Proof that the trace of a matrix is equal to the sum of its eigenvalues based on exercise 4.1.11 from the book “Linear Algebra Done Wrong” by Sergei Treil.

Step 1: Compute the coefficient of $ \lambda^{n-1} $

The characteristic polynomial of a matrix $ A $ is given by $ \det(A - \lambda I) $. The eigenvalues $ \lambda_1, \lambda_2, \dots, \lambda_n $ are the roots of this characteristic polynomial, so we can factor it as:

$$ \det(A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \dots (\lambda_n - \lambda) $$

We need to compute the coefficient of $ \lambda^{n-1} $. Expanding this product:

$$ (\lambda_1 - \lambda)(\lambda_2 - \lambda) \dots (\lambda_n - \lambda) = (-1)^n \lambda^n + (-1)^{n-1} (\lambda_1 + \lambda_2 + \dots + \lambda_n) \lambda^{n-1} + \dots $$

The coefficient of $ \lambda^{n-1} $ is $ (-1)^{n-1} (\lambda_1 + \lambda_2 + \dots + \lambda_n) $, which is just the sum of the eigenvalues multiplied by $ (-1)^{n-1} $.

Step 2: Show that $ \det(A - \lambda I) $ can be written in terms of the diagonal entries

For a matrix $ A = [a_{i,j}] $, we can expand $ \det(A - \lambda I) $ as follows:

$$ \det(A - \lambda I) = (a_{1,1} - \lambda)(a_{2,2} - \lambda) \dots (a_{n,n} - \lambda) + o(\lambda^{n-1}) $$

Here, $ o(\lambda^{n-1}) $ is a polynomial of degree at most $ n-2 $.

To show this is true, we can use the cofactor expansion of the determinant. We’ll use the following construction: suppose we start by crossing out the first row and column of the matrix $ A - \lambda I $. The important thing to notice is that the first term of this expansion is something of the form

$$ (a_{1,1} - \lambda)\det(\left( A - \lambda I \right)_{1,1}) $$

where $ \left( A - \lambda I \right)_{1,1} $ is the submatrix obtained by deleting the first row and column of $ A - \lambda I $. However, the other terms in this expansion involve crossing out the row $ 1 $ and column $ j $ for $ j \neq 1 $, which means that we will eliminate $ 2 $ different entries containing $ \lambda $. Therefore, the remaining terms result in a polynomial of at most degree $ n-2 $ (each coefficient does not contain $ \lambda $, and the minor determinants have degree at most $ n - 2 $).

We can repeat this procedure for row $ 2 $, $ 3 $, $ \dots $, and $ n $, which leaves us with the desired result.

Step 3: Compare coefficients of $ \lambda^{n-1} $

Now, we compare the coefficients of $ \lambda^{n-1} $ in both expressions.

  • From Step 1, we saw that the coefficient of $ \lambda^{n-1} $ is $ (-1)^{n-1} (\lambda_1 + \lambda_2 + \dots + \lambda_n) $, i.e., the sum of the eigenvalues.
  • From Step 2, expanding $ (a_{1,1} - \lambda)(a_{2,2} - \lambda) \dots (a_{n,n} - \lambda) $, we find that the coefficient of $ \lambda^{n-1} $ is $ (-1)^{n-1} (a_{1,1} + a_{2,2} + \dots + a_{n,n}) $, i.e., the trace of the matrix.

Since these two coefficients must be equal, we conclude that:

$$ \text{Tr}(A) = a_{1,1} + a_{2,2} + \dots + a_{n,n} = \lambda_1 + \lambda_2 + \dots + \lambda_n $$

Thus, the trace of the matrix equals the sum of the eigenvalues.

This completes the proof. $\blacksquare$