Why does $\det(A-\lambda I) = 0$

Question

I am looking at this website, and it says "therefore we will need to determine the values of $\lambda$ for which we get, $$\det(A-\lambda I) = 0."$$ Now I am a newbie to linear algebra, and I have looked at several other sources, but I still can't understand why $\det(A-\lambda I) = 0.$ I already understand why $(A-\lambda I)\vec{v} = 0.$Can anyone help me with this?

Answer 1

A matrix $B$ isn't invertible if and only if $\det(B)=0$ which is also equivalent to say that for some $v\ne0$ we have $Bv=0$ i.e. $B$ isn't injective. Now for $B=A-\lambda I$ we have $\det(A-\lambda I)=0$ iff for some $v$ we have $(A-\lambda I)v=0$ and so $Av=\lambda v$ i.e. $\lambda$ is an eigenvalue of $A$ and $v$ is an eigenvector associated to $\lambda$.

Answer 2

If you had $\det(A-\lambda I)\ne0$, the only solution of $(A-\lambda I)\vec v=0$ would be $\vec v=(A-\lambda I)^{-1}0=0$.

For a nontrivial solution to exist, the system must be rank-deficient.

Answer 3

If we have $$(A-I\lambda)\vec{v}=\vec{0}$$ and $(A-I\lambda)$ is invertible, which is equivalent to $\det(A-I\lambda)\neq 0$, then we can multiply the equation with the inverse $(A-I\lambda)^{-1}$ from the left to obtain:

$$(A-I\lambda)^{-1}(A-I\lambda)\vec{v}=(A-I\lambda)^{-1}\vec{0} \implies \vec{v}=\vec{0}.$$

So if $\det(A-I\lambda)\neq 0$, then we are only able to obtain the trivial solution $\vec{v}=\vec{0}$. In order to obtain a non-trivial solution we demand $\det(A-I\lambda) =0$.

Answer 4

If you write, and think, what for you have to do the above perhaps it'll be clearer: you want to find out eigenvalues of $\;A\;$ , i.t.: scalars $\;\lambda\;$for which there exists a nonzero vector $\;v\;$ s.t. $\;Av=\lambda v\;$, but this means:

$$Av=\lambda v\iff Av-\lambda v=0\iff (A-\lambda I)v=0\iff v\in\ker (A-\lambda I)\implies$$

and since $\;v\neq0\implies \ker(A-\lambda I)\neq\{0\}\iff A-\lambda I\;$ is singular $\;\iff \det (A-\lambda I)=0\;$