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I know a symmetric matrix is positive definite if and only if every eigenvalue is positive. However, is a matrix that is symmetric and has all positive eigenvalues always positive definite? More specifically, I have the following matrix:

$$\begin{bmatrix}3& -1 \\-1 & 3 \end{bmatrix}$$

Its eigenvalues are $4$ and $2$, and it is symmetric. Is it positive definite? Thanks.

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    $\begingroup$ Yes, it sure is. $\endgroup$ Mar 20, 2014 at 2:05
  • $\begingroup$ Use that the underlying space has a basis consisting of eigenvectors of the matrix. $\endgroup$ Mar 20, 2014 at 2:09
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    $\begingroup$ That's what "if and only if" means. That is, "a symmetric matrix is positive definite if and only if every eigenvalue is positive" means that "a symmetric matrix is positive definite if every eigenvalue is positive" and "a symmetric matrix has positive eigenvalues if it is positive definite". $\endgroup$ Mar 20, 2014 at 2:51

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Yes.

This follows from the if and only if relation.

Let $A$ is a symmetric matrix. We have:

$A$ is positive definite $\iff$ every eigenvalue of $A$ is positive

It is a two-sided implication.

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Yes in the real case.

Let $ A $ be a matrix that is symmetric and has all positive eigenvalues. Then, $\forall x\in V, x^TAx > 0 $. This is true for all eigenvectors $ x_i $ of $ A $. Hence $0<x^T_iAx_i=x_i\alpha_i x_i=||x_i||^2\alpha_i $, so $\alpha_i> 0 $.

Suppose $\alpha_i> 0\forall i $. Then, on eigenvectors, $0<x^T_iAx_i$. $ A $ is real and symmetric. Hence, in $ V $ there exists an orthonormal basis of $ A $, by the spectral theorem. In this basis, $\forall y\in V, y=\sum a_ix_i $. The above relation thus gives $ y^TAy=\sum a_i^2\alpha_i> 0 $. Q.E.D.

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