- #1
TranscendArcu
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Homework Statement
The Attempt at a Solution
T(1,0,0) = (3,-1,0)
T(0,1,0) = (0,1,0)
T(0,0,1) = (-1,2,4)
Thus, we have the matrix,
[itex]\left| \begin{array}{ccc}
3 &0&-1 \\
-1&1&2 \\
0&0&4 \end{array} \right|[/itex]
[itex]Δ_T (t) = det( \left| \begin{array}{ccc}
3 &0&-1 \\
-1&1&2 \\
0&0&4 \end{array} \right| - tI)[/itex]
I have this equaling: -(t-4)(t-3)(t-1), which is the characteristic polynomial. The roots are the eigenvalues, which are 4,3,1.
To compute the eigenvectors:
When t=4, we have,
-x-z=0
-x-3y+2z=0
0z=0
Which implies that eigenvectors are multiples of (-1,1,1).
When t=3, we have,
-z=0
-x-2y=0
z=0
Which implies that eigenvectors are multiples of (-1,2,0)
When t=1, we have,
2x-z=0
-x+2z=0
3z=0
Which implies that eigenvectors are multiples of (0,1,0).
T is diagonizable because (-1,1,1),(-1,2,0),(0,1,0) are lin. indep.
First of all, are these correct? Also, how does one determine eigenspaces? That's the part I feel that I really don't know.