→ D

1. ❌ true only if $x≠0$. If x = 0, then the equation is always satisfied regardless of $\lambda$, but it doesn’t make $\lambda$ an eigenvalue
2. ❌  $-v$ is still for eigenvalue $2$, because scalar multiples of eigenvectors are still eigenvectors for the same eigenvalues
3. ❌ $B^{-1}AB$ is similar to $A$, and similar matrices always have the same eigenvalues
4. ✅ If $Ax = \lambda x$, then $A^2 x = A(Ax) = \lambda (Ax) = \lambda(\lambda x) = \lambda^2 x$. So x is also an eigenvector for $A^2$ with eigenvalue $\lambda^2$
5. ❌ Having -5 as an eigenvalue means $det(B + 5I) = 0$, not $det(B - 5I)$.
   • Suppose $\lambda$ is an eigenvalue of B. Then $Bv = \lambda v \rightarrow (B-5I)v = (\lambda - 5)v$. So, the eigenvalues of $B-5I$ are $\lambda-5$. To be non-invertible, 0 must be an eigenvalue. This means $\lambda-5 = 0 \rightarrow \lambda = 5$

→ A

1. ✅ $p(0) = p(1)?$

• zero polynomial: $0 = 0$
• closed under addition:
  • if $p(0) = p(1)$ and $q(0) = q(1)$, then ($p+q)(0) = p(0) + q(0) = p(1) + q(1) = (p+q)(1)$
• closed under scalar multiplication
  • $(cp)(0) = cp(0) = cp(1) = (cp)(1)$

ii. ❌ $p(0)p(1) = 0$

• zero polynomial satisfies condition
• not closed under addition
  • take p(x) such that $p(0) = 0, p(1) ≠ 0$ and $q(0) ≠0, q(1) = 0$
  • then both satisfy $p(0)(p(1) = 0$, but $(p+q)(0)(p+q)(1) ≠ 0$

iii. ❌ integer coefficients

• not closed under scalar multiplication
  • multiply a polynomial with integer coefficients by $\frac{1}{2}$, not it has non-integer coefficients which are not in the set