Let Λ {\displaystyle \Lambda } be a symmetric, positive definite, non-singular square matrix. Then we have the following:
⟨ x − Λ − 1 y , Λ ( x − Λ − 1 y ) ⟩ = ⟨ x , Λ x ⟩ − ⟨ x , y ⟩ − ⟨ Λ − 1 y , Λ x ⟩ + ⟨ Λ − 1 y , y ⟩ {\displaystyle \langle x-\Lambda ^{-1}y,\Lambda (x-\Lambda ^{-1}y)\rangle =\langle x,\Lambda x\rangle -\langle x,y\rangle -\langle \Lambda ^{-1}y,\Lambda x\rangle +\langle \Lambda ^{-1}y,y\rangle } .
We have ⟨ Λ − 1 y , Λ x ⟩ = ⟨ x , y ⟩ {\displaystyle \langle \Lambda ^{-1}y,\Lambda x\rangle =\langle x,y\rangle } and ⟨ Λ − 1 y , y ⟩ = ⟨ y , Λ − 1 y ⟩ {\displaystyle \langle \Lambda ^{-1}y,y\rangle =\langle y,\Lambda ^{-1}y\rangle } since Λ {\displaystyle \Lambda } is symmetric.
From the above, we see that − 1 2 ⟨ x − Λ − 1 y , Λ ( x − Λ − 1 y ) ⟩ + 1 2 ⟨ y , Λ − 1 y ⟩ = − 1 2 ⟨ x , Λ x ⟩ + ⟨ x , y ⟩ {\displaystyle -{\frac {1}{2}}\langle x-\Lambda ^{-1}y,\Lambda (x-\Lambda ^{-1}y)\rangle +{\frac {1}{2}}\langle y,\Lambda ^{-1}y\rangle =-{\frac {1}{2}}\langle x,\Lambda x\rangle +\langle x,y\rangle }
be a symmetric, positive definite, non-singular square matrix. Then we have the following:
.
and 
since is symmetric.
