# Analyze the symmetric property of positive definite matrices [duplicate]

In the definition, a positive definite matrix is usually referred to symmetric expressed in quadratic form.

1. So I am confused about is it always symmetric?

2. Why do they refer to the symmetric property in its definition?

3. Please give me some examples and proof of this problem.

• A proof of what? That it is symmetric? Or that its eigenvalues are positive? – José Carlos Santos Jun 11 '18 at 13:55
• Let quadratic form $f$ be defined by $$f (\mathrm x) := \mathrm x^\top \mathrm A \,\mathrm x$$ where $\mathrm A \in \mathbb{R}^{n \times n}$. Since $\mathrm x^\top \mathrm A \,\mathrm x$ is a scalar, then $(\mathrm x^\top \mathrm A \,\mathrm x)^\top = \mathrm x^\top \mathrm A \,\mathrm x$, i.e., $\mathrm x^\top \mathrm A^\top \mathrm x = \mathrm x^\top \mathrm A \,\mathrm x$. Hence, $$\mathrm x^\top \left(\frac{\mathrm A - \mathrm A^\top}{2}\right) \mathrm x = 0$$ Thus, the skew-symmetric part of matrix $\mathrm A$ does not contribute anything to the quadratic form. – Rodrigo de Azevedo Jun 11 '18 at 14:31
• Please, define positive definiteness. AFAIK matrices are not positive definite, quadratic forms are. – Miguel Jun 11 '18 at 14:35
• @Miguel I think it's a little pedantic to make that distinction when this page exists: en.wikipedia.org/wiki/Positive-definite_matrix – N8tron Jun 11 '18 at 14:42
• @N8tron Well, that is the whole point of the OP: definite positiveness only makes sense for symmetric matrices, which can be identified with quadratic forms. – Miguel Jun 11 '18 at 15:35

Most authors define positive definite matrices as a subclass of symmetric matrices. This is not necessary, but then the usually equivalent definitions of positive definite matrices diverge.

Positive definite matrices by inner products

If you consider $\langle v, A v\rangle> 0,\forall v\in\Bbb R^n\setminus\{0\}$ as the defining property, then consider a matrix $A=S+T$, where $S$ is symmetric and positive definite (no definition problem because symmetric), and $T$ is a non-zero skew-symmetric matrix, i.e. $T^\top=-T$. Then $A$ is not symmetric, but positive definite because

$$\def\<{\langle} \def\>{\rangle}\< v, A v\>=\<v,(S+T)v\>=\underbrace{\<v,Sv\>}_{>\, 0}+\underbrace{\<v,Tv\>}_{=\,0}> 0,\qquad\text{for all v\in\Bbb R^n\setminus\{0\}}.$$

Note that $\<v,Tv\>=0$ because $\<v,Tv\>=\<T^\top v,v\>=\<-Tv,v\>=-\<v,Tv\>$.

Example. Take the positive definite identity matrix $I$ and the skew-symmetric matrix

$$S=\begin{pmatrix} \phantom+0 & 1 \\ -1 & 0 \end{pmatrix}.$$

From this we obtain the positive definite but not symmetric matrix

$$A=I+S=\begin{pmatrix} \phantom+1 & 1 \\ -1 & 1 \end{pmatrix}.$$

However, the eigenvalues are no longer real (they are $1\pm i$), hence no longer positive in the usual sense. They are however still located on the positive half-plane. Anyway, non-real eigenvalues are not a problem for the defining property, as we have $\<v,Av\>=v_1^2+v_2^2> 0$.

Positive definite matrices by eigenvalues

If you define positive definite matrix by having positive eigenvalues only, then there are also non-symmetric examples. E.g. take

$$A=\begin{pmatrix} \phantom+1 & 0 \\ -1 & 2 \end{pmatrix}$$

which is not symmetric, but has positive eigenvalues $1$ and $2$ only.

