Every linear operator in an n-dimensional vector space has n distinct eigen-values

To answer the question, we can break it down into steps:

Step 1: Understand the concept of eigenvalues and eigenvectors. In linear algebra, an eigenvalue is a scalar that represents how a linear transformation affects a vector. An eigenvector is a non-zero vector that remains in the same direction after the linear transformation.

Step 2: Recall that an n-dimensional vector space is a space that consists of vectors with n components. For example, a 2-dimensional vector space would have vectors with two components (x and y), while a 3-dimensional vector space would have vectors with three components (x, y, and z).

Step 3: Consider a linear operator in an n-dimensional vector space. A linear operator is a function that maps vectors from one vector space to another while preserving certain properties, such as linearity.

Step 4: The statement claims that every linear operator in an n-dimensional vector space has n distinct eigenvalues. This means that for any linear operator in an n-dimensional vector space, there will be n different eigenvalues associated with it.

Step 5: To prove this statement, we can use the fact that the characteristic polynomial of a linear operator in an n-dimensional vector space has n distinct roots, which correspond to the eigenvalues. The characteristic polynomial is obtained by subtracting the identity matrix multiplied by a scalar lambda from the linear operator and taking its determinant.

Step 6: By finding the roots of the characteristic polynomial, we can determine the eigenvalues of the linear operator. Since the characteristic polynomial has n distinct roots, it follows that the linear operator has n distinct eigenvalues.

Therefore, we can conclude that every linear operator in an n-dimensional vector space has n distinct eigenvalues.