Conditions for a certain parametric matrix to be positive definite

Question

The task:

Let's consider matrix $A=\begin{bmatrix}1&a&a\\a&1&a\\a&a&1 \end{bmatrix}$. Show that $A>0$ if and only if $-1<2a<2$.

Solution attempt: By definition $A$ is positive definite iff $\forall_{z\in\mathbb{R}^3}z^{\operatorname{T}}Az>0$.

For $z=\begin{bmatrix}z_1\\z_2\\z_3\end{bmatrix}$, $$\left(z^{\operatorname{T}}A\right)z=\left(z_1^2+az_1z_2+az_1z_3\right)+\left(az_1z_2+z_2^2+az_2z_3\right)+\left(az_1z_3+az_2z_3+z_3^2\right)\\=z_1^2+z_2^2+z_3^2+2a\left(z_1z_2+z_1z_3+z_2z_3\right)>0\\\iff\\z_1^2+z_2^2+z_3^2>-2a\left(z_1z_2+z_1z_3+z_2z_3\right)$$

This is trivially true for $z_1z_2+z_1z_3+z_2z_3\geq0$, so without loss of generality let's assume that $z_1z_2+z_1z_3+z_2z_3<0$, and thus $-(z_1z_2+z_1z_3+z_2z_3)>0$. Therefore we need to prove that:

$$\left({\Large\forall_{z_1,z_2,z_3}}-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a\right)\iff -1<2a<2$$

And this seems false to me. This is because $-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}$ is always $\geq0$. For this reason we have a trivial counterexample, eg when $2a=-1000$ then $2a\not\in(-1;2)$ but nonetheless $\forall_{z_1,z_2,z_3}-\frac{z_1^2+z_2^2+z_3^2}{z_1z_2+z_1z_3+z_2z_3}>2a$.

I can't find my error; could you kindly help me please?

Answer 1

Let's determine the eigenvalues of $A$. If we can find the conditions for which those eigenvalues are all positive, then we can solve the question.

Let $u \in \mathbb R^3$ be the vector with $1$ for all its components. Then note that $uu^T$ is the $3\times 3$ matrix with $1$ for all its entries. Therefore $$A=(1-a)I+auu^T$$ Now, notice that $$Au=(1-a)u+au (u^Tu)=(1-a)u+3au=(1+2a)u$$ So $u$ is an eigenvector of $A$ with eigenvalue $(1+2a)$.

Now, let's determine the other eigenvalues of $A$. Suppose $x\in\mathbb R^3$ is orthogonal to $u$, that is $u^Tx=0$, then $$Ax=(1-a)x+auu^Tx=(1-a)x$$ so $x$ is an eigenvector of $A$ with eigenvalue $(1-a)$. Since the space of vectors $x$ orthogonal to $u$ is of dimension 2, we have now determined that $(1-a)$ is an eigenvalue of multiplicity 2 for $A$.

In summary, the spectrum of $A$ is $(1+2a)$ (multiplicity 1) and $(1-a)$ (multiplicity 2). Consequently, $A$ is positive definite iff $$1+2a>0 \text{ and } 1-a >0$$ that is iff $-1 < 2a < 2$$

Answer 2

$z_1^2+z_2^2+z_3^2+2a\left(z_1z_2+z_1z_3+z_2z_3\right)>0\iff z_1^2+z_2^2+z_3^2>-2a\left(z_1z_2+z_1z_3+z_2z_3\right)$

This is trivially true for $z_1z_2+z_1z_3+z_2z_3\geq0$

Not, it is not, since $a$ might be negative.

I suppose showing $A$ is definite positive could be done your way, but it's way too exhausting for me. I'd go with using the fact that $$A\text{ definite positive } \iff \text{ its eigenvalues are } > 0$$

Now choose your favorite way to find the eigenvalues of $A$. Note that the sum of the rows are $2a+1$ so this is one of them ; which by the way leads to $-1<2a$. Now your turn...