Question

Let $A$ be a $3 \times 3$ real matrix having eigenvalues $1,2,3$. If \[ B = A^2 + 2A + I, \] where $I$ is the $3 \times 3$ identity matrix, then the eigenvalues of $B$ are:

• A. 4, 9, 16
• B. 1, 2, 3
• C. 1, 4, 9
• D. 4, 16, 25

Hint:

If $B=p(A)$ for a polynomial $p$, the eigenvalues of $B$ are simply $p(\lambda_i)$ where $\lambda_i$ are eigenvalues of $A$. This avoids direct matrix computation.

Answer 1

The Correct Option is A

Step 1: Recall eigenvalue property under polynomials.
If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, i.e. $Av=\lambda v$, then for any polynomial $p(x)$: \[ p(A)v = p(\lambda)v. \] Thus eigenvalues of $B=p(A)$ are obtained by applying $p(\lambda)$ to each eigenvalue $\lambda$ of $A$.
Step 2: Identify the polynomial $p(\lambda)$.
We are given $B=A^2+2A+I$. Hence the associated polynomial is \[ p(\lambda)=\lambda^2+2\lambda+1. \]
Step 3: Simplify $p(\lambda)$.
\[ p(\lambda)=\lambda^2+2\lambda+1=(\lambda+1)^2. \]
Step 4: Apply to the eigenvalues of $A$.
$A$ has eigenvalues $\lambda=1,2,3$. \[ p(1)=(1+1)^2=4, p(2)=(2+1)^2=9, p(3)=(3+1)^2=16. \] Final Answer: \[ \boxed{4,\ 9,\ 16} \]