How to take the determinant of $(A-\lambda I)X=0$?

Question

Consider the eigenvalue equation $AX=\lambda X$ where A is a $n\times n$ square matrix and $\lambda$ is the eigenvalue corresponding to the eigenvector $\lambda$. For two square matrices, A and $\det(AB)=\det A.\det B$. How to take the determinant of both sides of the equation $(A-\lambda I)X=0$ where $A-\lambda I$ is a $n\times n$ matrix and $X$ is a column vector.

My question is how to show that for nontrivial $X$, $\det(A-\lambda I)=0$? This is a matrix equation and not an algebraic equation of the form $ax=0$.

Answer 1

Observe that for any real or complex $\;\lambda\;$ and some square matrix $\;A\;$ , we have that $\;\lambda\;$ is an eigenvalue of $\;A\;$ iff $\;A-\lambda I\;$ is a regular matrix iff $\;\det (A-\lambda I)=0\;$ .

You can't take determinant of the neither side in $\;(A-\lambda I)X=0\;$ unless both sides sides are $\;1\times 1\;$ vectors. Determinant is defined only for square matrices .

Now, for an unknown $\;x\;$ , we have that $\;\det (A-xI)=p_A(x)\;$ is the characteristic polynomial of $\;A\;$ ,and we know that its roots are exactly the eigenvalues of $\;A\;$ , so for a number $\;\lambda\;$ we get what I wrote in the first parraph above.

Answer 2

Both sides of the equation are column matrices ($n\times 1$ matrices), so they have no determinant defined (except for the trivial case when $n=1$).

Instead, what is argued here is that the fact that $B\cdot X=0$ (with $B=A-\lambda I$) for some $X\neq 0$ (not equal to the null vector). This implies that $B$ is a singular matrix (otherwise $X=0$ would be the only solution) and being singular it has to be $\det(B)=0$.

That's why $\lambda$ is an eigenvector of $A$ $\iff$ $\det(A-\lambda I)=0$.

Answer 3

If I understand the question correctly, you are asking:

Why $(A-\lambda I)X=0$ for a nontrivial $X$ implies $\det(A-\lambda I)=0?$

Here is why:

If there is a nontrivial $X$ such that $(A-\lambda I)X=0$, then X is an eigenvector of the matrix $(A-\lambda I)$ with the eigenvalue zero. If one of the eigenvalues is zero, the product of the eigenvalues is also zero.

Now, we use the fact that the determinant of every matrix is the product of the eigenvalues of that matrix (see here). Applying this statement to the matrix $(A-\lambda I)$, we have, $\det(A-\lambda I)=0.$

Note: The above statement about determinant of a matrix being the product of its eigenvalues is my favorite definition of determinant. There are several definitions of determinant, and it is not trivial to see how they are equivalent. In my opinion, this statement is really about what determinant is as opposed to how it is calculated or what its properties are. I also find it the most helpful in proofs.