Is the spectral norm submultiplicative?

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I wonder if the $2$-norm or spectral norm is also submultiplicative for non-square matrices, i.e.,



$$| A B |_2 leq | A |_2 cdot | B |_2$$



if the number of columns of $A$ coincides with the number of rows of $B$. In the literature I can only find a statement about square matrices. Thanks a lot for any remarks.




matrices norm matrix-norms spectral-norm




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    up vote
    -1
    down vote

    favorite












    I wonder if the $2$-norm or spectral norm is also submultiplicative for non-square matrices, i.e.,



    $$| A B |_2 leq | A |_2 cdot | B |_2$$



    if the number of columns of $A$ coincides with the number of rows of $B$. In the literature I can only find a statement about square matrices. Thanks a lot for any remarks.




    matrices norm matrix-norms spectral-norm




    share|cite|improve this question
























      up vote
      -1
      down vote

      favorite









      up vote
      -1
      down vote

      favorite











      I wonder if the $2$-norm or spectral norm is also submultiplicative for non-square matrices, i.e.,



      $$| A B |_2 leq | A |_2 cdot | B |_2$$



      if the number of columns of $A$ coincides with the number of rows of $B$. In the literature I can only find a statement about square matrices. Thanks a lot for any remarks.




      matrices norm matrix-norms spectral-norm




      share|cite|improve this question














      I wonder if the $2$-norm or spectral norm is also submultiplicative for non-square matrices, i.e.,



      $$| A B |_2 leq | A |_2 cdot | B |_2$$



      if the number of columns of $A$ coincides with the number of rows of $B$. In the literature I can only find a statement about square matrices. Thanks a lot for any remarks.





      matrices norm matrix-norms spectral-norm





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      edited Aug 21 at 11:03









      Rodrigo de Azevedo

      12.6k41751




      12.6k41751










      asked Jan 10 at 13:03









      Tim vor der Brück

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      81




















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          If $A$ is not square I don't think it's proper to talk about the spectral norm of $A$. I take it that if $A$ is $mtimes n$ then $||A||_2$ is the norm of $A$ as a map from $mathbb R^m$ to $mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||Stx||le||S||,||Tx||le||S||,||T||,||x||,$$so $||ST||le||S||,||T||$.






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          • Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
            – Tim vor der Brück
            Jan 12 at 10:20











          • What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
            – David C. Ullrich
            Jan 12 at 13:17










          • $rho$ is the largest absolute eigenvalue
            – Tim vor der Brück
            Jan 14 at 14:45











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          If $A$ is not square I don't think it's proper to talk about the spectral norm of $A$. I take it that if $A$ is $mtimes n$ then $||A||_2$ is the norm of $A$ as a map from $mathbb R^m$ to $mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||Stx||le||S||,||Tx||le||S||,||T||,||x||,$$so $||ST||le||S||,||T||$.






          share|cite|improve this answer




















          • Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
            – Tim vor der Brück
            Jan 12 at 10:20











          • What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
            – David C. Ullrich
            Jan 12 at 13:17










          • $rho$ is the largest absolute eigenvalue
            – Tim vor der Brück
            Jan 14 at 14:45















          up vote
          1
          down vote



          accepted










          If $A$ is not square I don't think it's proper to talk about the spectral norm of $A$. I take it that if $A$ is $mtimes n$ then $||A||_2$ is the norm of $A$ as a map from $mathbb R^m$ to $mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||Stx||le||S||,||Tx||le||S||,||T||,||x||,$$so $||ST||le||S||,||T||$.






          share|cite|improve this answer




















          • Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
            – Tim vor der Brück
            Jan 12 at 10:20











          • What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
            – David C. Ullrich
            Jan 12 at 13:17










          • $rho$ is the largest absolute eigenvalue
            – Tim vor der Brück
            Jan 14 at 14:45













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          If $A$ is not square I don't think it's proper to talk about the spectral norm of $A$. I take it that if $A$ is $mtimes n$ then $||A||_2$ is the norm of $A$ as a map from $mathbb R^m$ to $mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||Stx||le||S||,||Tx||le||S||,||T||,||x||,$$so $||ST||le||S||,||T||$.






          share|cite|improve this answer












          If $A$ is not square I don't think it's proper to talk about the spectral norm of $A$. I take it that if $A$ is $mtimes n$ then $||A||_2$ is the norm of $A$ as a map from $mathbb R^m$ to $mathbb R^n$, where both spaces have the Euclidean norm? If so then this is obvious; it's trivial that operator norms are submultiplicative: $$||Stx||le||S||,||Tx||le||S||,||T||,||x||,$$so $||ST||le||S||,||T||$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 10 at 16:06









          David C. Ullrich

          55.1k33787




          55.1k33787











          • Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
            – Tim vor der Brück
            Jan 12 at 10:20











          • What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
            – David C. Ullrich
            Jan 12 at 13:17










          • $rho$ is the largest absolute eigenvalue
            – Tim vor der Brück
            Jan 14 at 14:45

















          • Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
            – Tim vor der Brück
            Jan 12 at 10:20











          • What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
            – David C. Ullrich
            Jan 12 at 13:17










          • $rho$ is the largest absolute eigenvalue
            – Tim vor der Brück
            Jan 14 at 14:45
















          Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
          – Tim vor der Brück
          Jan 12 at 10:20





          Thanks a lot, The usual definition: $lVert A rVert_2=sqrtrho(A^topA)$ applies to non-square matrices too in my opinion, doesnt it?
          – Tim vor der Brück
          Jan 12 at 10:20













          What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
          – David C. Ullrich
          Jan 12 at 13:17




          What is $rho$? Whether that formula makes sense for non-square matrices or not I don't think it's right to call it the spectral norm - that would be the modulus of the largest eigenvalue, and a non-square matrix doesn't have eigenvalues to begin with.
          – David C. Ullrich
          Jan 12 at 13:17












          $rho$ is the largest absolute eigenvalue
          – Tim vor der Brück
          Jan 14 at 14:45





          $rho$ is the largest absolute eigenvalue
          – Tim vor der Brück
          Jan 14 at 14:45













           

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