The eigenvalues of a symmetric positive definite matrix are all greater than zero. Prove this by considering the eigen-decomposition $A=QDQ^T$, with $Q$ orthogonal and $D$ diagonal. How exactly do I show this?
I'm not really sure how to approach this. I believe I should start by doing the following: $A=QDQ^T \implies Q^TA = DQ^T$ since $Q^{-1} = Q^T$. It follows then that $Q^TAQ =D$. The right side contains all the eigenvalues. If I could show that the left side becomes a diagonal with positive values, then I'd be done. However, I'm not sure how to do this.
Another inclination is to consider eigen-vectors of $A$, say $v_i$. Since $A$ is positive definite, $x^TAx>0$ for all $x\neq0$. So considering a normalized eigenvector $v_i$ and its associated eigenvalue $\lambda_i$, we have
$v_i^TQDQ^Tv_i>0$
But $Q^Tv_i = (v^Tv_i \hspace{1mm}\cdots\hspace{1mm}v^T_nv_i)^T = (0 \cdots0\hspace{1mm}1\hspace{1mm} 0\cdots0)^T$
and similarly,
$v_i^TQ = (v_i^Tv_1\hspace{2mm}\cdots\hspace{2mm}v_i^Tv_n) = (0 \cdots0\hspace{1mm}1\hspace{1mm} 0\cdots0)$
Letting $D = diag(\lambda_1,…,\lambda_i,…\lambda_n)$, we get
$(0 \cdots0\hspace{1mm}1\hspace{1mm} 0\cdots0)diag(\lambda_1,…,\lambda_i,…\lambda_n)(0 \cdots0\hspace{1mm}1\hspace{1mm} 0\cdots0)^T = \lambda_i > 0$.
Thus, this holds for the set $B=\{v_1,…,v_n\}$ of all normalized eigenvectors of $A$. So, is this a valid proof using eigendecomposition? If so, how does this change if we do not consider normalized eigenvectors?
Best Answer
An operator $A$ is positive-definite if, for every vector $v$, we have $\langle v, Av \rangle > 0$, where $\langle v, u \rangle = v^T u$ denotes the dot-product. If we have an eigenvector $Av = \lambda v$ of the positive-definite operator $A$, then $\langle v, Av \rangle = \lambda \langle v, v \rangle > 0$ and so $\lambda > 0$.
If the operator $A$ is also symmetric, then all of its eigenvalues are real, and hence all positive.