$Linear algebra$

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Let T:R2->R2 Linear Transformation such as T(1,1) = (8,8) and T(1,-1) = (24,8)

A) Find the matrix [T] = A from the linear transformation (in a canonical basis)
B)What are its Eigenvalues?
C)Find a basis for each eigenspace
d) is T diagonalizable? why?

Linear Algebra

Kastro

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Answer 1

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By linearity, \[ T(1,0) = \frac{1}{2}( T(1,1) + T(1,-1) ) = (16,8) \] and \[ T(0,1) = \frac{1}{2}( T(1,1) - T(1,-1) ) = (-8,0) \] so the matrix is \[ A = \begin{pmatrix} 16 & -8 \\ 8 & 0 \end{pmatrix}. \] The characteristic polynomial is \[ \det(xI-A) = x(x-16)+64 = x^2 - 16x + 64 = (x-8)^2 \] so the eigenvalues are both $8$. Since the eigenvalues are all equal but $A$ it isn't a multiple of the identity, it is not diagonalizable, it has a single eigenspace, and a basis is given by $(1,1)$ (which is an eigenvector) and any other linearly independent vector (e.g. $(1,-1)$).

Alessandro Iraci

$Linear algebra$

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