Question

hi, I am new here. In lesson 21, why \[A-\lambda I\] need to be singular in order for the equation \[(A-\lambda I)x=0\] to be true? Sorry if this is an obvious question..

Answer 1

May be I try to answer my own question, please let me know if I am right or not.
If A is non-singular/invertible, there would be only one trivial solution, i.e., x=0 to equation Ax=0, of which Professor Gilbert assume not to be the case in the first place.

Answer 2

If \(x_0\neq0\) is an eigenvector of the matrix \(A\), then theres is a real number \(\lambda\) such that
\[ Ax_0=\lambda x_0. \]
But this is equivalent to saying that
\[ 0=Ax_0-\lambda x_0=(A-\lambda I)x_0. \]
Since \(x_0\neq0\), the last equation actually says that the linear system
\[ (A-\lambda I)x=0 \]
has a non-trivial (hence infinitely many) solution, so the matrix \(A-\lambda I\) is singular.