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Linear transformation exercise

  1. Jun 5, 2012 #1
    Hello,

    I'm given this linear transformation and I'm asked to do the typical calculations (kernel, image, dimensions, etc.) but there's one thing I'm not sure I understand, here's the exercise:

    f(1,0,0)=(-1,-2,-3)
    f(0,1,0)=(2,2,2)
    f(0,0,1)=(0,1,2)

    a) Is f invertible?
    b)Find a basis of Ker(f) and a basis of Im(f)
    c)Find eigenvalues, eigenvectors. is f diagonalizable?
    d) Solve the system f^2(x)=0

    Can anybody point me in the right direction on how to solve d) ?

    Thanks a lot
     
  2. jcsd
  3. Jun 5, 2012 #2


    Write [itex]\,\mathbf{x}=(x_1,x_2,x_3)\,[/itex] , so that [tex]f^2(\mathbf{x})=f\left(f(x_1,x_2,x_3)\right)[/tex]
    But [itex]\,f(x_1,x_2,x_3)=f\left(x_1(1,0,0)+x_2(0,1,0)+x_3(0,0,1)\right)=x_1f(1,0,0)+x_2f(0,1,0)+x_3f(0,0,1)=\,[/itex]

    [itex]\,=(-x_1,-2x_1,-3_x1)+(2x_2,2x_2,2x_2)+(0,x_3,2x_3)=\,[/itex]...etc.

    Another, much easier way: write your lin. transf. as a matrix wrt the canonical basis:[tex]f\longrightarrow \left(\begin{array}{rrr}-1&-2&-3\\2&2&2\\0&1&2\end{array}\right)[/tex] so that [tex]f(\mathbf{x})\longrightarrow \left(\begin{array}{rrr}-1&-2&-3\\2&2&2\\0&1&2\end{array}\right)\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}[/tex] and then [tex]f^2\longrightarrow \left(\begin{array}{rrr}-1&-2&-3\\2&2&2\\0&1&2\end{array}\right)^2[/tex]

    DonAntonio
     
  4. Jun 5, 2012 #3
    Thank you for your reply DonAntonio, very helpful!
     
  5. Jun 6, 2012 #4
    A minor correction. That looks like the transpose of the matrix that you want. The columns should be the respective images of the standard basis vectors.
     
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