Linear transformation exercise

[knightmetal]

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

 

[DonAntonio]

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

Write \,\mathbf{x}=(x_1,x_2,x_3)\, , so that f^2(\mathbf{x})=f\left(f(x_1,x_2,x_3)\right)
But \,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)=\,

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

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

DonAntonio

 

[knightmetal]

Thank you for your reply DonAntonio, very helpful!

 

[Vargo]

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.