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: Build a concrete matrix with the given eigenvalues.
Since the trace only depends on the eigenvalues, not on which specific matrix has them, pick the simplest matrix with eigenvalues $1,2,3$: the diagonal matrix
\[ A=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix} \]

Step 2: Compute $A^2$ directly by squaring this matrix.
For a diagonal matrix, squaring it just squares each diagonal entry:
\[ A^2=\begin{bmatrix}1^2&0&0\\0&2^2&0\\0&0&3^2\end{bmatrix}=\begin{bmatrix}1&0&0\\0&4&0\\0&0&9\end{bmatrix} \]

Step 3: Read off the trace from this matrix.
\[ \text{Trace}(A^2)=1+4+9=14 \]

Step 4: Justify why this particular matrix is enough, even though the actual $A$ is not given.
Any matrix with eigenvalues $1,2,3$ is similar to this diagonal matrix, meaning $A=PDP^{-1}$ for some invertible $P$, where $D$ is the diagonal matrix above. Then $A^2=PD^2P^{-1}$, and similar matrices always share the same trace, since $\text{trace}(PXP^{-1})=\text{trace}(X)$. So regardless of the actual entries of $A$, $\text{trace}(A^2)$ must equal $\text{trace}(D^2)=14$.

Final Answer:
$\text{Trace}(A^2)=14$, option (B). \[ \boxed{14} \]