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Prove that if $A$ is a symmetric matrix with all eigenvalues greater than $0$, then it is positive definite.

If $A$ is symmetric then there exists an orthogonal matrix $S$, such that $S^TAS$ is a diagonal matrix. I'm not sure about this, but if a matrix is diagonalizable, then the eigenvalues of that matrix are equal to the corresponding entries in the diagonal of its diagonalized form, correct? So if $S^TAS$ is diagonal, then the entries in the diagonal equal to the eigenvalues?

Either way, even if they were and assuming they were positive, I could only show that $x^TAx>0$ if $x$ is an eigenvector of $A$, but how would one prove this for all $x \neq 0$?

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    $\begingroup$ You can express vectors as linear combinations of eigenvectors. $\endgroup$ – Gerry Myerson Jun 23 '14 at 7:33
  • $\begingroup$ In the last statement, do you mean for all $x\ne 0$. $\endgroup$ – L.K. Jun 23 '14 at 7:40
  • $\begingroup$ Yes I meant to write $\forall x \neq 0$. $\endgroup$ – eager2learn Jun 23 '14 at 7:42
  • $\begingroup$ Take a non null vector $x$ and note that $x^TAx=x^TU^TDUx$, where $U=S^T$. Now let $y=Ux$ and rewrite $x^TU^TDUx$ as $y^TDy$. Now think about what $y^TDy$ looks like. $\endgroup$ – Git Gud Jun 23 '14 at 7:42
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Since a symmetric matrix is always diagonalizable, its eigenvectors form a basis. Moreover there there exists an orthonormal basis of eigen vectors corresponding to the eigen values. Let $\{e_1,e_2,..e_n\}$ be the ortonormal basis of eigen vectors corresponding to the eigenvalues $\{\lambda_1,\lambda_2,..\lambda_n\}$. For $x \ne 0$, $x =c_1e_1+c_2e_2+..c_ne_n$. So $x^{T}Ax=\{c_1e_1^{T}+...c_ne_n^T\}\{c_1Ae_1+..c_nAe_n\}=\{c_1e_1^{T}+...c_ne_n^T\}\{c_1\lambda_1e_1+..c_n\lambda_ne_n\}=c_1^2\lambda_1+...c_n^2\lambda_n \ge 0$

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From a different view, as we already know that A is symmetric and has only positive eigenvalues, then we know it contains only positive pivots. If we have vector $x = [x_1 x_2 ... x_n]^T$, then $x^T Ax$ can be expressed by a polynomial function $\sum_{i=1}^n a_{ii}{x_i}^2 + \sum_{i=1}^n \sum_{j=i+1}^n 2a_{ij}x_ix_j$, where $a_{ij}$ denotes the element in $i-th$ row and $j-th$ column of matrix A. Then we express polynomial function in a form as $pivot_k \times (ax_i + bx_j)^2$, then it is clear that the function is always positive (except $x$ is zero vector), this means $x^TAx$ is always positive, and $A$ is positive definite.

For example, $A = \begin{bmatrix} a_{11} & a_{12}\\[0.3em] a_{21} & a_{22} \end{bmatrix}$ and x = $\begin{bmatrix} x_{1} \\[0.3em] x_{2} \end{bmatrix}$, then we can express $x^TAx=f(x_1, x_2)=a_{11}x_1^2 + 2a_{12}x_1x_2 + a_{22}x_2^2= a_{11}(x_1+\sqrt{\frac{a_{12}}{a_{11}}x_2})^2+(a_{22}-\frac{a_{12}^2}{a_{11}})x_2^2$, where $a_{11}, a_{22}-\frac{a_{12}^2}{a_{11}}$ are the pivots of matrix $A$. Function $f(x_1, x_2)$ has minimum value 0 when $x$ is zero vector.

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