Define the trace of a matrix with entries in $\mathbb C$ to be the sum of its eigenvalues, counted with multiplicity. It is a standard (but I think extremely surprising) fact that this is the sum of the elements along the diagonal. One proof of this is as follows:

Define $Tr'(A)$ to be the sum of the entries along the diagonal of $A$. If $A$ is an $n\times m$ matrix and $B$ and $m\times n$ matrix, we have $$Tr'(AB)=\sum_{i=1}^n\sum_{j=1}^m a_{ij}b_{ji}=\sum_{j=1}^m\sum_{i=1}^n b_{ji}a_{ij}=Tr'(BA)$$ and thus for any invertible matrix $P$ we have $Tr'(PAP^{-1})=Tr'(P^{-1}PA)=Tr'(A)$, i.e. $Tr'$ is independent of basis. Thus it suffices to note that when $A$ is in Jordan Normal Form, $Tr'(A)$ is the trace of $A$.

I find this proof pretty unsatisfying, mainly because I don't see any reason I would expect the sum along the diagonal to be basis-independent. Is there a more illuminating proof of this out there?

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    $\begingroup$ This isn't a strictly rigorous proof, but here's an intuitive argument. For some matrix $M$, $\frac{d}{dt}$ of $\det(I+tM)$ at $t=0$ is the trace of $M$. If you imagine deforming the unit hypercube a bit by the linear transformation $M$, the infinitesimal change in volume can be seen to be the sum of the infinitesimal change of each unit vector in its own direction, i.e. the sum of the diagonal elements of $M$ times $dt$. $\endgroup$
    – Malper
    Oct 24, 2013 at 0:23
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    $\begingroup$ @Stefan: that works if you are thinking in terms of matrices. But one considers matrices initially because they are a way to represent a linear operator on a finite-dimensional space. As such, one would want a basis-independent definition, which can be achieved by defining the trace as the sum of the eigenvalues (counting multiplicity). $\endgroup$ Oct 24, 2013 at 0:31
  • $\begingroup$ @StefanSmith In what way is my definition of trace nonstandard? $\endgroup$ Oct 24, 2013 at 1:16
  • $\begingroup$ @AlexBecker : if you pick up a book on linear algebra, and look up "trace" of a matrix, the definition will almost certainly be the sum of the diagonal elements. This is the definition that I have always seen, and is obviously simpler than any definition that involves eigenvalues (you don't even need to know what an eigenvalue is). Your definition is also correct, of course, because you get the same number. The advantage of your definition is that you can apply it to any linear map from a finite-dimensional vector space to itself... $\endgroup$ Oct 24, 2013 at 2:52
  • $\begingroup$ "extremely surprising fact" ... well, not for a diagonal matrix, no? :-) $\endgroup$
    – leonbloy
    Oct 24, 2013 at 2:52

4 Answers 4


Let us start with another basis-independent yet more tractable (as it does not require the characteristic polynomial to split) definition of the trace. We will check in the end that it coincides with your definition, and with the sum of the diagonal coefficients with respect to any basis.

Let $V$ be an $n$-dimensional vector space over a field $F$. And let $L(V)$ be the algebra of $F$-linear maps from $V$ to $V$.

Note that we have a canonical isomorphism

$$ L(V)\simeq V\otimes V^* $$

via $v\otimes w^* \simeq w^*(\cdot)v$. In other words, $L(V)$ is a natural incarnation of the tensor product of $V$ with its dual $V^*$, with rank-one operators as elementary tensors.

Observe that the bilinear map $(v,w^*)\longmapsto w^*(v)$ factors uniquely through the tensor product.

That's the trace, which is therefore characterized by $$ \mathrm{tr}:V\otimes V^*\longrightarrow F\qquad \mathrm{tr}(v\otimes w^*)=w^*(v). $$

Now choose any basis $\{e_i\}$ for $V$ and denote its dual basis by $\{e_i^*\}$. We have $\mathrm{tr}(e_i\otimes e_j^*)=\delta_{ij}$. Therefore, for every $x=\sum x_{ij}e_i\otimes e_j^*\in L(V)$, we have $$ \mathrm{tr} (x)=\sum_{i=1}^n x_{ii}. $$

Conclusion When given a matrix $x$ in $M_n(F)$, think of it as an operator in $L(F^n)$ via the canonical basis of $F^n$. Its trace is then defined canonically as above. And whatever basis you choose for $F^n$, the sum of the diagonal coefficients will be equal to $\mathrm{tr}(x)$. In particular, it is also equal to the sum of the eigenvalues counted with multiplicities when the characteristic polynomial of $x$ splits.

Note It also helps understand why $\mathrm{tr} (ab)=\mathrm{tr}(ba)$, beyond the calculation you mentioned. Indeed $$ \mathrm{tr}((v_1\otimes w_1^*)(v_2\otimes w_2^*))=w_1^*(v_2)\mathrm{tr}(v_1\otimes w_2^*)=w_1^*(v_2)w_2^*(v_1) $$ $$ =w_2^*(v_1)w_1^*(v_2)=w_2^*(v_1)\mathrm{tr}(v_2\otimes w_1^*)=\mathrm{tr}((v_2\otimes w_2^*)(v_1\otimes w_1^*)) $$

