To determine when the matrix $S = \begin{bmatrix} c & 1 & 1 \\ 1 & c & 1 \\ 1 & 1 & c \end{bmatrix}$ is positive definite using the trace, determinant, and eigenvalues, we follow these steps:

Step 1: Positive Definite Condition via Eigenvalues

A matrix is positive definite if and only if all its eigenvalues are positive. To find the eigenvalues, we'll start by solving the characteristic equation:

$\text{det}(S - \lambda I) = 0$

Here, $\lambda$ represents the eigenvalues, and $I$ is the identity matrix. So, the characteristic matrix is:

$S - \lambda I = \begin{bmatrix} c-\lambda & 1 & 1 \\ 1 & c-\lambda & 1 \\ 1 & 1 & c-\lambda \end{bmatrix}$

The determinant of this matrix is:

$\text{det}(S - \lambda I) = \text{det}\left(\begin{bmatrix} c-\lambda & 1 & 1 \\ 1 & c-\lambda & 1 \\ 1 & 1 & c-\lambda \end{bmatrix}\right)$

Expanding the determinant:

[ \text{det}(S - \lambda I) = (c - \lambda)\left[(c - \lambda)^2 - 1\right] - 1\left[(c - \lambda) - 1\right] + 1\left[(c - \lambda) - 1\right] ] Simplifying:

$\text{det}(S - \lambda I) = (c - \lambda)\left[(c - \lambda)^2 - 1\right] - 1 + 1$ $= (c - \lambda)\left[(c - \lambda)^2 - 2\right]$

So, the characteristic equation is:

$(c - \lambda)\left[(c - \lambda)^2 - 2\right] = 0$

This gives us the eigenvalues:

$\lambda_1 = c$ $\lambda_2, \lambda_3 = c \pm \sqrt{2}$

Step 2: Conditions for Positive Definite Matrix

For the matrix $S$ to be positive definite, all eigenvalues must be positive:

1. $\lambda_1 = c > 0$
2. $\lambda_2 = c + \sqrt{2} > 0$
3. $\lambda_3 = c - \sqrt{2} > 0$

Step 3: Analyze the Conditions

• From $\lambda_1 = c > 0$, we have $c > 0$.
• From $\lambda_2 = c + \sqrt{2} > 0$, we have $c > -\sqrt{2}$, but since $c > 0$, this condition is automatically satisfied.
• From $\lambda_3 = c - \sqrt{2} > 0$, we have $c > \sqrt{2}$.

Conclusion:

Thus, combining all the conditions, the matrix $S$ is positive definite if and only if:

$c > \sqrt{2}$

Step 4: Using Trace and Determinant

The trace and determinant of a matrix provide additional insights:

• Trace of $S$ (sum of eigenvalues): $\text{Tr}(S) = \lambda_1 + \lambda_2 + \lambda_3 = 3c$.
• Determinant of $S$: This is the product of the eigenvalues, $\text{det}(S) = c(c^2 - 2)$.

For positive definiteness:

• The trace must be positive: $3c > 0$, confirming $c > 0$.
• The determinant must be positive: $c(c^2 - 2) > 0$, which simplifies to $c > \sqrt{2}$ or $c < -\sqrt{2}$. However, since $c > 0$, we focus on $c > \sqrt{2}$.

Thus, using the trace and determinant confirms the condition $c > \sqrt{2}$ derived from eigenvalue analysis.

Would you like further details or have any questions?

Here are five related questions to explore further:

1. How does the trace of a matrix relate to its eigenvalues?
2. Can you determine if a matrix is positive definite using only its trace?
3. What role does the determinant play in determining matrix definiteness?
4. How would the conditions change if the matrix had negative entries?
5. How do eigenvalues affect the stability of a system in control theory?

Tip: Remember, for a matrix to be positive definite, all eigenvalues must be strictly positive, which can be checked through the trace, determinant, and direct computation of eigenvalues.