You can find more such matrices in the following ways:

• Method 1. Choose a basis $v_1,...,v_n$ of $\Bbb R^n$, but no orthogonal basis. In the above example I chose $(1,0)$ and $(1,1)$. Then choose $n$ different eigenvalues $\lambda_1,...,\lambda_n> 0$. Let $V=(v_1,...,v_n)^\top$ be the matrix with the $v_i$ as its rows, and $D=\mathrm{diag}(\lambda_1,...,\lambda_n)$. Then the matrix $$A=VDV^{-1}$$ is non-symmetric and has only positive eigenvalues $\lambda_1,...,\lambda_n$. That it is not be symmetric can be seen as follows: the $v_i$ will be the eigenvectors of your matrix $A$. But symmetric matrices will have an orthogonal basis of eigenvectors, while our matrix will not (because we have chosen them this way). So it cannot be symmetric.

• Method 2. Choose a non-diagonal upper/lower triangluar matrix (only non-zero values above/below the diagonal). If you put only positive values on the diagonal, then this matrix will have only positive eigenvalues (as these are exactly the values on the diagonal), but it is obviously not symmetric. By putting sufficiently large values off-diagonal, we can have positive eigenvalues, while not satisfying $\<v,Av\>>0$ for all $v\in\Bbb R^n\setminus\{0\}$. Choose e.g. $$A=\begin{pmatrix} \phantom{+10}1 & 0 \\ -100 & 2 \end{pmatrix}.$$ This matrix has again eigenvalues $1$ and $2$, but $\<v,Av\>=v_1^2+2v_2^2-100v_1v_2$, hence $v=(1,1)$ gives $\<v,Av\>=-97<0$.

• Can you give a counterexample? – Karim Jun 11 '18 at 14:19
• @greenworld Counterexample to what? A positive semi-definite non-symmetric matrix? A symmetric but indefinite matrix? – M. Winter Jun 11 '18 at 14:20
• It is a positive definite matrix but it is not symmetric. By the way, can you explain why do they mention about its symmetric property in the definition? Why do they not give a general definition for all case? – Karim Jun 11 '18 at 14:21
• It's kind of overdramatic to say "the eigenvalues are no longer positive, not even real"... they're as positive (i.e. in the right half-plane) as you could possibly expect a number in the complex plane to be. That last counterexample is quite a gem though. +1 – user541686 Jun 11 '18 at 17:44
• @Mehrdad Thanks for this remark. I have not looked at it from this angle. I might include this in my answer when I have time. – M. Winter Jun 11 '18 at 20:26

Recall that any matrix can be expressed as the sum of a symmetric part and a skew part as follow

$$A=\overbrace{\frac12(A+A^T)}^{symmetric}+\overbrace{\frac12(A-A^T)}^{skew}$$

and since the quadratic form for the skew part is equal to $0$ we have that the positive definiteness depends solely upon the symmetric part.

• My question is for a general case when we do not consider quadratic form. Can we say that a positive definite matrix is always symmetric? – Karim Jun 11 '18 at 14:15
• @greenworld The definition of positive definitiveness for a matrix $A$ is given by the sign of the associated quadratic form $x^TAx$ and then the conclusion is quite general. Notably, since the positive definiteness depends solely upon the symmetric part, of course we can have $A$ positive definite but not symmetric. – user Jun 11 '18 at 14:22
• Ah yes, "recall" this fact! I totally recall this fact that nobody ever mentioned :-) +1 – user541686 Jun 11 '18 at 17:33

Sometimes symmetry is required in the definition. If one removes the symmetry requirement, then $$\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$$ is positive definite, but not symmetric.

• Its eigenvalues are complex numbers, so we cannot conclude that it is positive definite. – Karim Jun 11 '18 at 14:10
• @greenworld Compute the quadratic form: $\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = x^2 + y^2 \geq 0$. The eigenvalue characterization only applies to the symmetric part of the matrix. – Bungo Jun 11 '18 at 14:33
• So the definition in wiki page has a problem: en.wikipedia.org/wiki/… When they stated: "All its eigenvalues are positive". right? – Karim Jun 11 '18 at 14:47
• @greenworld No, look at the first sentence in that section: "Let $M$ be an $n \times n$ Hermitian matrix." The property is true for a Hermitian (including symmetric) positive definite matrix. – Bungo Jun 11 '18 at 14:53