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    $\begingroup$ The OP has chosen to define the trace of a matrix as the sum of its eigenvalues. Your alternative definition may be better than the OP's, but it doesn't answer their question. $\endgroup$
    – Rob Arthan
    Oct 24, 2013 at 2:46
  • $\begingroup$ @RobArthan You can define $f'(x)$ as $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$. But if you want to understand the product rule, it is helpful to note that this definition is equivalent to $f(x+h)=f(x)+hf'(x)+o(h)$. $\endgroup$
    – Julien
    Oct 24, 2013 at 2:52
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    $\begingroup$ @juliien: duh??? $\endgroup$
    – Rob Arthan
    Oct 24, 2013 at 3:01
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    $\begingroup$ I think the connection here to eigenvalues is obvious enough; the only eigenvector of $w^*(\cdot)v$ is $v$ with eigenvalue $w^*(v)$. $\endgroup$ Oct 24, 2013 at 4:26
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    $\begingroup$ Good points. :+) $\endgroup$
    – Mikasa
    Oct 26, 2013 at 18:35

Are you surprised that if a polynomial $f(x) = x^n + a_{n-1}x^{n-1} + \ldots$ of degree $n$ has roots $r_1, \ldots r_n$, then $a_{n-1} = - (r_1 + \ldots + r_n)$? Now think about how the coefficients of $x^{n-1}$ arise in the characteristic polynomial of a matrix $M$.

(And the characteristic polynomial is basis-independent, because the eigenspaces and corresponding eigenvalues are basis-independent and determine the characteristic polynomial).

  • $\begingroup$ I like this. I'll have to think on it a little more before I'm sure it makes intuitive sense, but it's a good start. $\endgroup$ Oct 24, 2013 at 1:17
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    $\begingroup$ +1 BTW, as a subproduct of this way, you get the result that the product of the eigenvalues is the determinant (looking at the independent term) $\endgroup$
    – leonbloy
    Oct 24, 2013 at 2:53
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    $\begingroup$ I don't follow. What is your definition of characteristic polynomial $p_A(X)$? If it is $\det(xI-A)$, then how to explain, in an intuitive way, that this is basis-independent? This seems to me just a tradeoff between using $\mathrm{Tr}'(PAP^{-1})=\mathrm{A}$ and using $\det(xI-A)=\det(xI-PAP^{-1})$. If $p_A(X)$ is defined as $\prod_i (x-\lambda_i)$, where the multipliplicity of each eigenvalue is determined by the generalised eigenspaces, then again, how do you explain that the coefficient of $x^{n-1}$ equal to $-\mathrm{Tr}'(A)$? $\endgroup$
    – user1551
    Oct 24, 2013 at 9:45
  • $\begingroup$ The notions of eigenvalue and eigenvector of $A$ are basis-independent. The coefficients of the characteristic polynomial $p_A$ (defined as $\det(xI- A)$) depend on the basis but $p_A(x)$ vanishes iff $x$ is an eigenvalue, so the set of roots of the characteristic polynomial is basis-independent. $\endgroup$
    – Rob Arthan
    Oct 25, 2013 at 2:25
  • $\begingroup$ Hmm, but when we calculate the trace, we need the algebraic multiplicities of the eigenvalues as well. How do we know that the algebraic multiplicities are basis-independent? If we resort to generalised eigenspaces, how is such an argument different from one using Jordan nromal form? $\endgroup$
    – user1551
    Oct 25, 2013 at 10:50

If you work over an algebraically closed field (i.e. if you are over ${\mathbb R}$, just think it all over ${\mathbb C}$), then you can write your matrix in a triangular form.

In that case, you have that the elements in the diagonal are your eigenvalues. By the computation you showed the sum of the elements in the diagonal is independent of the basis chosen ($tr(PAP^{-1})=tr(AP^{-1}P)=tr(A)$), equality follows.

The rest of the answer has already been written in this posting.


The general definition is that the trace is the sum of the diagonal elements in any orthonormal basis, regardless of the choice of a specific orthonormal basis. If you take the eigenbasis of $A$ as your orthonormal basis, then the matrix is represented by a diagonal matrix containing the eigenvalues, hence their sum is the trace. If you take the canonical unit vectors, to be the basis, then the trace is the sum of the diagonal elements.

Seeing that they both equal the same is not that hard.

Let $A \in \mathbb{C}^{n \times n}$ be a matrix. Then, by the spectral theorem, we have an orthonormal set of vectors $\{v_i\}_i$ satisfying

$$A = \sum_i \lambda_i v_i v_i^*$$ where $(\lambda_i, v_i)$ is the i-th eigenpair.

Then, we have the equality

$$\text{Tr}(A) = \sum_i \lambda_i = \sum_i v_i^* A v_i$$

Now let, $\{w_i\}_i$ be any collection of orthonormal vectors, we want to show $$\text{Tr}(A) = \sum_i w_i^* A w_i$$

To this end we represent $v_i$ in the basis $w_i$, i.e.

$$v_i = \sum_j \langle v_i, w_j\rangle w_j$$

Then, by plugging this in the sum

$$\sum_i v_i^* A v_i = \sum_{ij} \langle v_i, w_j \rangle \langle w_j, v_i\rangle w_j^* A w_j$$

We note that

$$\sum_i \langle v_i, w_j\rangle \langle w_j, v_i\rangle = \sum_i |\langle v_i, w_j \rangle|^2 = |v_i|^2 = 1$$

where the first equality follows as the two inner products are conjugated, and the second because $w_j$ is an orthonormal basis, i.e. the base transformation is an isometry, hence the length in the base $w_i$ is the same as the base in the base $v_i$.

If you set $w_i = u_i$, i.e. the $i$-th canonical unit vector, it's not hard to see that we have $$u_i^* A u_i = a_{ii}$$


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