For solving $(A - \lambda I)x = 0 $ why do we take $det(A - \lambda I) = 0$ ?
wikipedia: Consider a matrix A and a nonzero vector 𝑣. If applying A to 𝑣 (denoted by 𝐴𝑣) simply scales 𝑣 by a factor of λ, where λ is a scalar, then 𝑣 is an eigenvector of A, and λ is the corresponding eigenvalue. This relationship can be expressed as: 𝐴𝑣=𝜆𝑣.
$Av = \lambda v$
$Av - \lambda v = 0$
$(A - \lambda I) v = 0$
this equation looks like $Bx = 0$ (homogenous system of linear equation) and from the definition of eigen vector we know that $v$ is a non-zero vector. here $v$ acts as a non-trivial solution to this homogenous system of linear equation.
here we are assuming, $B = A - \lambda I$ and $x = v$
when we have a non-trivial solution to $Bx = 0$ (homogenous system of linear equation) then we can write the linear combination of columns ($Bx$) of matrix $B$ like this:
$c_1.v_1 + c_2.v_2 + c_3.v_3 + ......... = 0$
as we have a non-trivial solution $x = \begin{bmatrix}c_1\\c_2\\c_3\\.\\.\\.\end{bmatrix}$ that means we have atleat one $c_i$ that is non-zero. suppose that non-zero $c_i$ is $c_2$ then we can write the equation as:
$v_2 = -1/c_2 (c_1.v_1 + c_3.v_3 + c_4.v_4 + .........)$
now it means there is atleast one column $v_2$ which is linear combination of other columns and having even atleast one L.D. column means $det(B) = 0$.
$det(B) = 0$
$det(A - \lambda I) = 0$
