How to show for any orthogonal vector set and any matrix $sum_u |Au|_2^2 leq |A|_F^2$?












3














The following paper on page 16, in line 17
Online Principal Component Analysis
says for any orthogonal vector set and any matrix $sum_u |Au|_2^2 leq |A|_F^2$ is true. However, if we let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$ we have



$$sum_u |Au|_2^2=|AU|_F^2$$
where $U$ is the matrix that has $u_i$ as its columns and $|cdot|_F$ is Frobenius norm. Using Cauchy-Schwarz inequality



$$|AU|_F^2 leq |A|_F^2|U|_F^2$$



My question is how we can ignore $|U|_F^2$?






linear-algebra cauchy-schwarz-inequality







share|cite|improve this question


















  • 1




    $|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
    – A.Γ.
    2 days ago








  • 1




    But they are not orthanormal, it says they are orthogonal.
    – Saeed
    2 days ago










  • Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
    – Adrian Keister
    2 days ago






  • 2




    It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
    – A.Γ.
    2 days ago


















3














The following paper on page 16, in line 17
Online Principal Component Analysis
says for any orthogonal vector set and any matrix $sum_u |Au|_2^2 leq |A|_F^2$ is true. However, if we let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$ we have



$$sum_u |Au|_2^2=|AU|_F^2$$
where $U$ is the matrix that has $u_i$ as its columns and $|cdot|_F$ is Frobenius norm. Using Cauchy-Schwarz inequality



$$|AU|_F^2 leq |A|_F^2|U|_F^2$$



My question is how we can ignore $|U|_F^2$?






linear-algebra cauchy-schwarz-inequality







share|cite|improve this question


















  • 1




    $|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
    – A.Γ.
    2 days ago








  • 1




    But they are not orthanormal, it says they are orthogonal.
    – Saeed
    2 days ago










  • Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
    – Adrian Keister
    2 days ago






  • 2




    It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
    – A.Γ.
    2 days ago
















3












3








3







The following paper on page 16, in line 17
Online Principal Component Analysis
says for any orthogonal vector set and any matrix $sum_u |Au|_2^2 leq |A|_F^2$ is true. However, if we let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$ we have



$$sum_u |Au|_2^2=|AU|_F^2$$
where $U$ is the matrix that has $u_i$ as its columns and $|cdot|_F$ is Frobenius norm. Using Cauchy-Schwarz inequality



$$|AU|_F^2 leq |A|_F^2|U|_F^2$$



My question is how we can ignore $|U|_F^2$?






linear-algebra cauchy-schwarz-inequality







share|cite|improve this question













The following paper on page 16, in line 17
Online Principal Component Analysis
says for any orthogonal vector set and any matrix $sum_u |Au|_2^2 leq |A|_F^2$ is true. However, if we let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$ we have



$$sum_u |Au|_2^2=|AU|_F^2$$
where $U$ is the matrix that has $u_i$ as its columns and $|cdot|_F$ is Frobenius norm. Using Cauchy-Schwarz inequality



$$|AU|_F^2 leq |A|_F^2|U|_F^2$$



My question is how we can ignore $|U|_F^2$?






linear-algebra cauchy-schwarz-inequality




linear-algebra cauchy-schwarz-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 days ago









Saeed

698310




698310








  • 1




    $|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
    – A.Γ.
    2 days ago








  • 1




    But they are not orthanormal, it says they are orthogonal.
    – Saeed
    2 days ago










  • Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
    – Adrian Keister
    2 days ago






  • 2




    It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
    – A.Γ.
    2 days ago
















  • 1




    $|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
    – A.Γ.
    2 days ago








  • 1




    But they are not orthanormal, it says they are orthogonal.
    – Saeed
    2 days ago










  • Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
    – Adrian Keister
    2 days ago






  • 2




    It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
    – A.Γ.
    2 days ago










1




1




$|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
– A.Γ.
2 days ago






$|AU|_F=|A|_F$ which follows from the trace property and $UU^*=I$.
– A.Γ.
2 days ago






1




1




But they are not orthanormal, it says they are orthogonal.
– Saeed
2 days ago




But they are not orthanormal, it says they are orthogonal.
– Saeed
2 days ago












Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
– Adrian Keister
2 days ago




Some authors are sloppy about the distinction between "orthonormal" and "orthogonal". It's pretty straight-forward to normalize orthogonal vectors (orthogonalizing them using, say, Gram-Schmidt, is what takes all the effort!), so some authors don't trouble about it. Technically, you're correct, of course.
– Adrian Keister
2 days ago




2




2




It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
– A.Γ.
2 days ago






It is sloppy written, but $U$ matrix they work with in this proof has orthonormal columns (Observation 14, page 14). Otherwise it is not true (take one vector and multiply it by a huge number).
– A.Γ.
2 days ago












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Note that $A^*A$ is a square matrix and $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_m}$ for $mathbb{R}^m$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.



