4

It's pretty basic and I'm sure I'm missing something dumb here, but I'd like to know why $||A+B||_F \leq ||A||_F+||B||_F$

The way I understand it,

$||A+B||^2_F=tr((A+B)^T(A+B))=tr((A^T+B^T)(A+B))=tr(A^TA+A^TB+B^TA+B^TB)$

Now using the property the trace is linear:

$tr(A^TA+A^TB+B^TA+B^TB) = tr(A^TA)+tr(B^TB)+tr(A^TB+B^TA)=||A||^2_F+||B||^2_F+tr(A^TB+B^TA)$

Now if we were to prove that $2||A||_F||B||_F \geq tr(A^TB+B^TA)$ that would solve the question. But I don't see how that's trivial, and generally multiplying $\sum$s together is something I avoid like the plague. Is this indeed the way? would someone help me with this last step?

Oria Gruber
  • 12,739
  • 1
    Use Cauchy-Schwarz http://math.stackexchange.com/questions/564873/frobenius-norm-triangle-inequality – parsiad May 27 '14 at 21:15
  • 1
    The Frobenius norm of a matrix is just the Euclidean norm of the vector obtained by stacking the matrix columns on top of one another. So, the reason why the triangle inequality holds for Frobenius norm is exactly the same as the reason why the triangle holds for Euclidean norm. – user1551 May 27 '14 at 21:15
  • 1
    Cauchy-Schwarz! ofcourse! – Oria Gruber May 27 '14 at 21:18

1 Answers1

2

Notice that $\text{tr}(A^TB)=\sum_{i,j}a_{ij}b_{ij}$, so by the Cauchy-Schwarz applied to $(a_{ij})$ and $(b_{ij})$ treated as vectors $|\text{tr}(A^TB)|\leq\text{tr}(A^TA)^{1/2}\text{tr}\,(B^TB)^{1/2}=\|A\|_F\|B\|_F$, and switching $|\text{tr}(B^TA)|\leq\text{tr}(B^TB)^{1/2}\text{tr}\,(A^TA)^{1/2}=\|A\|_F\|B\|_F$. Therefore, $$|\text{tr}(A^TB+B^TA)|\leq|\text{tr}(A^TB)|+|\text{tr}(B^TA)|\leq 2\|A\|_F\|B\|_F.$$

Conifold
  • 11,756