I have to show that given $G = A^{T}A$ and $E = B^{T}B$. the matrix $G$ or $E$ is semi-definite positive and also $G + E$. Do you think this is correct?
Because the dot product is a definite positive bilinear form, we can show that: $$\langle v,Gv\rangle = \langle v,A^{T}Av\rangle = \langle Av,Av\rangle \geq 0$$ And for $E + G$ we have using the formula above: $$\langle (G+E)v,v\rangle = \langle Gv+Ev,v\rangle = \langle Gv,v\rangle + \langle Ev,v\rangle \geq 0$$


1 Answer 1


It's correct . To show that E+G is positive semidefinite you can also use the easy to show fact that the sum of two positive semidefinite matrices is also positive semidefinite .

When dealing with finite dimensional spaces , using the matrix notation $<u,v> == u^{T}v$ can make things a bit more clear.

  • $\begingroup$ Using the properly-typeset matrix notation $\langle u, v\rangle = u^T v$ will make things even more clear than that. $\endgroup$ Oct 8, 2018 at 19:51
  • $\begingroup$ Well maybe is geographical question, but here, in switzerland, we use to use "brackets" notation and in mechanics the dot. Even if we understand the other ones, is easier for me to read it as I did. $\endgroup$
    – Moleson
    Oct 8, 2018 at 20:31
  • $\begingroup$ Misha Lavrov - if you bake me some cookies , I will write as you please :)). $\endgroup$ Oct 9, 2018 at 6:22

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