We have
$$sum_{i=1}^m |Au_i|_2^2 = sum_{i=1}^m langle Au_i, Au_irangle = sum_{i=1}^m langle A^*Au_i, u_irangle = sum_{i=1}^m lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$



The interesting part is that the sum $sum_{i=1}^m |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_m}$. Indeed, if ${v_1, ldots, v_m}$ is some other orthonormal basis for $mathbb{R}^m$, we have
begin{align}
sum_{i=1}^m |Au_i|_2^2 &= sum_{i=1}^m langle A^*Au_i, u_irangle\
&= sum_{i=1}^m leftlangle sum_{j=1}^mlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^mlangle u_i,v_krangle v_krightrangle\
&= sum_{j=1}^m sum_{k=1}^m left(sum_{i=1}^mlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
&= sum_{j=1}^m sum_{k=1}^m langle v_j,v_kranglelangle A^*A v_j,v_krangle\
&= sum_{j=1}^m langle A^*A v_j,v_jrangle\
&= sum_{j=1}^m |Av_j|_2^2
end{align}



Hence we can extend any orthonormal set ${v_1, ldots, v_k} subseteq mathbb{R}^m$ to an orthonormal basis ${v_1, ldots, v_m}$ for $mathbb{R}^m$ to obtain
$$sum_{i=1}^k |Av_i|_2^2 lesum_{i=1}^m |Av_i|_2^2 = sum_{i=1}^m |Au_i|_2^2 = |A|_F^2 $$






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    Note that $A^*A$ is a square matrix and $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_m}$ for $mathbb{R}^m$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.



    We have
    $$sum_{i=1}^m |Au_i|_2^2 = sum_{i=1}^m langle Au_i, Au_irangle = sum_{i=1}^m langle A^*Au_i, u_irangle = sum_{i=1}^m lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$



    The interesting part is that the sum $sum_{i=1}^m |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_m}$. Indeed, if ${v_1, ldots, v_m}$ is some other orthonormal basis for $mathbb{R}^m$, we have
    begin{align}
    sum_{i=1}^m |Au_i|_2^2 &= sum_{i=1}^m langle A^*Au_i, u_irangle\
    &= sum_{i=1}^m leftlangle sum_{j=1}^mlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^mlangle u_i,v_krangle v_krightrangle\
    &= sum_{j=1}^m sum_{k=1}^m left(sum_{i=1}^mlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
    &= sum_{j=1}^m sum_{k=1}^m langle v_j,v_kranglelangle A^*A v_j,v_krangle\
    &= sum_{j=1}^m langle A^*A v_j,v_jrangle\
    &= sum_{j=1}^m |Av_j|_2^2
    end{align}



    Hence we can extend any orthonormal set ${v_1, ldots, v_k} subseteq mathbb{R}^m$ to an orthonormal basis ${v_1, ldots, v_m}$ for $mathbb{R}^m$ to obtain
    $$sum_{i=1}^k |Av_i|_2^2 lesum_{i=1}^m |Av_i|_2^2 = sum_{i=1}^m |Au_i|_2^2 = |A|_F^2 $$






    share|cite|improve this answer


























      1














      Note that $A^*A$ is a square matrix and $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_m}$ for $mathbb{R}^m$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.



      We have
      $$sum_{i=1}^m |Au_i|_2^2 = sum_{i=1}^m langle Au_i, Au_irangle = sum_{i=1}^m langle A^*Au_i, u_irangle = sum_{i=1}^m lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$



      The interesting part is that the sum $sum_{i=1}^m |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_m}$. Indeed, if ${v_1, ldots, v_m}$ is some other orthonormal basis for $mathbb{R}^m$, we have
      begin{align}
      sum_{i=1}^m |Au_i|_2^2 &= sum_{i=1}^m langle A^*Au_i, u_irangle\
      &= sum_{i=1}^m leftlangle sum_{j=1}^mlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^mlangle u_i,v_krangle v_krightrangle\
      &= sum_{j=1}^m sum_{k=1}^m left(sum_{i=1}^mlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
      &= sum_{j=1}^m sum_{k=1}^m langle v_j,v_kranglelangle A^*A v_j,v_krangle\
      &= sum_{j=1}^m langle A^*A v_j,v_jrangle\
      &= sum_{j=1}^m |Av_j|_2^2
      end{align}



