Prove another matrix is positive definite given that A is a Hermitian matrix Suppose that $A$ is a Hermitian symmetric $n\times n$ matrix of complex numbers all of whose eigenvalues lie inside the interval $(-1,1).$ Prove that the matrix $A^3+Id$ is positive definite.

An Hermitian matrix is defined as $A^*=A.$ I guess the finite-dimensional spectral theorem should be applied here but how?
 A: Hint: $A$ is diagonalizable, meaning $A=PDP^{-1}$, $D$ diagonal, for some $P$. Now try to diagonalize $A^3+I$
A: I assume that by the phrase "Hermitian symmetric" our OP lovelesswang means what is usually called simply Hermitian, and by the notation $A^\ast$ is meant what is usually denoted by $A^\dagger$, that is, the Hermitian adjoint of $A$, or, in matrix, terms, its conjugate transpose, $A^\dagger = \bar A^T$, where $(\bar A)_{ij} = \bar A_{ij}$.  This interpretation is consistent with the OP's assertions that $A$ is a complex $n \times n$ matrix and that $A^\ast = A$; if lovelesswang simply meant the complex conjugate (what I call $\bar A$) by $A^\ast$, then $A^\ast = A$ would imply that $A$ is real, whereas lovelesswang says $A$ may be complex.  So . . . I assume we are talking about what I am used to calling Hermtitian adjoints here, and I shall use $A^\dagger$, not $A^\ast$, to denote the Hermitian adjoint of $A$ throughout this answer.
These things being said:
The hypothesis that $A$ is Hermitian, $A = A^\dagger$, implies that there exists a unitary matrix $U$, $UU^\dagger = U^\dagger U = I$, which diagonalizes $A$:
$U^\dagger AU = \Lambda = [\lambda_{ij}], \tag{1}$
where $\Lambda$ is a diagonal matrix such that the $\lambda_i = \lambda_{ii}$ are the $n$ eigenvalues of $A$ and $\lambda_{ij} = 0$ when $i \ne j$.  If we left multiply (1) by $U$ we find
$AU = UU^\dagger AU = U\Lambda, \tag{2}$
and careful scrutiny of (2) reveals that the columns of $U$ are in fact an orthonormal set of eigenvectors of $A$; indeed, writing $U$ in columnar form
$U = [\mathbf u_1, \mathbf u_2, \ldots, \mathbf u_n], \tag{3}$
we have
$AU = [A\mathbf u_1, A\mathbf u_2, \ldots, A\mathbf u_n] \tag{3}$
and
$U\Lambda = [\lambda_1 \mathbf u_1, \lambda_2 \mathbf u_2, \ldots, \lambda_n \mathbf u_n]; \tag{4}$
comparison of (3) and (4) shows that
$A\mathbf u_i = \lambda_i \mathbf u_i \tag{5}$
for $1 \le i \le n$; that the $\mathbf u_i$ are orthonormal, $\langle \mathbf u_i, \mathbf u_j \rangle = \delta_{ij}$, follows from $U^\dagger U = I$.  Since $A$ is Hermitian, $\lambda_i \in \Bbb R$ for all $i$ and furthermore we are given that all $\lambda_i \in (-1, 1)$, that is
$-1 < \lambda_i < 1. \tag{6}$
(6) clearly implies that
$-1 < \lambda_i^3 < 1 \tag{7}$
for all $i$, and hence that
$0 < \lambda_i^3 + 1 < 2 \tag{8}$
for all $i$ as well.  From (5) we have
$A^3 \mathbf u_i = A^2(A \mathbf u_i) = A^2 (\lambda_i \mathbf u_i) = \lambda_i A (A \mathbf u_i) = \lambda_i^2 A \mathbf u_i = \lambda_i^3 \mathbf u_i, \tag{9}$
and thus
$(A^3 + I) \mathbf u_i = A^3 \mathbf u_i + \mathbf u_i = \lambda_i^3 \mathbf u_i + \mathbf u_i = (\lambda_i^3 + 1) \mathbf u_i. \tag{10}$
Since the $\mathbf u_i$ are orthonormal, they form a linearly independent set of vectors; since there are $n$ $\mathbf u_i$, the set of $\mathbf u_i$ forms a basis.  Thus any vector $\mathbf x$ may be written
$\mathbf x = \sum_1^n x_i \mathbf u_i \tag{11}$
where the $x_i \in \Bbb C$, $1 \le i \le n$.  Using (10) and (11), we evaluate $\langle \mathbf x, (A^3 + I)\mathbf x \rangle$ for $\mathbf x \ne 0$:
$\langle \mathbf x, (A^3 + I)\mathbf x \rangle = \langle \sum_1^n x_i \mathbf u_i, (A^3 + I)(\sum_1^n x_j \mathbf u_j) \rangle = \langle \sum_1^n x_i \mathbf u_i, \sum_1^n x_j (A^3 + I)\mathbf u_j \rangle$
$= \langle \sum_1^n x_i \mathbf u_i, \sum_1^n x_j (\lambda_j^3 + 1)\mathbf u_j \rangle = \sum_{i,j = 1}^n \bar x_i x_j (\lambda_j^3 + 1) \langle \mathbf u_i, \mathbf u_j \rangle$ $= \sum_{i,j = 1}^n \bar x_i x_j (\lambda_j^3 + 1) \delta_{ij} = \sum_1^n (\lambda_i^3 + 1)\bar x_i x_i > 0 \tag{12}$
by virtue of (8) and the fact that $\mathbf x \ne 0$ implies $x_i \ne 0$ for at least one value of the index $i$.  (12) says that $\langle \mathbf x, (A^3 + I)\mathbf x \rangle > 0$ for nonzero $\mathbf x$; that is, that $A$ is positive definite.  QED.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
