Does $\forall x\ne 0: x^TAx>0$ means all eigenvalues of $A$ are real?

Question

Let $A\in\mathbb{R}^{n\times n}$

Does $\forall x\ne 0,x\in \mathbb{R}^n: x^TAx>0$ means $A$ has only real eigenvalues (roots of the characteristic polynomial are all real)?

Answer 1

No. Take $A:\mathbb{R}^2\to\mathbb{R}^2$ to be a rotation by a small angle.

Answer 2

Consider the matrix $A = \begin{bmatrix}a & -b\\ b & a\end{bmatrix}$ with $a > 0$ and $b \neq 0$.

Then, $x^TAx = \begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}a & -b\\ b & a\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix} = a(x_1^2+x_2^2) > 0$ for all $x \neq 0$.

But the eigenvalues of $A$ are $a \pm bi$, which are not real.

Answer 3

No. Consider a $2 \times 2$ real matrix $A$ such that

$A = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \tag{1}$

with $a > 0$, $b \ne 0$. If $x = (y, z)^T$ then

$x^TAx = (y, z)\begin{bmatrix} a & b \\ -b & a \end{bmatrix}(y, z)^T = (y,z)(ay + bz, -by + az)^T = a(y^2 + z^2) > 0 \tag{2}$

if $x \ne 0$. But the characteristic polynomial $p_A(\lambda)$ of $A$ is evidently

$p_A(\lambda) = \lambda^2 - 2a\lambda + (a^2 + b^2); \tag{3}$

the roots of (3) are $a \pm bi$, not real since $b \ne 0$.

It is also possible to construct examples of greater size than $2$ by placing $2 \times 2$ matrices of the same general form as $A$ as diagonal blocks in a larger block diagonal matrix.

Note added Monday 1 September 2014 2:13 PM PST: Here is another class of matrices which, though positive definite, has complex eigenvalues; it represents a significant generalization of the above examples. Let $K$ be a non-zero, real, skew-symmetric matrix of size $n$; $K^T = -K$. It is well-known that the non-zero eigenvalues of such a $K$ are purely imaginary. Choosing real $\lambda > 0$, set $A = \lambda I + K$. Then for any nonzero $x \in \Bbb R^n$,

$x^TAx = x^T (\lambda I + K)x = x^T(\lambda I)x + x^TKx = \lambda x^Tx > 0, \tag{4}$

since $x^TKx = 0$ for skew-symmetric $K$ ($(x^TKx)^T = x^TK^Tx = -x^TKx$; but $x^TKx$ is real scalar so we have $(x^TKx)^T = x^TKx$ and thus $x^TKx = - x^TKx \Rightarrow x^TKx = 0$). Thus $A$ is positive definite. However, the eigenvalues of $A$ are of the form $\lambda + i\omega$, where $i\omega$ is an eigenvalue of $K$: if $Kz = i\omega z$, then $Az = (\lambda + i\omega) z$ and if $Az = (\lambda I + K)z = (\lambda + i\omega)z$ then $Kz = i\omega z$ etc. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 4

It is not true in general,as counterexample given by @robert lewis. But if A is a real symmetric matrix then not only eigen values will be real but positive also.