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I am currently working on a question about proving the sum of eigenvalues and I have been searching for the solution from YouTube.

However, I do not understand why the teacher uses the diagonal method to show that the sum of eigenvalues equals the trace of matrix $A$. Doesn't the diagonal method only apply on $3 \times 3$ or less dimensional matrix?

Thank you so much.

Rócherz
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Www
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    What do you mean by "diagonal method" ? Never heard of it in this context... – DonAntonio Mar 03 '19 at 09:44
  • Also, you cannot prove a sum, you can prove a claim (which then becomes a theorem). What is the claim? – Hagen von Eitzen Mar 03 '19 at 09:45
  • Show that the Trace of a Matrix is the sum of its Eigenvalues. Is it better? – Www Mar 03 '19 at 09:47
  • In the video, the guy times all diagonal elements in that matrix. But the matrix is not a triangular matrix. How can he operates like that? – Www Mar 03 '19 at 09:50
  • @Ricky He's not doing that at all. He even remarks that besides the elements on the main diagonal there are all the other entries off that diagonal which must also be taken into account. He just wrote that determinant in that way to try to make his point clearer. – DonAntonio Mar 03 '19 at 09:59
  • Oh right. I think I made some stupid mistakes. Thank You so much. – Www Mar 03 '19 at 10:05

2 Answers2

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If we already know that the determinant of a square matrix is , up to sign, the sum of some products of the matrix's entry, all of which are formed by taking exactly one entry from each row and eactly one entry from each column, then it is not that hard: observe the coefficient of $\;\lambda^{n-1}\;$ in the char. polynomial $\;p_A(\lambda):=| \lambda I - A | \;$ of the matrix $\;A\;$ .

Observe what my definition of the char. pol. is: in the video they use $\; |A- \lambda I|\;$ , which always yields the "problem" of having either $\;1\;$ or $\;-1\;$ as the main coefficient of the pol., and this is avoided with the definition I use.

In order to get the coefficient of $\;\lambda^{n-1}\;$ in that determinant's expansion, we must first observe that we can write the char. pol. as

$$p_A(\lambda)=\prod_{k=1}^n(\lambda-\lambda_k)=(\lambda-\lambda_1)(\lambda-\lambda_2)\cdot\ldots\cdot(\lambda-\lambda_n)$$

where the $\;\lambda_k\;$ are the matrix's eigenvalues (there may well be repeated values. We don't claim all the $\;\lambda_k\,'$ s are different, of course) . Warning: the above expression happens in a field that contains all the eigenvalues of the given matrix, so it may not be the field over which our matrix is defined! If working over the rational or real fields, the above expression is always true when working over the complex $\;\Bbb C\;$ .

Now, in the above expression evaluate the coefficient of $\;\lambda^{n-1}\;$ : you must take $\;n-1\;$ times $\;\lambda\;$ and once some $\;\lambda_i\;$ . At the end you get $\;-(\lambda_1+\lambda_2+\ldots+\lambda_n)\lambda^{n-1}\;$ . Observe the sign: it is always $\;-$Tr.$\,A\;$ with my definition. With the one used in the video it is always Tr.$\,A\,$ . There you go...

DonAntonio
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There is an known easy proof for that. Indeed, let A be an $n \times n$ matrix, due Schur decomposition, there exists an orthogonal matrix $Q$ such that $ A = Q T Q^{H}$, with the eigenvalues of $A$, say $\lambda_1,\lambda_2,\cdots, \lambda_n$, in the diagonal of $T$. Since the product can commute in the argument of the trace without changing its value, i.e., $\text{tr}(X Y) = \text{tr}(Y X) $, for all $X, Y$, then $\text{tr}(A)= \text{tr} ( Q T Q^{H}) = \text{tr} ( Q^{H} Q T ) = \text{tr} ( T ) = \sum^{n}_{i=1} \lambda_i$ the result follows.

R. W. Prado
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