Why is $\det(A - \lambda I)$ zero?

Question

I was reading over some notes of mine from the Linear Algebra class I took a while ago in order to prepare myself for Computer Graphics course I'll be taking this Fall and I ran across a sentence that I can't quite grasp. It states that

Since we have a non-zero vector $x$ in the nullspace of $A - \lambda I$, then $\det(A - \lambda I) = 0$.

From reading Wikipedia page on determinants I got that

... the system has a unique solution exactly when the determinant is nonzero; when the determinant is zero there are either no solutions or many solutions.

So if we have vector $x$ in the nullspace of $A-\lambda I$, then we have either 0 solutions or many solutions. But I still don't understand why $\det(A - \lambda I)$ has to be equal zero?

Answer 1

If $x \ne 0$ is in the null space of $A - \lambda I$, i.e., $(A - \lambda I)x = 0$, that means that $A - \lambda I$ is singular (noninvertible), which is exactly the same as saying that its determinant is zero.

Answer 2

There is nothing at all special here about $A-\lambda I$.

If $M$ is any square matrix with a nonzero vector in its null space, then $det(M)=0$. Do you know that any (real) $n\times n$ matrix $M$ induces a linear map from $\mathbb R^n \to \mathbb R^n$? Matrix $M$ is non-singular if and only if the induced map is one-to-one. But if $x$ is a nonzero vector in the null space of $M$, this is violated since $M$ also sends the zero vector to itself. Thus $M$ is singular in the case you are considering.

Answer 3

Thinking of it this way helped me to grasp the idea.

$(A-\lambda I)x = Bx = 0$

A homogeneous systems of linear equations $Bx=0$ has non zero solutions only if it has dependent rows (equations). If a matrix has dependent rows its determinant is zero.

wiki: https://en.wikipedia.org/wiki/System_of_linear_equations