I'm studying eigenvalues and stumbled upon this:
The system
$(A-\lambda_i \mathbb{1}_n)\cdot X = 0$
only if
$\det(A-\lambda_i \mathbb{1}_n)=0$
I understand that this is a homogeneous linear system, and I found that they need to have determinant zero in order to have non-trivial solutions. But why exactly is this?
I'm trying to understand this intuitively instead of just taking it as it is, so any explanation is welcome.
The system
$(A-\lambda_i \mathbb{1}_n)\cdot X = 0$
only if
$\det(A-\lambda_i \mathbb{1}_n)=0$
A homogeneous system always has the zero solution $X = 0$ and for a square system ($n \times n$), this is the only solution if the determinant of the coefficient matrix is non-zero. So the system only has non-zero solutions $X$, the eigenvector(s), if the determinant is in fact $0$.
If $AX = 0$, and $\det A\neq 0$, then $A^{-1}$ is defined, so $(A^{-1}A)X = A^{-1}0$, so $X = 0$.
If a matrix has determinant different from 0, it is an invertible matrix. If the matrix is invertible, the system
$$Ax = b $$
Has one unique solution because one can multiply both sides by $A^{-1} $.
But we want the homogeneous system to have infinite solutions; hence the matrix cannot have determinant $\not= 0 $