Question

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Let V be a vector space with dim(V)=3. Explain why any T: V → V has at least one real eigenvalue.

Answer 1

In linear algebra, any T: V → V has at least one real eigenvalue when dim(V) = 3.

Step-by-step explanation:

In linear algebra, a real eigenvalue of a linear transformation is a scalar λ such that there exists a non-zero vector xf in the vector space V that satisfies the equation T(xf) = λxf. To show that any T: V → V has at least one real eigenvalue when dim(V) = 3, we can use the fact that the characteristic polynomial of T always has at least one real root. The characteristic polynomial is defined as det(T - λI), where T is the matrix representation of the linear transformation and I is the identity matrix.

Since dim(V) = 3, the matrix representation of T will be a 3x3 matrix. The characteristic polynomial of a 3x3 matrix is a cubic polynomial, which means it will always have at least one real root. Therefore, any T: V → V has at least one real eigenvalue.