1
$\begingroup$

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:

enter image description here


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?

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.