# proof that a matrix $M$ is positive definite iff all its eigenvalues are strictly positive.

By definition, an $n\times n$ matrix $M$ is positive definite iff $$\forall z\in\mathbb R^n:z^TAz>0$$

We want to show that this is satisfied iff for all eigenvalues $\lambda _i$ of $M$, we have $$\lambda_i >0$$

We can show this by decomposing $M$ into a diagonal matrix: $$M=P^{-1}DP$$ where $P$ is the matrix built up by the eigenvectors of $M$, and where $D$ is the diagonal matrix with the eigenvalues on the main diagonal.

Then if we dfine $y=Pz$, we can rewrite the criterion as:

$$z^TP^{-1}Dy>0$$

However, in the text I found on wikipedia, they go one step further, and rewrite it as $$\forall y\in \mathbb R^n: y^TDy>0$$ From that condition it is easy to see that all eigenvalues must be strictly positive, but I don't know how this conclusion is reached since $y^T=z^TP^T\neq z^TP^{-1}$

ps. here is the source:

Edit: I now understand that $z^TP^T=z^TP^{-1}$ is based on the assumption that $M$ is symmetric and therefore $P$ is orthogonal. However, I am wondering, can we still say that a non-symmetric $M$ is positive definite iff all its eigenvalues are strictly positive?

• You can choose the eigenvectors to form an orthonormal basis. Then your matrix $P$ is unitary (or orthogonal), meaning that $P^{-1} = P^T$. – amsmath Oct 10 '17 at 17:29
• You should assume that $M$ is symmetric – egreg Oct 10 '17 at 17:30
• @amsmath If $M$ is not symmetric, it's not necessarily diagonalizable. – egreg Oct 10 '17 at 17:34
• @Programmer2134 The matrix $\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$ has rank two and is not diagonalizable. It's quite a deep result that a symmetric matrix can be diagonalized with an orthogonal matrix. – egreg Oct 10 '17 at 17:53
• The source does start with "Let $M$ be an $n \times n$ Hermitian matrix." – Daniel Schepler Oct 10 '17 at 19:16

Your source talks about matrices over the complex numbers and $M$ is supposed to be Hermitian, that is, equal to its “conjugate transpose”.

I'll denote by $M^*$ the conjugate transpose of the (generic) matrix $M$. A couple of definitions; let $M$ be a square matrix:

• $M$ is Hermitian when $M=M^*$

• $M$ is normal when $MM^*=M^*M$

The source also talks about the spectral theorem, which says that every normal matrix $M$ can be written as $$M=\sum_{k=1}^r \lambda_k P_k\tag{*}$$ where $\lambda_1,\dots,\lambda_r$ are the distinct eigenvalues of $M$ and $P_1,\dots,P_r$ are the projections matrices to the eigenspaces; in particular they satisfy $P_k^2=P_k$ and $P_k^*=P_k$. Moreover $P_kP_l$ is the null matrix when $k\ne l$.

Clearly any Hermitian matrix is normal and it's easy to deduce that if $M$ is Hermitian then its eigenvalues are real: indeed, using its spectral decomposition (*), we have, from $M=M^*$, $$\sum_{k=1}^r \lambda_kP_k=\sum_{k=1}^r \lambda_k^*P_k$$ and upon multiplying this by $P_l$ we get $\lambda_l=\lambda_l^*$.

The statement that any Hermitian matrix has real eigenvalues can be proved also without the spectral theorem, by directly using the definition: let $v$ be an eigenvector relative to the eigenvalue $\lambda$; then $$\lambda(v^*v)=v^*(\lambda v)=v^*Av=v^*A^*v=(Av)^*v=(\lambda v)^*v=\lambda^*v^*v$$ Since $v^*v\ne0$, we get $\lambda=\lambda^*$.

Another important fact about normal matrices is that they can be diagonalized with a unitary matrix; this is actually an equivalent condition.

A square matrix $M$ is normal if and only if there exists a unitary matrix $U$ (that is, $U^{-1}=U^*$) such that $D=U^{-1}MU$ is diagonal.

Now, suppose $M$ is (Hermitian) and positive definite, that is, for $v\ne0$, $v^*Mv>0$. Consider $e_k$, the $k$-th vector of the canonical basis and consider $v_k=Ue_k$; then $$0<v_k^*Mv_k=e_k^*U^*UDU^*Ue_k=e_k^*De_k$$ and, clearly, $e_k^*De_k$ is the entry at place $(k,k)$ in $D$. Since $D$ has on its diagonal the eigenvalues of $M$, we are done.

Conversely, if $D$ has all its diagonal entries positive, then, for every $v\ne0$ we have $$v^*Mv=(Uv)^*D(Uv)>0$$ because clearly $D$ is positive definite.