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Hi could you help me with following

$$ \begin{pmatrix} 1 & a & a \\ a & 1 & a \\ a & a & 1 \end{pmatrix} $$

is a $3 \times 3$ matrix.

Find the largest interval for a such that this matrix is positive definite.

How to do this??

Thanks a lot!

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What have you tried? Where did you get stuck? Do you have any thoughts on the problem? – Jonas Meyer Dec 28 '12 at 18:10

The matrix $\begin{bmatrix}-b&a&a\\a&-b&a\\a&a&-b\end{bmatrix}$ is invertible except when $b=2a$ or $b=-a$. Therefore the matrix $A=\begin{bmatrix}0&a&a\\a&0&a\\a&a&0\end{bmatrix}$ has eigenvalues $2a$ and $-a$. It follows that $I+A =\begin{bmatrix}1&a&a\\a&1&a\\a&a&1\end{bmatrix}$ has eigenvalues $2a+1$ and $1-a$. A real symmetric matrix is positive definite if and only if its eigenvalues are (strictly) positive.

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Hint: a matrix is positive definite if and only if its leading principal minors are all positive. In your case, this means your matrix is positive definite iff the three quantities $$ 1,\ \det\begin{pmatrix}1&a\\a&1\end{pmatrix},\ \det\begin{pmatrix}1&a&a\\a&1&a\\a&a&1\end{pmatrix} $$ are all positive.

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You need to take the determinant if its leading principle minors and set them greater than $0$. (The matrix is hermitian).

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