Let $T$ be a self-adjoint operator in V and $A=[T]_B$ where $B$ s an orthonormal basis of $V$.

So, I have to prove that $T$ is positive definite if and only if $L_A$ is positive definite.

I haven't tried much since I don't really known where to start. I've proven that $T$ is definite positive if and only if all its eigen values are positive, but I don't know if that will help.

  • $\begingroup$ Can you say what $T_B$ is. Also, what is $L_A$. Is T linear? You haven't said much about V -- can we assume it is not finite dimensional? It is true that any operator is positive definite iff all its eigenvalues are positive; it's one of several equivalent definitions of positive definite in a finite dimensional space. In an infinite dim space there may not be any eigenvalues; you would need T bounded. $\endgroup$
    – Betty Mock
    Oct 24, 2013 at 19:06
  • $\begingroup$ $[T]_B$ is the matrix asociated to T in the basis B. $L_A(x)=Ax$. T is linear and V is finite dimensional. $\endgroup$
    – Cath
    Oct 24, 2013 at 19:12

1 Answer 1


Suppose $A$ is a positive definite operator. Fix an orthonormal basis $\left\{ e_{j}\right\} $ in V. For all $v=\sum v_{j}e_{j}$, $$ \sum_{ij}v_{i}\overline{v_{j}}\left\langle Ae_{i},e_{j}\right\rangle =\left\langle Av,v\right\rangle \geq0. $$ Thus, the matrix of $A$, i.e., $\left(\left\langle Ae_{i},e_{j}\right\rangle \right)_{ij}$, is positive definite.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .