Question

Let \(A\) be a \(2\times2\) real symmetric matrix such that the trace of \(A\) is \(6\) and the determinant of \(A\) is \(5\). Which one of the following statements is true?

• A. \(A-I_2\) is positive definite
• B. \(A-3I_2\) is negative definite
• C. \(A^2+A-3I_2\) is positive definite
• D. \(A^2-6A\) is negative definite

Hint:

For symmetric matrices, definiteness can be checked directly using eigenvalues: positive definite means all eigenvalues are positive, and negative definite means all eigenvalues are negative.

Answer 1

The Correct Option is C

Step 1: Get the eigenvalues from trace and determinant.
The two eigenvalues add to $6$ and multiply to $5$, so they solve $\lambda^2-6\lambda+5=0$, giving $\lambda=1$ and $\lambda=5$.

Step 2: Use the shifting rule.
For any polynomial $p$, the matrix $p(A)$ has eigenvalues $p(1)$ and $p(5)$. A matrix is positive definite when both are $>0$ and negative definite when both are $<0$.

Step 3: Rule out the first three.
$A-I_2$ gives $0,4$, so only semidefinite, (A) out. $A-3I_2$ gives $-2,2$, indefinite, (B) out. For $\lambda^2+\lambda-3$ we get $-1$ and $27$, mixed signs, (C) out.

Step 4: Check the last one.
For $\lambda^2-6\lambda=\lambda(\lambda-6)$: at $1$ it is $-5$, at $5$ it is $-5$. Both negative, so $A^2-6A$ is negative definite.

Step 5: Conclude.
The true statement is option (D).
\[ \boxed{(D)} \]