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So I was reading up on the Wikipedia page on eigenvalues and eigenvectors (which I fondly call eigencrap ;) ) and I was confused by one paragraph in particular.

Let $A$ be an arbitrary $n$ by $n$ matrix of complex numbers with eigenvalues $\lambda_1,\lambda_2, \ldots, \lambda_n$. Each eigenvalue appears $\mu_A(\lambda_i)$ times in this list, where $\mu_A(\lambda_i)$ is the eigenvalue's algebraic multiplicity. The following are the properties of this matrix and its eigenvalues:

  • The trace of $A$, defined as the sum of its diagonal elements, is also the sum of all eigenvalues, $$ \text{tr}(A)=\sum_{i=1}^{n}A_{i,i}=\sum_{i=1}^{n}\lambda_i=\lambda_1+\lambda_2+\ldots+\lambda_n. $$
  • The determinant of $A$ is the product of all its eigenvalues, $$ \det(A)=\prod_{i=1}^{n}\lambda_i=\lambda_1\lambda_2\cdots\lambda_n. $$ First off, $\mathbb{R}\subset\mathbb{C}$, correct? Therefore the properties listed should hold for any $n\times n$ square matrix -- namely that $\operatorname{tr}(A)$, the sum of the diagonal elements, is equal to the sum of all eigenvalues, and that $\operatorname{det}(A)$ is equal to the product of such.

This does not strike me as intuitively true, why should this be the case? Is there a condition missing from the theorem -- perhaps that the matrix must be orthogonally diagonalized, or symmetric?

Just a thought, but is there a relation to Vieta's theorems? The sum/product identities seem indicative of such.

Any thoughts, intuitions, and explanations are appreciated, thank you!

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  • $\begingroup$ For diagonal matrices, these properties are obvious consequences of the Vieta's formulas. Then diagonalization preserves the determinant $|PDP^{-1}|=|P||D||P|^{-1}$ and the trace $\text{Tr}(PDP^{-1})=\text{Tr}(DP^{-1}P)=\text{Tr}(D)$. $\endgroup$ – Yves Daoust Mar 15 '18 at 14:30
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    $\begingroup$ The Eigencomponents are not crap. $\endgroup$ – Yves Daoust Mar 15 '18 at 14:35
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Another well known property of the Trace operator is that it is invariant over commutation of matrices, that means: $$ Tr(AB)=Tr(BA) $$ So, if $A$ is diagonalized by: $$ A=V D V^{-1} $$ With $D$ a diagonal matrix, then: $$ Tr(A) = Tr(V D V^{-1})=Tr(V V^{-1} D) = Tr(D)= \lambda_1+ \cdots +\lambda_n $$ Where the invariance over commutation was used.

Regarding the second property, another well known result is Binet's theorem, that states: $$ Det(AB)= Det(A)Det(B) $$ Thus, upon diagonalization: $$ Det(A)= Det(V D V^{-1}) = Det(V)Det(D)Det(V^{-1}) $$ And of course this also implies: $$ Det(I)= Det(V V^{-1}) = Det(V)Det(V^{-1}) =1 $$ Thus: $$ Det(A)= Det(D) = \lambda_1 \times \cdots \times\lambda_n $$

Also, yes, there are some relation to Viète's theorems (or Girard Relations), that come from the characteristic polynomial: $$ p(\lambda) = Det(A-\lambda I) $$ Since for instance this implies: $$ p(0)=Det(A) $$

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(Small correction: You mean $\mathbb{R} \subset \mathbb{C}$, yes.) First, yes, these properties of trace and determinant are true, and they come from the fact that trace and determinant are independent of the chosen basis of the vector space. So you can always choose a basis transformation which transforms the matrix in its Jordan normal form (it need not be diagonalizable, but every matrix has a Jordan normal form), which might look e.g. like this: $$ \left( \begin{array}{ccc} \lambda_1 & 1 & 0 \\ 0 &\lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{array} \right)$$ You see that the eigenvalues are on the diagonal, and no matter how many "1"s you have off-diagonal, you will always have that the trace is the sum of eigenvalues and the determinant the product.

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It is not true that $\mathbb{R} \in \mathbb{C}$, but it is true that $\mathbb{R} \subseteq \mathbb{C}$, which is what I think you meant.

The theorem is always true; there are no extra requirements. It is true for matrices where all of the entries are real numbers, as long as you include the eigenvalues which are complex.

There is in fact a relation to Vieta's formulae. The eigenvalues of $A$ are the roots (counting multiplicity) of the polynomial $$ \det(xI - A). $$

If you think about it for a bit, and use the definition of the determinant as the sums of products of entries of the matrix where each product contains exactly one element from each row and one element from each column, then you can convince yourself that the coefficient of $x^{n-1}$ in $\det(xI - A)$ is just the coefficient of $x^{n-1}$ in the product of the diagonal terms of $xI - A$, and that this is equal to the minus the trace of $A$.

(i.e. We use the definition of the determinant where $$ \det(B) = \sum_{\sigma \in \operatorname{Sym}(n)} \prod_{i=1}^{n} B_{i,\sigma(i)}. $$ We then notice that to get a term containing $x^{n-1}$ in $\det(xI - A)$, the product in the sum has to include $(n - 1)$ of the diagonal terms of $xI - A$, and so it also includes the last remaining diagonal term. It follows that the only term in the sum that contributes a $x^{n-1}$ is the one which is the product of the diagonal elements of $xI - A$, and so the coefficient of $x^{n-1}$ in $\det(xI - A)$ is the same as the coefficient of $x^{n-1}$ in $$ \prod_{i=1}^{n} (x - A_{i, i}), $$ which we can see is equal to $$ -\sum_{i=1}^{n} A_{i,i} $$ (also by Vieta's formulae), and this is equal to $-\operatorname{Tr}(A)$.)

Vieta's formulae also tell us that the product of the roots of $\det(xI - A)$ is $(-1)^n$ times the constant term of $\det(xI - A)$, which is equal to $(-1)^n \det(0\cdot I - A) = (-1)^n \det(-A) = \det(A)$.

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