Characterizing the trace of a matrix

I have been studying for my finals lately, and so I decided to put together a proof of a nice exercise I found in some book. The trace function, given by \(tr : \mathbb{K}^{n \times n} \to \mathbb{K}\), is defined as

$$tr(A) = \sum_{i=1}^n A_{ii}$$

First of all, the proof of additivity

$$\begin{equation} \begin{split} tr(A + B) &= \sum_{i=1}^n (A+B)_{ii} \\ &= \sum_{i=1}^n (A)_{ii} + (B)_{ii} \\ &= \sum_{i=1}^n (A)_{ii} + \sum_{i=1}^n (B)_{ii} \\ &= tr(A) + tr(B) \end{split} \end{equation}$$

Afterwards, the proof of homogeneity

$$\begin{equation} \begin{split} tr(\lambda A) &= \sum_{i=1}^n (\lambda A)_{ii}\\ &= \sum_{i=1}^n \lambda A_{ii}\\ &= \lambda \sum_{i=1}^n A_{ii}\\ &= \lambda tr(A) \end{split} \end{equation}$$

Hence, \(tr\) is a linear transform from the vector space \(\mathbb{K}^{n \times n}\) into \(\mathbb{K}\). The cool thing about the trace is that it has many more interesting properties which are not difficult to prove. First of all, that it is invariant under transposition

$$\begin{equation} \begin{split} tr(A^T) &= \sum_{i=1}^n (A^T)_{ii}\\ &= \sum_{i=1}^n A_{ii}\\ &= tr(A) \end{split} \end{equation}$$

And that it doesn’t change when a product commutes

$$\begin{equation} \begin{split} tr(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &= \sum_{i=1}^n \sum_{k=1}^n A_{ik} B_{ki}\\ &= \sum_{k=1}^n \sum_{i=1}^n B_{ki} A_{ik}\\ &= \sum_{k=1}^n (BA)_{kk}\\ &= tr(BA) \end{split} \end{equation}$$

Therefore, we get a pretty non-intuitive property of the trace, we have that

$$tr(ABC) = tr(BCA) = tr(CAB)$$

Which comes from applying the previous property to different parenthesizations of the matrix product (notice that this is doable for any number of matrices). Going further, suppose that \(P \in \mathbb{K}^{n \times n}\) is an invertible matrix, then

$$tr(PAP^{-1}) = tr(AP^{-1}P) = tr(A)$$

Meaning that if two matrices \(A\) and \(B\) are similar, their traces are the same!. Additionally, writing \(A\) in any given basis will not change its trace. Last of all, the reason I decided to write this post:

Suppose that we have a linear functional \(f : \mathbb{K}^{n \times n} \to \mathbb{K}\) such that \(\forall A, B : f(AB) = f(BA)\), then \(\exists \lambda \in \mathbb{K} / f(A) = \lambda tr(A)\). That is, linearity and the product property completely determine the trace function, up to a constant factor. The proof is fairly easy. Let

$$\beta = {E^{ij}}_{1 \leq i, j \leq n}$$

be the canonical ordered basis for \(\mathbb{K}^{n \times n}\). Notice that:

$$\begin{equation} \begin{split} E^{ij} E^{kl} &= \sum_{m=1}^n E^{ij}_{i'm} E^{kl}_{mj'}\\ &= \sum_{m=1}^n \delta_{ii'} \delta_{jm} \delta_{km} \delta_{lj'}\\ &= \delta_{ii'} \delta_{lj'} \delta_{jk} \end{split} \end{equation}$$

Since \(f\) is a linear functional, we can write it as \(f(A) = \sum_{i, j = 1}^n \alpha_{ij} \phi_{ij}(A)\), where \(\phi_{ij}\) are the corresponding functions from the dual basis, meaning that \(\phi_{ij}(E^{i’j'}) = \delta_{ii'} \delta_{jj'}\). Thus, we get that

$$f(E^{ij} E^{lm}) = \alpha_{im} \delta_{jl}$$

And

$$f(E^{lm} E^{ij}) = \alpha_{mi} \delta_{lj}$$

The additional condition that \(f(AB) = f(BA)\) means that \(\alpha_{im} \delta_{jl} = \alpha_{mi} \delta_{lj} \forall i, j, m, l \in {1, …, n}\). In particular, if we take \(i = m, j = l\), this turns into \(\alpha_{ii} \delta_{jj} = \alpha_{ii} \delta_{jj}\), which implies that \(\delta_{ii} = \delta_{jj} \forall i, j \in {1, …, n}\), we can call this value \(\lambda\). Finally, taking \(l \neq j, m = i\), we get that \(\alpha_{lj} = 0 \forall l, j \in {1, …, n}, l \neq j\). This completely determines every single one of the \(\alpha\)s. Hence:

$$f(A) = \lambda \sum_{i = 1}^n \phi_{ii}(A) = \lambda \sum_{i = 1}^n A_{ii} = \lambda tr(A)$$

