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The trace of a matrix: Courses and corrected exercises

This article aims to present the trace of a matrix through its definition, examples and corrected exercises. It is good to have the basic knowledge of what a square matrix is.

Definition

Let A be a square matrix of size n defined by its coefficients (aij). The trace of A, denoted Tr(A) is the sum of the diagonal coefficients of this matrix. With the help of a sum, it is therefore written:

Tr(A) = \sum_{k=1}^n a_{ii}

It is therefore an application of

M_n(\mathbb{K}) \mapsto \mathbb{K}

where K is a field.

Examples

Take the following matrix:

A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6\\ 7 & 8 & 9 \end{pmatrix}

The trace of A is then

1 + 5 + 9 = 15

Manufacturing

With the help of a corrected exercise of which here is the statement, we will show the most important properties of matrix traces:

Matrix trace exercise

We are going to do these proofs on the field of real numbers but they are more generally valid in a field K.

So let's start by showing the linearity of the trace. Let λ be a real number and A a square matrix of size n.

\begin{array}{ll} Tr(\lambda A) &=\displaystyle \sum_{i=1}^n (\lambda a)_{ii}\\ & =\displaystyle \sum_{i=1}^ n \lambda a_{ii}\\ &=\displaystyle \lambda \sum_{i=1}^n a_{ii}\\ & =\lambda Tr(A) \end{array}

Let B be a square matrix of size n, we have:

\begin{array}{ll} Tr(A+B) &=\displaystyle \sum_{i=1}^n (a+b)_{ii}\\ & =\displaystyle \sum_{i=1}^ n a_{ii}+b_{ii}\\ &=\displaystyle \sum_{i=1}^n a_{ii}+ \sum_{k=1}^n b_{ii}\\ & =Tr(A ) +Tr(B) \end{array}

This shows us the linearity of the trace function. It is therefore a linear form!

Let us show that we can commute two terms:

\begin{array}{ll} Tr(AB) &=\displaystyle \sum_{i=1}^n (ab)_{ii}\\ & =\displaystyle \sum_{i=1}^n\sum_{ j=1}^n a_{ij}b_{ji}\\ & =\displaystyle \sum_{j=1}^n\sum_{i=1}^n a_{ij}b_{ji}\\ &= \displaystyle \sum_{j=1}^n\sum_{i=1}^n b_{ji}a_{ij}\\ & =\displaystyle \sum_{j=1}^n (ba)_{ji} \\ & =Tr(BA) \end{array}

Result : The trace is a similarity invariant. One

B = PAP^{-1}

So we have:

\begin{array}{ll} Tr(B) &= Tr(PAP^{-1})\\ &= Tr(P(AP^{-1}))\\ &= Tr((AP^{- 1})P)\\ &= Tr(A(P^{-1}P))\\ & = Tr(A) \end{array}

For the last property, which is not as classic as the previous ones. If such a matrix of size n existed, we would have:

Tr(AB-BA) = Tr(I_n) 

Yet

Tr(AB-BA) = Tr(AB) - Tr(BA) 

Et

Tr(AB) = Tr(BA) 

So

0 = Tr(I_n) = n

This allows us to arrive at a contradiction!

To note more : The trace is invariant by the transpose:

Tr({}^t A) = Tr(A)

corrected exercise

trace property

To do this exercise, we will use a formula established above:

Tr(AM) = \sum_{i=1}^n \sum_{j=1}^n a_{ij}m_{ji}

So we have :

\sum_{i=1}^n \sum_{j=1}^n a_{ij}m_{ji} = \sum_{i=1}^n \sum_{j=1}^n b_{ij}m_ {ji}

Thanks to the matrix M = Elk elementary matrix, we directly obtain

a_{kl} = b_{kl}

because Ilk is the only nonzero term of the sum and equals 1. We therefore have, by varying k and l:

\forall k,l \in \{1, \ldots, n \},a_{kl} = b_{kl}

Hence A = B

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