Question

Matrix A has the eigenvalues 1, 2, and 3. The Trace of \(A^2\) is

• A. 6
• B. 14
• C. 20
• D. 8

Hint:

The eigenvalues of \(A^2\) are the squares of the eigenvalues of \(A\), and the trace of any matrix equals the sum of its eigenvalues.

Answer 1

The Correct Option is B

Step 1: Recall the property connecting trace and eigenvalues.
For any square matrix, the trace (the sum of the diagonal entries) always equals the sum of its eigenvalues, counted with multiplicity. This holds no matter which basis the matrix is written in.

Step 2: Recall how eigenvalues behave when a matrix is raised to a power.
If \(\lambda\) is an eigenvalue of \(A\) with eigenvector \(v\), then \(Av=\lambda v\), so \(A^2v=A(\lambda v)=\lambda(Av)=\lambda^2 v\). So \(\lambda^2\) is an eigenvalue of \(A^2\); the eigenvalues of \(A^2\) are just the squares of the eigenvalues of \(A\).

Step 3: Square each given eigenvalue.
The eigenvalues of \(A\) are \(1,2,3\). Squaring each:
\[ 1^2=1,\quad 2^2=4,\quad 3^2=9 \]
These are the eigenvalues of \(A^2\).

Step 4: Add them to get the trace of \(A^2\).
\[ \text{Trace}(A^2)=1+4+9=14 \]

Step 5: Rule out the other options.
Option (A), 6, is just \(1+2+3\), the trace of \(A\) itself, not of \(A^2\), a common slip of forgetting to square before adding. Option (C), 20, would come from mixing up unrelated combinations of the eigenvalues instead of simply summing their squares. Option (D), 8, does not correspond to any consistent operation on \(1,2,3\) and is just a distractor.

Final Answer:
The trace of \(A^2\) is 14, option (B). \[ \boxed{14} \]