By this point, we have pretty much characterized the trace function, in the sense that we know that any function that is a linear transform and does not change when the matrix product order changes, then it is a scalar multiple of the trace. There is only one last important property to uniquely determine the trace. Given \(f\) with the previous properties, \(f(I) = n \iff f = tr\):

$$f(I) = \lambda tr(I) = \lambda n = n$$

Which happens if and only if \(\lambda = 1\).I have been studying for my finals lately, and so I decided to put together a proof of a nice exercise I found in some book. The trace function, given by \(tr : \mathbb{K}^{n \times n} \to \mathbb{K}\), is defined as

$$tr(A) = \sum_{i=1}^n A_{ii}$$

First of all, the proof of additivity

$$\begin{equation} \begin{split} tr(A + B) &= \sum_{i=1}^n (A+B)_{ii} \\ &= \sum_{i=1}^n (A)_{ii} + (B)_{ii} \\ &= \sum_{i=1}^n (A)_{ii} + \sum_{i=1}^n (B)_{ii} \\ &= tr(A) + tr(B) \end{split} \end{equation}$$

Afterwards, the proof of homogeneity

$$\begin{equation} \begin{split} tr(\lambda A) &= \sum_{i=1}^n (\lambda A)_{ii}\\ &= \sum_{i=1}^n \lambda A_{ii}\\ &= \lambda \sum_{i=1}^n A_{ii}\\ &= \lambda tr(A) \end{split} \end{equation}$$

Hence, \(tr\) is a linear transform from the vector space \(\mathbb{K}^{n \times n}\) into \(\mathbb{K}\). The cool thing about the trace is that it has many more interesting properties which are not difficult to prove. First of all, that it is invariant under transposition

$$\begin{equation} \begin{split} tr(A^T) &= \sum_{i=1}^n (A^T)_{ii}\\ &= \sum_{i=1}^n A_{ii}\\ &= tr(A) \end{split} \end{equation}$$

And that it doesn’t change when a product commutes

$$\begin{equation} \begin{split} tr(AB) &= \sum_{i=1}^n (AB)_{ii}\\ &= \sum_{i=1}^n \sum_{k=1}^n A_{ik} B_{ki}\\ &= \sum_{k=1}^n \sum_{i=1}^n B_{ki} A_{ik}\\ &= \sum_{k=1}^n (BA)_{kk}\\ &= tr(BA) \end{split} \end{equation}$$

Therefore, we get a pretty non-intuitive property of the trace, we have that

$$tr(ABC) = tr(BCA) = tr(CAB)$$

Which comes from applying the previous property to different parenthesizations of the matrix product (notice that this is doable for any number of matrices). Going further, suppose that \(P \in \mathbb{K}^{n \times n}\) is an invertible matrix, then

$$tr(PAP^{-1}) = tr(AP^{-1}P) = tr(A)$$

Meaning that if two matrices \(A\) and \(B\) are similar, their traces are the same!. Additionally, writing \(A\) in any given basis will not change its trace. Last of all, the reason I decided to write this post:

Suppose that we have a linear functional \(f : \mathbb{K}^{n \times n} \to \mathbb{K}\) such that \(\forall A, B : f(AB) = f(BA)\), then \(\exists \lambda \in \mathbb{K} / f(A) = \lambda tr(A)\). That is, linearity and the product property completely determine the trace function, up to a constant factor. The proof is fairly easy. Let

$$\beta = {E^{ij}}_{1 \leq i, j \leq n}$$

be the canonical ordered basis for \(\mathbb{K}^{n \times n}\). Notice that:

$$\begin{equation} \begin{split} E^{ij} E^{kl} &= \sum_{m=1}^n E^{ij}_{i'm} E^{kl}_{mj'}\\ &= \sum_{m=1}^n \delta_{ii'} \delta_{jm} \delta_{km} \delta_{lj'}\\ &= \delta_{ii'} \delta_{lj'} \delta_{jk} \end{split} \end{equation}$$

Since \(f\) is a linear functional, we can write it as \(f(A) = \sum_{i, j = 1}^n \alpha_{ij} \phi_{ij}(A)\), where \(\phi_{ij}\) are the corresponding functions from the dual basis, meaning that \(\phi_{ij}(E^{i’j'}) = \delta_{ii'} \delta_{jj'}\). Thus, we get that

$$f(E^{ij} E^{lm}) = \alpha_{im} \delta_{jl}$$

And

$$f(E^{lm} E^{ij}) = \alpha_{mi} \delta_{lj}$$

The additional condition that \(f(AB) = f(BA)\) means that \(\alpha_{im} \delta_{jl} = \alpha_{mi} \delta_{lj} \forall i, j, m, l \in {1, …, n}\). In particular, if we take \(i = m, j = l\), this turns into \(\alpha_{ii} \delta_{jj} = \alpha_{ii} \delta_{jj}\), which implies that \(\delta_{ii} = \delta_{jj} \forall i, j \in {1, …, n}\), we can call this value \(\lambda\). Finally, taking \(l \neq j, m = i\), we get that \(\alpha_{lj} = 0 \forall l, j \in {1, …, n}, l \neq j\). This completely determines every single one of the \(\alpha\)s. Hence:

$$f(A) = \lambda \sum_{i = 1}^n \phi_{ii}(A) = \lambda \sum_{i = 1}^n A_{ii} = \lambda tr(A)$$

By this point, we have pretty much characterized the trace function, in the sense that we know that any function that is a linear transform and does not change when the matrix product order changes, then it is a scalar multiple of the trace. There is only one last important property to uniquely determine the trace. Given \(f\) with the previous properties, \(f(I) = n \iff f = tr\):

$$f(I) = \lambda tr(I) = \lambda n = n$$

Which happens if and only if \(\lambda = 1\).