# Proof that a positive definite matrix of size $d$ by $d$ has $d$ distinct eigenvalues

How to prove that a positive definite matrix of size $d$ by $d$ has $d$ distinct eigenvalues? I have the following statement I want to prove. Positive definite matrix of size $d$ by $d$ has $d$ distinct real eigenvalues and its $d$ eigenvectors are orthogonal.

I know how to prove that the eigenvalues are real (I prove this from the symmetric property only) and I know how to prove that for two eigenvectors corresponding to two different eigenvalues their inner product equal zero and thus they orthogonal (I prove this from the symmetric property only too). I asking about how to prove that a positive definite matrix $A$ of size $d \times d$ ($A$ symmetric ($A^T=A$) and $x^TAx>0$ for all $x \in R^N$) has exactly $d$ distinct eigenvalues.

EDIT:

This is a statement written in lecture notes I have without a proof. (In the lecture notes its written about the covariance matrix of multivariate Gaussian). Is there a statement about when $d$ by $d$ positive definite matrix has d distinct eigenvalues and when not? There are conditions on positive definite matrix that if they are applied then there are $d$ distinct eigenvalues?

• Are you sure you have the problem down correctly? The $d\times d$ identity matrix is positive definite and all $d$ eigenvalues are $1$. – Casteels Feb 14 '15 at 21:45
• @Casteels: my thoughts exactly! – Robert Lewis Feb 14 '15 at 21:46
• This is a statement written in lecture notes I have without a proof. (In the lecture notes its written about the covariance matrix of multivariate Gaussian). Is there a statement about when $d$ by $d$ positive definite matrix has $d$ distinct eigenvalues and when not? There are conditions on positive definite matrix that if they are applied then there are $d$ distinct eigenvalues? – guest Feb 14 '15 at 21:50
• A correct statement is that a $d\times d$ symmetric matrix has a set of $d$ mutually orthogonal eigenvectors (whether the matrix is positive definite or not). But it takes a fair bit of work to prove this. – Gerry Myerson Feb 14 '15 at 22:08

The $$n\times n$$ -identity matrix is symmetric and positive definite, since it is the representative matrix of canonical inner product in $$\mathbb{R}^{n}$$ :$$v^{T}I_{n}v=\sum_{i=1}^{n}v_{i}^{2}=\left\langle v,v\right\rangle$$ for all $$v=\left(v_{1},\ldots,v_{n}\right)\in\mathbb{R}^{n}$$. But only $$1$$ is an eigenvalue for $$I_{n}$$ : indeed, its characteristic polynomial is $$\left(X-1\right)^{n}$$ and $$\mathbb{R}^{n}$$ is the eigenspace associated to $$1$$.