Let \(M_{3\times3}(\mathbb R)\) be the vector space of real \(3\times3\) matrices. For \(A\in M_{3\times3}(\mathbb R)\) define \[ L_A(B)=AB. \] Suppose \[ \det A=32 \] and the minimal polynomial of \(A\) is \[ (t-4)(t-2). \] What is \(\operatorname{tr}(L_A)\)?
Solution
1. The matrix \(A\) is diagonalizable over \(\RR\), and its eigenvaluesare \(2\) and \(4\).
The minimal polynomial \[ (t-4)(t-2) \] splits over \(\RR\) into distinct linear factors, so \(A\) is diagonalizable. For a diagonalizable operator, the minimal polynomial is the product of the distinct factors \(t-\lambda\) as \(\lambda\) ranges over its eigenvalues. Since both factors occur here, the eigenvalues are precisely \(2\) and \(4\).
2. The eigenvalue \(4\) has multiplicity \(2\), the eigenvalue \(2\) hasmultiplicity \(1\), and therefore\[ \operatorname{tr}(A)=10. \]
Let \(m\) be the multiplicity of the eigenvalue \(4\). By step 1, both eigenvalues occur, so \(m\in\{1,2\}\) and the multiplicity of \(2\) is \(3-m\). Because \(A\) is diagonalizable, \[ 32 = \det A = 4^m2^{3-m} = 2^{m+3}. \] Hence \(m+3=5\), so \(m=2\). The trace is therefore \[ \operatorname{tr}(A)=2\cdot4+1\cdot2=10. \]
3. The operator \(L_A\) has trace\[ \operatorname{tr}(L_A)=3\operatorname{tr}(A). \]
Write a matrix \(B\in M_{3\times3}(\RR)\) by its columns as \[ B=[b_1\ b_2\ b_3]. \] Then \[ L_A(B)=AB=[Ab_1\ Ab_2\ Ab_3]. \] Thus, under the column-space identification \[ M_{3\times3}(\RR) \cong \RR^3\oplus\RR^3\oplus\RR^3, \] the operator \(L_A\) is \(A\oplus A\oplus A\). Its matrix is block diagonal with three copies of \(A\), so its trace is the sum of the traces of those three blocks.
4. The requested trace is\[ \boxed{\operatorname{tr}(L_A)=30}. \]
By step 2, \(\operatorname{tr}(A)=10\), and step 3 gives \(\operatorname{tr}(L_A)=3\operatorname{tr}(A)=30\).
5. Q.E.D.
Step 4 gives the requested value.