Real Symmetric Positive Definite Matrix has Positive Eigenvalues

Theorem

Let $A$ be a symmetric positive definite matrix over $\mathbb R$.

Let $\lambda$ be an eigenvalue of $A$.

Then $\lambda$ is real with $\lambda > 0$.

Proof

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf v$ be a corresponding eigenvector.

From Real Symmetric Matrix has Real Eigenvalues, $\lambda$ is real.

From the definition of a positive definite matrix, we have:

$\mathbf v^\intercal A \mathbf v > 0$

That is:

\(\ds 0\)

\(<\)

\(\ds \mathbf v^\intercal A \mathbf v\)

\(\ds \)

\(=\)

\(\ds \mathbf v^\intercal \paren {\lambda \mathbf v}\)

Definition of Eigenvector of Real Square Matrix

\(\ds \)

\(=\)

\(\ds \lambda \paren {\mathbf v^\intercal \mathbf v}\)

\(\ds \)

\(=\)

\(\ds \lambda \paren {\mathbf v \cdot \mathbf v}\)

Definition of Dot Product

\(\ds \)

\(=\)

\(\ds \lambda \norm {\mathbf v}^2\)

Dot Product of Vector with Itself

From Euclidean Space is Normed Vector Space, we have:

$\norm {\mathbf v}^2 > 0$

so:

$\lambda > 0$

$\blacksquare$