      Hence we can extend any orthonormal set ${v_1, ldots, v_k} subseteq mathbb{R}^m$ to an orthonormal basis ${v_1, ldots, v_m}$ for $mathbb{R}^m$ to obtain
      $$sum_{i=1}^k |Av_i|_2^2 lesum_{i=1}^m |Av_i|_2^2 = sum_{i=1}^m |Au_i|_2^2 = |A|_F^2 $$






      share|cite|improve this answer
























        1












        1








        1






        Note that $A^*A$ is a square matrix and $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_m}$ for $mathbb{R}^m$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.



        We have
        $$sum_{i=1}^m |Au_i|_2^2 = sum_{i=1}^m langle Au_i, Au_irangle = sum_{i=1}^m langle A^*Au_i, u_irangle = sum_{i=1}^m lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$



        The interesting part is that the sum $sum_{i=1}^m |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_m}$. Indeed, if ${v_1, ldots, v_m}$ is some other orthonormal basis for $mathbb{R}^m$, we have
        begin{align}
        sum_{i=1}^m |Au_i|_2^2 &= sum_{i=1}^m langle A^*Au_i, u_irangle\
        &= sum_{i=1}^m leftlangle sum_{j=1}^mlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^mlangle u_i,v_krangle v_krightrangle\
        &= sum_{j=1}^m sum_{k=1}^m left(sum_{i=1}^mlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
        &= sum_{j=1}^m sum_{k=1}^m langle v_j,v_kranglelangle A^*A v_j,v_krangle\
        &= sum_{j=1}^m langle A^*A v_j,v_jrangle\
        &= sum_{j=1}^m |Av_j|_2^2
        end{align}



        Hence we can extend any orthonormal set ${v_1, ldots, v_k} subseteq mathbb{R}^m$ to an orthonormal basis ${v_1, ldots, v_m}$ for $mathbb{R}^m$ to obtain
        $$sum_{i=1}^k |Av_i|_2^2 lesum_{i=1}^m |Av_i|_2^2 = sum_{i=1}^m |Au_i|_2^2 = |A|_F^2 $$






        share|cite|improve this answer












        Note that $A^*A$ is a square matrix and $A^*A ge 0$ so there exists an orthonormal basis ${u_1, ldots, u_m}$ for $mathbb{R}^m$ such that $A^*A u_i = lambda_i u_i$ for some $lambda ge 0$.



        We have
        $$sum_{i=1}^m |Au_i|_2^2 = sum_{i=1}^m langle Au_i, Au_irangle = sum_{i=1}^m langle A^*Au_i, u_irangle = sum_{i=1}^m lambda_i =operatorname{Tr}(A^*A) = |A|_F^2$$



        The interesting part is that the sum $sum_{i=1}^m |Au_i|_2^2$ is actually independent of the choice of the orthonormal basis ${u_1, ldots, u_m}$. Indeed, if ${v_1, ldots, v_m}$ is some other orthonormal basis for $mathbb{R}^m$, we have
        begin{align}
        sum_{i=1}^m |Au_i|_2^2 &= sum_{i=1}^m langle A^*Au_i, u_irangle\
        &= sum_{i=1}^m leftlangle sum_{j=1}^mlangle u_i,v_jrangle A^*A v_j , sum_{k=1}^mlangle u_i,v_krangle v_krightrangle\
        &= sum_{j=1}^m sum_{k=1}^m left(sum_{i=1}^mlangle u_i,v_jrangle langle v_k,u_irangleright)langle A^*A v_j,v_krangle\
        &= sum_{j=1}^m sum_{k=1}^m langle v_j,v_kranglelangle A^*A v_j,v_krangle\
        &= sum_{j=1}^m langle A^*A v_j,v_jrangle\
        &= sum_{j=1}^m |Av_j|_2^2
        end{align}



        Hence we can extend any orthonormal set ${v_1, ldots, v_k} subseteq mathbb{R}^m$ to an orthonormal basis ${v_1, ldots, v_m}$ for $mathbb{R}^m$ to obtain
        $$sum_{i=1}^k |Av_i|_2^2 lesum_{i=1}^m |Av_i|_2^2 = sum_{i=1}^m |Au_i|_2^2 = |A|_F^2 $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










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