Scalar transformation criterion

Let $𝐯 0, \dots, 𝐯 𝑛$ form a basis of a vector space $𝕂 𝑛 + 1$ , and let $𝐯 𝑛 + 1 = \sum 𝑛 𝑖 = 0 𝐯 𝑖$ . Then $Φ \in G L (𝑛 + 1, 𝕂)$ is a scalar transformation iff every $𝐯 𝑖$ for $𝑖 = 0, \dots, 𝑛 + 1$ is an eigenvector. #m/thm/linalg Thus, if all $𝐯 𝑖$ are eigenvectors then they all have the same nonzero eigenvalue.

Proof

Let $𝜆 𝑖$ be the eigenvalue corresponding to $𝐯 𝑖$ . Then

𝑛 \sum 𝑖 = 0 𝜆 𝑛 + 1 𝐯 𝑖 = 𝜆 𝑛 + 1 𝐯 𝑛 + 1 = Φ (𝐯 𝑛 + 1) = Φ (𝑛 \sum 𝑖 = 0 𝐯 𝑖) = 𝑛 \sum 𝑖 = 0 𝜆 𝑖 𝐯 𝐢

and since the decomposition of a vector into basis vectors is unique, it follows $𝜆 𝑖 = 𝜆 𝑛 + 1$ for all $𝑖 = 0, \dots, 𝑛 + 1$ . The converse is trivial, since every nonzero vector is an eigenvalue of a scalar transformation.

#state/tidy | #lang/en | #SemBr