16.-Linear-Algebra-by-Kenneth-Hoffman · Problem 6.2.1

Elementary Canonical Forms

Mathematics — Linear Algebra · characteristic-values

Problem

In each of the following cases, let TT be the linear operator on R2R^2 which is represented by the matrix AA in the standard ordered basis for R2R^2, and let UU be the linear operator on C2C^2 represented by AA in the standard ordered basis. Find the characteristic polynomial for TT and that for UU, find the characteristic values of each operator, and for each such characteristic value cc find a basis for the corresponding space of characteristic vectors. A=[1000],A=[23−11],A=[1111].A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad A = \begin{bmatrix} 2 & 3 \\ -1 & 1 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}.
Intuitive explanation

The characteristic polynomial comes from det⁡(A−λI)\det(A - \lambda I), which tells you what scalar multiples λ\lambda leave the transformation "singular" — these are your eigenvalues. Once you find them, plug each back into (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} to find the eigenvectors that actually get scaled by that λ\lambda. The key insight: over R\mathbb{R} you may get fewer eigenvalues than over C\mathbb{C} (complex roots appear), and some matrices might have repeated eigenvalues with fewer independent eigenvectors than you'd expect.

Intuitive explanation

The characteristic polynomial comes from det⁡(A−λI)\det(A - \lambda I), which tells you what scalar multiples λ\lambda leave the transformation "singular" — these are your eigenvalues. Once you find them, plug each back into (A−λI)v=0(A - \lambda I)\mathbf{v} = \mathbf{0} to find the eigenvectors that actually get scaled by that λ\lambda. The key insight: over R\mathbb{R} you may get fewer eigenvalues than over C\mathbb{C} (complex roots appear), and some matrices might have repeated eigenvalues with fewer independent eigenvectors than you'd expect.

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