Question

Given\[A=\left[\begin{matrix}2 & 1 & 3 \\ 0 & -1 & 4 \\ 0&0&0\end{matrix}\right]\]Determine the eigenvalues and who that the corresponding eigenspaces are dimension 1 and are generated by the eigenvectors\[\left(\begin{matrix}1 \\ 0\\0\end{matrix}\right),\left(\begin{matrix}1 \\ -3\\0\end{matrix}\right),\left(\begin{matrix}-7 \\ 8\\2\end{matrix}\right)\]

Answer 1

Find determinate of (A-lambda) and set equal to 0
\[\det\left[\begin{matrix}2-L& 1&3 \\ 0 & -1-L&4\\0&0&-L\end{matrix}\right] = L(2-L)(1-L)=0\]
\[L = 0,1,2\]

Answer 2

for eigenvalue, L=0
\[\left[\begin{matrix}2& 1&3 \\ 0 & -1&4\\0&0&0\end{matrix}\right]\left[\begin{matrix} x \\ y\\z\end{matrix}\right]= \left[\begin{matrix} 0 \\ 0\\0\end{matrix}\right]\]

Answer 3

solving system yields...
\[x = -(7/2)z\]
\[y = 4z\]
If z = 2, then x = -7, y = 8
therefore eigenvector (-7,8,2) goes with eigenvalue L=0
do the other 2 eigenvalues the same way

Answer 4

here is a reference

http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors