Let \(V\) be the space of \(n \times n\) matrices over \(F\). Let \(A\) be a fixed \(n \times n\) matrix over \(F\). Let \(T\) be the linear operator ‘left multiplication by \(A\)’ on \(V\). Is it true that \(A\) and \(T\) have the same characteristic values?
Solution
1. The answer is yes: \(A\) and \(T\) have the same characteristic values in \(F\).
statement of the claim.
2. If \(\lambda \in F\) is a characteristic value of \(A\), then \(\lambda\) is a characteristic value of \(T\):
2.1. Since \(\lambda\) is an eigenvalue of \(A\), there exists a non-zero column vector \(v \in F^n\) such that \(Av = \lambda v\).
definition of characteristic value for \(A\).
2.2. Define the matrix \(B \in M_n(F)\) whose first column is \(v\) and all other columns are zero.
construction of matrix.
2.3. Since \(v \neq 0\), \(B \neq 0\) in \(V = M_n(F)\).
non-zero entry in the first column.
2.4. Compute the matrix product \(T(B) = AB\): The \(j\)-th column of \(AB\) is \(A\) times the \(j\)-th column of \(B\), so:\[ \operatorname{col}_1(AB) = Av = \lambda v = \lambda \operatorname{col}_1(B), \quad \operatorname{col}_j(AB) = A(0) = 0 = \lambda \operatorname{col}_j(B) \text{ for } j \ge 2. \]
column-by-column matrix multiplication.
2.5. Thus \(T(B) = AB = \lambda B\).
2.4.
2.6. Since \(B \neq 0\), \(B\) is an eigenvector of \(T\) with eigenvalue \(\lambda\), so \(\lambda\) is a characteristic value of \(T\).
3 and 5.
3. If \(\lambda \in F\) is a characteristic value of \(T\), then \(\lambda\) is a characteristic value of \(A\):
3.1. Since \(\lambda\) is an eigenvalue of \(T\), there exists a non-zero matrix \(B \in V\) such that \(T(B) = \lambda B\).
definition of characteristic value for the operator \(T\).
3.2. This means \(AB = \lambda B \iff (A - \lambda I)B = 0\).
definition of \(T(B) = AB\).
3.3. Since \(B \neq 0\), there exists some column index \(j \in \{1, \dots, n\}\) such that the \(j\)-th column \(v = \operatorname{col}_j(B) \in F^n\) is non-zero.
a matrix is non-zero if and only if at least one column is non-zero.
3.4. Multiplying \((A - \lambda I)\) by the \(j\)-th column gives \((A - \lambda I)v = 0\), so \(Av = \lambda v\).
column extraction of the matrix equation \((A - \lambda I)B = 0\).
3.5. Since \(v \neq 0\), \(v\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), so \(\lambda\) is a characteristic value of \(A\).
3 and 4.
4. Conclusion: The set of characteristic values of \(T\) in \(F\) is identical to the set of characteristic values of \(A\) in \(F\).
2 and 3.
Q.E.D.