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Exercise:
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.

Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
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Exercise:
If langle rangle_A is a scalar product Longrightarrow A is positive definite. If langle rangle_A is a Hermitian product iff A is positive definite.

Solution:
Proof. We'll prove it for complex matrices and Hermitian products the case of symmetrical matrices & scalar products is very similar and even simpler. Let Ain M_ntimes nmathbbC and suppose that langle vwrangle_A : v^TAoverlinew is a Hermitian product. &Longrightarrow forall vwin mathbbC^n textwe have langle vwrangle_A langle overlinewvrangle_A &Longrightarrow v^TAoverlinewoverlinew^TAoverlinev Longrightarrow v^TAoverlinew overlinew^ToverlineAv Longrightarrow v^tAoverlinewoverlinew^ToverlineAv^T &Longrightarrow v^TAoverlinewv^ToverlineA^Toverlinew. Take ve_i we_j standard basis. Write A pmatrix a_ & hdots & a_j & hdots & a_n vdots & vdots & vdots & vdots & vdots a_n & hdots & a_nj & hdots & a_nn pmatrix. e_i^T A e_j pmatrix s & e_i & s pmatrix A pmatrix vdots e_j vdots pmatrix pmatrix s & e_i & s pmatrix pmatrix a_j vdots a_nj pmatrix Similarly e_i^ToverlineA^Te_j overlinea_ji Longrightarrow overlinea_ji forall ij Longrightarrow overlineA^TA. The fact that A is positive definite is precisely the condition that langle vvrangle_A forall vneq . The opposite statement: A positive definite Longrightarrow langle rangle_A is a Hermitian product.
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eth, fs23, linalg ii, proof, scalar product
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