T:M₂(ℝ)→M₂(ℝ) sends X to AXB, with $A=\begin{pmatrix}1&-2\\1&4\end{pmatrix}$, $B=\begin{pmatrix}6&5\\-2&-1\end{pmatrix}$. P is its matrix in the standard basis. Which statements hold?
Correct answer: (A) P is invertible.; (B) trace(P)=25.
Explanation
A has eigenvalues 2,3; B has eigenvalues 1,4. Eigenvalues of X→AXB are pairwise products 2,8,3,12. None is zero, so P is invertible; trace is 25. Eigenvalue 2 makes P²−4I singular and P−2I have nonzero nullity. **A,B**.
More Eigenvalues and Canonical Forms questions
GATE 2024 MA Q19– A=pmatrix0&2\\2&0pmatrix. On M₂(ℂ), T(B)=AB. Find T’s characteristic polynomial.
GATE 2024 MA Q20– A is a complex normal matrix. I: real eigenvalues imply A is Hermitian. II: unit-modulus…
GATE 2024 MA Q51– T,S:ℝ⁴→ℝ⁴ are nonzero and nonidentity; T²=T. Which statements hold?
GATE 2024 MA Q56– T:ℝ⁴→ℝ⁴ has only eigenvalues 1,2, dim ker(T−I)=1 and rank(T−2I)=2. Which Jordan forms are…
GATE 2023 MA Q19– M=pmatrix4&-3\\1&0pmatrix. P:M⁸+M¹² is diagonalizable. Q:M⁷+M⁹ is diagonalizable. Choose…
GATE 2023 MA Q22– T:ℝ⁴→ℝ⁴ has kernel 4x₁+3x₂+2x₃+x₄=0 and rank(T−3I)=3. Find its minimal polynomial.