Why does the sum of eigenvalues equal to trace in terms of linear transformations? While studying eigenvectors, I was confronted with two statements:

*

*The product of the eigenvalues of some matrix $A$ is equal to the determinant of $A$


*The trace of $A$ is equal to the sum of its eigenvalues
The thing is I am trying to understand every topic I learn in terms of linear transformations. For example, the first statement made sense to me because since the determinant is how much area scaled after a linear transformation and we are stretching $2$ vectors by their eigenvalues therefore determinant equals the product of eigenvalues.
But when I try to understand the second statement I can not relate the trace of a matrix and its eigenvalues because I can't also understand what the trace of a matrix tells us about a linear transformation.
I would love it if you can give me an intuitive explanation or let me know if there are topics that I haven't studied that prevent me from understanding this.
 A: Let $F$ be an algebraicall y closed field. Let $C$ and $D$ be $n \times n$ matrices over $F$. Then,by multilinearity of the determinant function, $$\det (C+D)=\sum_{i=0}^n \text{ all determinants formed by taking n-i rows from $C$ & i rows from $D$  }$$ Let $A=\lambda I, D=-A$. We deine a principal minor of order $i$ of a square matrix $M$ ito be the determinant of an $i \times i$ sub-matrix of $M$ formed from the same $i$ rows and columns of $M$. Then $$\det (\lambda I-A)=\lambda ^n+\sum_{i=1}^n\beta_i\lambda^{n-i}$$ $$\text { where } \beta_i=(-1)^i\sum \text {  all principal minors of order $i$ of $A$}. $$ In particular, $\beta_1=- \text {trace}(A)$ and $\beta_n=(-1)^n\det(A)$.   Since $\det(\lambda I-A)$ is a polynomial in $\lambda$ with roots $\lambda_1,...,\lambda_n,$ $$\det(\lambda I-A)=\prod_{i=1}^n(\lambda-\lambda_i)$$ $$=\lambda^n+\sum_{i=1}^n\gamma_i\lambda^{n-i}$$ $$\text { where }\gamma_i=(-1)^i\sum \text {all products of $i$ of $\lambda_1,...,\lambda_n$}.$$ In particularthe trace of $A$ is the sum of the eigenvalues and the determinant of $A$ is the product of the eigenvalues.
A: You can think of the trace geometrically as the derivative of the determinant at the identity. What this means is that for a fixed matrix $A$ and for small $\varepsilon > 0$ we have
$$\det(I + \varepsilon A) = 1 + \varepsilon \text{ tr}(A) + O(\varepsilon^2).$$
This isn't hard to prove directly from the Leibniz formula for the determinant. On the other hand, if $A$ has eigenvalues $\lambda_i$, then $I + \varepsilon A$ has eigenvalues $1 + \varepsilon \lambda_i$, and the product of these is
$$\det(I + \varepsilon A) = \prod_{i=1}^n (1 + \varepsilon \lambda_i) = 1 + \varepsilon \sum_{i=1}^n \lambda_i + O(\varepsilon^2).$$
So we see that $\text{tr}(A) = \sum_{i=1}^n \lambda_i$ is the sum of the eigenvalues as desired. In other words, the fact that the trace is the sum of the eigenvalues is the derivative of the fact that the determinant is the product of the eigenvalues.
