Is there any straightforward intuition for why the trace equals the sum of the matrix eigenvalues?

I've looked through various mathematical proofs, as well as discussions in other threads such as Proof that the trace of a matrix is the sum of its eigenvalues

I've tried looking at matrices which are diagonalizable: $$\text{tr}(A)=\text{tr}(PDP^{-1})=\text{tr}(P^{-1}PD)=\text{tr}(ID)=\text{tr}(D)$$ But this isn't intuitive to me because the cyclic trace property itself is not intuitive (I can understand the cyclic property mathematically and know various proofs of it, but I can't intuitively feel it). More importantly, even if I found the cyclic property intuitive, this still wouldn't answer why the trace is equal to the sum of the matrix eigenvalues in some intuitive way (e.g. it doesn't provide some geometric intuition).

Maybe there isn't a simple intuitive explanation for the equality. That's fine too. But I'd like to know if someone knows of one. When I say 'simple', I mean easy to understand for someone with let's say no more than a semester of first-year linear algebra.

  • $\begingroup$ Which cyclic property ? $\endgroup$
    – lhf
    Jul 22, 2022 at 23:13
  • $\begingroup$ @lhf surely, that $\operatorname{tr}(AB) = \operatorname{tr}(BA)$? $\endgroup$
    – user7530
    Jul 22, 2022 at 23:19
  • $\begingroup$ Yes it's exactly what @user7530 says. I edited the question to make it clearer. $\endgroup$
    – Garp
    Jul 22, 2022 at 23:31
  • $\begingroup$ You don't need to assume that a matrix is diagonalizable. Just use the Jordan decomposition instead of the eigendecomposition. $\endgroup$ Jul 23, 2022 at 10:53

3 Answers 3


For symmetric matrices $M$, there is the following geometric picture: the trace is proportional to the average squared norm of all vectors on the unit sphere, with respect to inner product $M$:

$$\operatorname{tr}(M) \propto \int_{\|\mathbf{v}\|=1} \mathbf{v}^T M \mathbf{v}\,dA,$$ where the proportionality constant depends on the dimension of $M$. You can prove this formula by noticing that when $M = e_i e_j^T$ for $i\neq j$, the integral is zero by symmetry arguments; for general $M$, since the formula is linear in $M$, you can break $M$ up into a sum of matrices each containing exactly one nonzero.

Now for any rotation $R$, $\operatorname{tr}(R^TMR) = \operatorname{tr}(M)$, since the conjugation by $R$ simply "spins the unit sphere." So you might as well take $R$ to be the matrix diagonalizing $M$.


This mostly comes from the Dunford (also called Jordan-Chevalley) decomposition.

i.e you can write $A=D+N$ with $D$ diagonalisable and $N$ nilpotent.

Note: the so called normal Jordan form is a particular such decomposition where $D$ is diagonal and $N$ has all zero entries except maybe on the diagonal line above the central diagonal where entries can be either $0$ or $1$.

The trace being linear and nilpotent operator having trace zero, we have $\operatorname{Tr}(A)=\operatorname{Tr}(D)+\underbrace{\operatorname{Tr}(N)}_0=\sum eigenv$.

Note: for $D=P\Delta P^{-1}$ diagonalisable you proved yourself that trace $\operatorname{Tr}(D)=\operatorname{Tr}(\Delta)$ where $\Delta$ diagonal, and thus the sum of eigenvalues.

In $\mathbb C$ we can always triangularize a matrix because the characteristic polynomial is split. But a triangular matrix can also be written $T=D+U$ with $D$ diagonal and $U$ upper triangular with zeroes on the diagonal, and this $U$ is then nilpotent (all eigenvalues zero).

So everything is connected.

  • 1
    $\begingroup$ I think that $D$ is diagonalizable in the Jordan-Chevalley decomposition, not necessarily diagonal, right? So how do you know the trace of $D$ is the sum of the eigenvalues? We can't assume a priori that the trace is invariant under similarity. $\endgroup$ Jul 23, 2022 at 0:39
  • $\begingroup$ ok, but OP proved it himself (that trace is invariant by similarity), but I'll change the wording. $\endgroup$
    – zwim
    Jul 23, 2022 at 0:43
  • $\begingroup$ Sure but OP also said that was unintuitive and so was looking for a different approach. $\endgroup$ Jul 23, 2022 at 0:47

Consider the characteristic polynomial $\det(xI - A)$ for an $n\times n$ matrix $A$ with entries $(a_{ij})$. When computing the determinant using the usual algorithm expanding along the top row, say, every term in the sum uses every row and column exactly once. This is the permutation definition of determinant. In particular, if you've used $n-1$ factors from the diagonal then you must use the $n$th factor from the diagonal as well, since there are no more rows or columns left to choose. It follows that

$$\det(xI - A) = (x - a_{11})(x - a_{22})\cdots (x - a_{nn}) + O(x^{n-2}).$$

Factoring this polynomial over $\mathbb{C}$ gives $$\det(xI - A) = (x- \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)$$

where $\lambda_i$ are the eigenvalues, and so $$(x- \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n) = (x - a_{11})(x - a_{22})\cdots (x - a_{nn}) + O(x^{n-2}).$$

Now consider the $x^{n-1}$ term on both sides; you can ignore the $O(x^{n-2})$ term when you do so.

tl;dr Just remember that

i) The $x^{n-1}$ term of a polynomial is the sum of its zeros and

ii) It's good enough to only consider the diagonal of $xI - A$ when taking the $x^{n-1}$ term of $\det(x I - A)$.


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