)$ should hold in that case as well.



It can be noted that $M$ might add more degrees of freedom than all $alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    active

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    active

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    1












    $begingroup$

    There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^top X$. These can be calculated using



    $$
    Y_i = X,e_i
    $$



    with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^top Y_i$ is the $i$th diagonal term of $X^top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^top Y_i leq alpha_i$, namely



    $$
    begin{bmatrix}
    I & Y_i \ Y_i^top & alpha_i
    end{bmatrix} succeq 0,
    $$



    with $alpha_i in mathbb{R}$. Now a bound for $|X|_F^2$ can be found by summing all $alpha_i$, which should be smaller or equal to $t$



    $$
    sum alpha_i leq t,
    $$



    which is also a linear inequality.




    By using an intermediate LMI for $X^top X$ you might also be able to write $X^top X preceq M$, with $M = M^top$, as



    $$
    begin{bmatrix}
    I & X \ X^top & M
    end{bmatrix} succeq 0.
    $$



    An upper bound for $|X|_F^2$ would then be $text{Tr}(M)$, so adding the linear inequality $text{Tr}(M) leq t$ would make this system of LMI's equivalent to your problem. To show that $X^top X preceq M$ also implies that $text{Tr}(X^top X) leq text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^top X preceq M$ is equivalent to $M - X^top X succeq 0$, thus $M - X^top X$ can only have non-negative eigenvalues and therefore $text{Tr}(M - X^top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $text{Tr}(X^top X) leq text{Tr}(M)$ is equivalent to $text{Tr}(M - X^top X) geq 0$ and in the previous sentence it was shown that it holds when $X^top X preceq M$, thus $text{Tr}(X^top X) leq text{Tr}(M)$ should hold in that case as well.



    It can be noted that $M$ might add more degrees of freedom than all $alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^top X$. These can be calculated using



      $$
      Y_i = X,e_i
      $$



      with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^top Y_i$ is the $i$th diagonal term of $X^top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^top Y_i leq alpha_i$, namely



      $$
      begin{bmatrix}
      I & Y_i \ Y_i^top & alpha_i
      end{bmatrix} succeq 0,
      $$



      with $alpha_i in mathbb{R}$. Now a bound for $|X|_F^2$ can be found by summing all $alpha_i$, which should be smaller or equal to $t$



      $$
      sum alpha_i leq t,
      $$



      which is also a linear inequality.




      By using an intermediate LMI for $X^top X$ you might also be able to write $X^top X preceq M$, with $M = M^top$, as



      $$
      begin{bmatrix}
      I & X \ X^top & M
      end{bmatrix} succeq 0.
      $$



      An upper bound for $|X|_F^2$ would then be $text{Tr}(M)$, so adding the linear inequality $text{Tr}(M) leq t$ would make this system of LMI's equivalent to your problem. To show that $X^top X preceq M$ also implies that $text{Tr}(X^top X) leq text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^top X preceq M$ is equivalent to $M - X^top X succeq 0$, thus $M - X^top X$ can only have non-negative eigenvalues and therefore $text{Tr}(M - X^top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $text{Tr}(X^top X) leq text{Tr}(M)$ is equivalent to $text{Tr}(M - X^top X) geq 0$ and in the previous sentence it was shown that it holds when $X^top X preceq M$, thus $text{Tr}(X^top X) leq text{Tr}(M)$ should hold in that case as well.



      It can be noted that $M$ might add more degrees of freedom than all $alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^top X$. These can be calculated using



        $$
        Y_i = X,e_i
        $$



        with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^top Y_i$ is the $i$th diagonal term of $X^top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^top Y_i leq alpha_i$, namely



        $$
        begin{bmatrix}
        I & Y_i \ Y_i^top & alpha_i
        end{bmatrix} succeq 0,
        $$



        with $alpha_i in mathbb{R}$. Now a bound for $|X|_F^2$ can be found by summing all $alpha_i$, which should be smaller or equal to $t$



        $$
        sum alpha_i leq t,
        $$



        which is also a linear inequality.




        By using an intermediate LMI for $X^top X$ you might also be able to write $X^top X preceq M$, with $M = M^top$, as



        $$
        begin{bmatrix}
        I & X \ X^top & M
        end{bmatrix} succeq 0.
        $$



        An upper bound for $|X|_F^2$ would then be $text{Tr}(M)$, so adding the linear inequality $text{Tr}(M) leq t$ would make this system of LMI's equivalent to your problem. To show that $X^top X preceq M$ also implies that $text{Tr}(X^top X) leq text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^top X preceq M$ is equivalent to $M - X^top X succeq 0$, thus $M - X^top X$ can only have non-negative eigenvalues and therefore $text{Tr}(M - X^top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $text{Tr}(X^top X) leq text{Tr}(M)$ is equivalent to $text{Tr}(M - X^top X) geq 0$ and in the previous sentence it was shown that it holds when $X^top X preceq M$, thus $text{Tr}(X^top X) leq text{Tr}(M)$ should hold in that case as well.



        It can be noted that $M$ might add more degrees of freedom than all $alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.






        share|cite|improve this answer











        $endgroup$



        There might be a compacter and more elegant way, but one way you can represent it as LMI's is by using intermediate values for each diagonal term of $X^top X$. These can be calculated using



        $$
        Y_i = X,e_i
        $$



        with $e_i$ a vector with the $i$th element equal to one and the rest zeros (so $Y_i$ is the $i$th column of $X$), such that $Y_i^top Y_i$ is the $i$th diagonal term of $X^top X$. Then using the Schur complement you can write for every diagonal term an LMI for $Y_i^top Y_i leq alpha_i$, namely



        $$
        begin{bmatrix}
        I & Y_i \ Y_i^top & alpha_i
        end{bmatrix} succeq 0,
        $$



        with $alpha_i in mathbb{R}$. Now a bound for $|X|_F^2$ can be found by summing all $alpha_i$, which should be smaller or equal to $t$



        $$
        sum alpha_i leq t,
        $$



        which is also a linear inequality.




        By using an intermediate LMI for $X^top X$ you might also be able to write $X^top X preceq M$, with $M = M^top$, as



        $$
        begin{bmatrix}
        I & X \ X^top & M
        end{bmatrix} succeq 0.
        $$



        An upper bound for $|X|_F^2$ would then be $text{Tr}(M)$, so adding the linear inequality $text{Tr}(M) leq t$ would make this system of LMI's equivalent to your problem. To show that $X^top X preceq M$ also implies that $text{Tr}(X^top X) leq text{Tr}(M)$ you can use that the trace of a matrix is equal to the sum of all its eigenvalues. Namely $X^top X preceq M$ is equivalent to $M - X^top X succeq 0$, thus $M - X^top X$ can only have non-negative eigenvalues and therefore $text{Tr}(M - X^top X)$ is the sum of these non-negative eigenvalues, which is also non-negative. The trace inequality $text{Tr}(X^top X) leq text{Tr}(M)$ is equivalent to $text{Tr}(M - X^top X) geq 0$ and in the previous sentence it was shown that it holds when $X^top X preceq M$, thus $text{Tr}(X^top X) leq text{Tr}(M)$ should hold in that case as well.



        It can be noted that $M$ might add more degrees of freedom than all $alpha_i$ so might or might not be an attractive alternative of writing the problem as LMI's.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 10 at 16:54

























        answered Jan 10 at 15:26









        Kwin van der VeenKwin van der Veen

        5,6402828




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            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
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