If the matrix is positive definite, then its similar matrix is also positive definite?

Question

If $A$ is positive definite and $B$ is similar to $A$. Can we say that $B$ is also positive definite? I guess it is true since two matrices have same eigenvalues, and if $\sigma(A) > 0$, and so is $\sigma(B)$.

Answer 1

AS the comments suggest, the definition of positive definite requires that the matrix be Hermitian - at least in all the sources I consulted. See for example Linear Algebra by Friedberg, Insel and Spence - p.377 and Matrices and Linear Transformations by Cullen - p.251.

Then a very simple example: $\begin{bmatrix} 1&0\\0&2 \end{bmatrix}$ is positive definite, and is similar to $\begin{bmatrix} 1&1\\0&2 \end{bmatrix}$, which is clearly not Hermitian.

Answer 2

In several sources they ask for the quadratic form defined via the matrix to be positive definite, i.e. we study the function $(u,v)\to (Bu,v)$. In this case, one can say that matrix is positive definite even it's not Hermitian. Note that the matrices $B$ and $\frac{1}{2}(B+B^\ast)$ generate the same quadratic form, but their eigenvalues can differ significantly.