DISCLAIMER ! THIS IS NOT A DUPLICATE OF MY OTHER QUESTIONS. THE QUESTION I AM ASKING NOW ONLY TAKES INTO CONSIDERATION THE SAME MATRIX I HAVE USED IN OTHER QUESTIONS. THAT'S IT.
So i have this matrix
A= $$ \begin{pmatrix} -2 & 0 & 0 \\ 3 & 1 & -6 \\ 0 & 0 & -2 \\ \end{pmatrix} $$
We compute the characteristic polynomial
p(λ) = det(A − λId) = (λ + 2)²· (1 − λ)
which gives me two roots namely λ1 = −2 of multiplicity 2, and λ2 = 1 of multiplicity 1.
First of all , I solve the homogeneous system associated with A + 2Id which gives us
Nul(A+2Id)=Span $$ \begin{pmatrix} 2 & -1 \\ 0 & 1 \\ 1 & 0 \\ \end{pmatrix} $$
I know how to solve the system associated with A+2Id since the matrix i get by doing (A+2Id) is pretty simple (there are two 0 rows) and the homogeneous equation associated with it is x+y-2z=0
However my confusion arises when , in order to find a basis of eigenvectors, I need to compute E(λ2), so we need to solve the homogeneous system associated with(A − Id).
I do not now how to go on from here.
Computing A-Id (please correct me if im wrong, since A-1(Id)=A-Id), gives me the matrix
A-Id= $$ \begin{pmatrix} -3 & 0 & 0 \\ 3 & 0 & -6 \\ 0 & 0 & -3 \\ \end{pmatrix} $$
How do i proceed from here? On the answers sheet it says that Nul(A − Id) = Span (column vector)= (0,1,0).
Should i compute the characteristic polynomial of the (A-Id) matrix? How do i do that ? Is that the right way to do that?
Please if anybody can show me some simple and clear steps on how to get
Nul(A − Id) = Span (column vector)= (0,1,0) and tell me if what I'm doing is right! Thanks!
You put $A-I$ in reduced row echelon form: $$\begin{pmatrix}-3&0&0\\3&0&-6\\0&0&-3\end{pmatrix}\rightsquigarrow\begin{pmatrix}1&0&0\\1&0&-2\\0&0&1\end{pmatrix}\rightsquigarrow\begin{pmatrix}1&0&0\\0&0&-2\\0&0&1\end{pmatrix}\rightsquigarrow\begin{pmatrix}1&0&0\\0&0&1\\0&0&0\end{pmatrix}$$ so we have the equations $$x=0,\quad z=0.$$