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: Find the eigenvalues of \(A\).
Since \(A\) is a real symmetric matrix, all eigenvalues of \(A\) are real.
Let the eigenvalues of \(A\) be
\[ \lambda_1,\lambda_2 \]
We are given
\[ \lambda_1+\lambda_2=\text{trace}(A)=6 \] and
\[ \lambda_1\lambda_2=\det(A)=5 \]
Thus, the characteristic equation is
\[ \lambda^2-6\lambda+5=0 \]
Factorizing,
\[ (\lambda-1)(\lambda-5)=0 \]
Hence, the eigenvalues are
\[ \lambda_1=1,\quad \lambda_2=5 \]

Step 2: Check option (A).
The eigenvalues of
\[ A-I_2 \] are obtained by subtracting \(1\) from each eigenvalue of \(A\).
Thus, the eigenvalues become
\[ 1-1=0 \] and
\[ 5-1=4 \]
So, the eigenvalues are
\[ 0,\ 4 \]
A matrix is positive definite only if all eigenvalues are strictly positive.
Since one eigenvalue is \(0\), the matrix is only positive semidefinite, not positive definite.
Hence, option (A) is false.

Step 3: Check option (B).
The eigenvalues of
\[ A-3I_2 \] are
\[ 1-3=-2 \] and
\[ 5-3=2 \]
Thus, the eigenvalues are
\[ -2,\ 2 \]
Since one eigenvalue is positive and the other is negative, the matrix is indefinite.
Therefore, it is not negative definite.
Hence, option (B) is false.

Step 4: Check option (C).
The eigenvalues of
\[ A^2+A-3I_2 \] are obtained by substituting each eigenvalue \(\lambda\) into the polynomial
\[ \lambda^2+\lambda-3 \]
For
\[ \lambda=1, \] we get
\[ 1^2+1-3 = 1+1-3 = -1 \]
For
\[ \lambda=5, \] we get
\[ 5^2+5-3 = 25+5-3 = 27 \]
The eigenvalues are
\[ -1,\ 27 \]
Since one eigenvalue is negative, this matrix is not positive definite.
Thus, option (C) appears false.

Step 5: Check option (D).
The eigenvalues of
\[ A^2-6A \] are
\[ \lambda^2-6\lambda = \lambda(\lambda-6) \]
For
\[ \lambda=1, \] we get
\[ 1(1-6)=-5 \]
For
\[ \lambda=5, \] we get
\[ 5(5-6)=-5 \]
Thus, both eigenvalues are
\[ -5,\ -5 \]
Since all eigenvalues are strictly negative, the matrix is negative definite.
Hence, option (D) is true.

Step 6: Final conclusion.
Therefore, the correct statement is
\[ \boxed{(D)} \]