Eigenvalue of Matrix: Determinant equals to the product of all its eigenvalue

Question

Question:

The solution of (a) let the $\lambda = 0$, l do not understand why. Isn't that $\lambda$ can only have the value which is the same as each eigenvalue of the matrix? why $\lambda$ can be substituted with $0$ ? Does $\lambda$ still have the meaning of eigenvalue?

Answer 1

For any square matrix $A$, we define $$f(x)=|xI-A|$$ also according to definition of eigenvalue$$Av=\lambda v\to (A-\lambda I)v=0$$this means that there is a linear dependence between the rows of $A-\lambda I$ (with the coefficients being the entries of $v$) therefore $|A-\lambda I|=0$. Conversely, if for some $\lambda$ we have $|A-\lambda I|=0$ then there exists some linear dependence as described before for which$$(A-\lambda I)v=0$$and $\lambda$ and $v$ become eigenvalue and eigenvector respectively. Therefore the equation $|A-\lambda I|=0$ yields to all eigenvalues. Equivalently (and fortunately!) all the roots of $f(\lambda)=0$ are the eigenvalues. This means that we can express $f(\lambda)$ as following$$f(\lambda)=|\lambda I-A|=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdots (\lambda-\lambda_n)$$by substituting $0$ we obtain $$|-A|=(-1)^n |A|=(-1)^n \lambda_1\lambda_2\cdots \lambda_n$$which leads to $$|A|= \lambda_1\lambda_2\cdots \lambda_n$$

Answer 2

Any matrix $A$ has a characteristic polynomial associated to it, call it $f_A(x)$. Yes, the eigenvalues of $A$ are defined to be the values of $x$ for which $f_A(x) = 0$. However, the characteristic polynomial is still just a polynomial, and it is completely indifferent to the existence of $A$ or its eigenvalues. I can plug in any number I want into $f_A(x)$. I can plug in $2$, $3.5$, $-1$, or anything else. The polynomial doesn't know where it came from, and there is no law that says "you can only insert eigenvalues here".

As an analogy, I can buy a burger from my favorite restaurant, and once I get home, I take out the buns and make a peanut butter sandwich with them. The buns have no clue they came from a burger, and they can be used for any purpose I want. They may taste best in a burger, but I don't have to use them in one.

Now, even though you can plug in anything you want into the polynomial, there is still a connection between $A$ and $f_A(x)$. The first one, that you are already aware of, is that $f_A(eigenvalue) = 0$. Now, the problem is asking you to prove another connection, namely that $f_A(0) = \lambda_1\lambda_2\lambda_3 = det(A)$.