I am given a matrix $A= \bigg({} \matrix{10 & 7 \\-14 &-11} \bigg{)}$ and eigenvalue $3$. My elite mission is to find the treacherous basis for the eigenspace.
I used the $(A -eI)=v$ where $e$ is the eigenvalue and $I$ is the identity matrix. I then obtained the reduced row echelon form or the resulting matrix to find non-zero solutions for $v$. However the matrix came back with no free variables so all solutions are $0$. The book I am reading from says the answer is $\{[-1;1]\}$. How is that possible.
Doesn't make sense to me.
The system you should solve to find the eigenvectors is $$ (A - 3 I)v = 0 \implies\\ \pmatrix{7&7\\-14&-14} \pmatrix{v_1\\v_2} = \pmatrix{0\\0} $$ What is the solution to this system?