Find linear transformation using diagonal matrix

[csc2iffy]

Homework Statement

Define L:R3-->R3 by L(x,y,z)=(y-z,x+z,-x+y).
A. Show that L is self-adjoint using the standard orthonormal basis B of R3.
B. Diagonalize L and find and orthogonal basis B of R3 of eigenvectors of L and the diagonal matrix.
C. Considering only the eigenvalues of L, determine if L is an isomorphism.
D. Find L(1,0,0) using the diagonal matrix of L.

Homework Equations

L(x,y,z)=(y-z,x+z,-x+y)

Matrix of L with respect to orthonormal basis:
0 1 -1
1 0 1
-1 1 0

Diagonal matrix of L:
1 0 0
0 1 0
0 0 -2

The Attempt at a Solution

I already answered A and B. For C, I said L is not an isomorphism because of repeated eigenvalues. For D, I am not sure how to find the linear transformation using only the diagonal matrix, but I know the answer is (0,1,-1).

 

[HallsofIvy]

A linear transformation is an isomorphism as long as it is invertible. What does that tell you about its eigenvalues? I'm not sure how to interpret "using the diagonal matrix". You cannot do this using only the diagonal matrix (the eigenvalues), you have to use the eigenvectors as well. This matrix is diagonalizable because there exist a basis for the space consisting of eigenvectors. Write (1, 0, 0) as a linear combination of eigenvectors.

 

[csc2iffy]

thanks for your response! I'm pretty much done with the question besides part D.)
It says to find L(1,0,0) using the matrix in B.)
I'm not really sure how to do this.. I've been looking through my textbook for hours. Any extra help would be very much appreciated!

I tried writing it as a linear combination
|1 -1 1| |a| |1|
|1 0 -1| |b| = |0|
|0 1 1| |c| |0|
and I get
|a| | .333 |
|b| = |-.333 |
|c| | .333 |

and idk I'm getting lost... someone please help