Linear Algebra

366 Chapter Five. Similaritythen 2 is an eigenvalue of the matrix T, associated with these eigenvectors.{( ) (c 0 ∣∣ 2 0c 1 0 0) ()c 0=c 1( ) ( )2c 0 c 0 ∣∣} = { c0 ∈ C, c 0 ≠ 0}2c 1 0On the other hand, if we represent t with respect to D = 〈2 + 1x, 1 + 0x〉( )3 1S = Rep D,D (t) =−3 −1then the eigenvectors associated with the eigenvalue 2 are these.{( ) (c 0 ∣∣ 3 1c 1 −3 −1) ()c 0=c 1( ) ( )2c 0 0 ∣∣} = { c1 ∈ C, c 1 ≠ 0}2c 1 c 13.8 Remark Here is an informal description of the reason for the difference.The underlying transformation doubles the eigenvectors ⃗v ↦→ 2 · ⃗v. But whenthe matrix representing the transformation is T = Rep B,B (t) then the matrix“assumes” that column vectors are representations with respect to B. In contrast,S = Rep D,D (t) “assumes” that column vectors are representations with respectto D. So the column vector representations that get doubled by each matrix aredifferent.The next example shows the basic tool for finding eigenvectors and eigenvalues.3.9 Example If⎛⎞1 2 1⎜⎟T = ⎝ 2 0 −2⎠−1 2 3then to find the scalars x such that T⃗ζ = x⃗ζ for nonzero eigenvectors ⃗ζ, bringeverything to the left-hand side⎛⎞ ⎛1 2 1⎜⎟ ⎜z ⎞ ⎛1⎟ ⎜z ⎞1⎟⎝ 2 0 −2⎠⎝z 2 ⎠ − x ⎝z 2 ⎠ = ⃗0−1 2 3 z 3 z 3and factor (T − xI)⃗ζ = ⃗0. (Note that it says T − xI. The expression T − x doesn’tmake sense because T is a matrix while x is a scalar.) This homogeneous linearsystem⎛⎞ ⎛ ⎛ ⎞⎜⎝1 − x 2 1⎟ ⎜z ⎞1 0⎟ ⎜ ⎟2 0 − x −2 ⎠ ⎝z 2 ⎠ = ⎝0⎠−1 2 3 − x z 3 0has a nonzero solution ⃗z if and only if the matrix is singular. We can determine