# Positive definite operator

If a self adjoint operator over an $n$ dimensional (real/complex) inner product space V has all its eigenvalues positive then the operator is positive definite.

Definition: Operator $T$ is positive definite if inner product of $Ax$ and $x$ is always positive.

I am comfortable when $T$ is real symmetric. But don't know how to proceed for general self adjoint operator.

• That is true. But my question is the other way. That is if eigenvalues are always positive for self adjoint T , then T is positive definite.
– Hili
Apr 24, 2017 at 7:57
• So you want to prove that if the inner product of $Ax$ and $x$ is always positive then the eigenvalues of $A$ are positive. Is that right?
– yoyo
Apr 24, 2017 at 8:06
• No. The other way actually. If the eigenvalues are positive then the inner product is positive( given the fact that A is self adjoint)
– Hili
Apr 24, 2017 at 8:09

Since $A$ is self-adjoint, then $A$ is orthogonally diagonalizable. So any vector $x$ can express as the linear combination of eigenvectors $\{x_i\}_{i=1}^n$, say $x=\sum_{i=1}^n a_ix_i$. Note that we can assume that $\{x_i\}_{i=1}^n$ form a orthonormal basis and $Ax_i=\lambda_i x_i$ for some $\lambda_i>0$. Then

$(Ax,x)=(\sum_{i=1}^n a_iAx_i,\sum_{j=1}^n a_jx_j)=\sum_{i,j}a_i \overline{a_j}(Ax_i, x_j)=\sum_{i,j}a_i \overline{a_j}(\lambda_ix_i, x_j)$

by $\{x_i\}_{i=1}^n$ form a orthonormal basis $\sum_{i,j}a_i \overline{a_j}(\lambda_ix_i, x_j)=\sum_{i}|a_i|^2\lambda_i(x_i, x_i)=\sum_{i}|a_i|^2\lambda_i||x_i||^2>0$.

• I was thinking the result about orthonormal basis that you have used works when the underlying field for V is the real line. Correct me if I'm wrong.
– Hili
Apr 24, 2017 at 8:50
• en.wikipedia.org/wiki/Diagonalizable_matrix#Diagonalization "When the matrix $A$ is a Hermitian matrix (resp. symmetric matrix), eigenvectors of $A$ can be chosen to form an orthonormal basis of $\mathbb{C}^n$ (resp. $\mathbb{R}^n$)."
– yoyo
Apr 24, 2017 at 9:12
• Okay. Thanks a lot for all your help.
– Hili
Apr 24, 2017 at 9:14