Trace of a matrix and properties

Before we start with the concept of a trace of matrices, let’s first understand what is a matrix. A rectangular arrangement of objects (numbers or symbols or any other objects) is called a matrix (plural: matrices). A matrix is only a representation of the symbol, number, or object. It does not have any value. Usually, a matrix is denoted by capital letters. Matrix of order m × n, (read as m by n matrix) means that the matrix has m number of rows and n number of columns. In real life, we can use a trace of the matrix to construct Hamiltonians for quantum systems with discrete and finite sets of energy equivalence.

In this article, we will cover the concept trace of matrices. This category falls under the broader category of Matrices, which is a crucial Chapter in class 12 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of eight questions have been asked on this topic in Jee mains (2013 to 2023), one in 2013, one in 2021.

Square matrix

The square matrix is the matrix in which the number of rows $=$ number of columns. So a matrix $A=\left[a_{i j}\right]_{m \times n}$ is said to be a square matrix when $m=n$. E.g.
$\left[\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right]_{3 \times 3}$ or, $\left[\begin{array}{cc}2 & -4 \\ 7 & 3\end{array}\right]_{2 \times 2}$

Trace of the matrix:

The sum of all diagonal elements of a square matrix is called the trace of a matrix. Lying along the principal diagonal is called the trace of A.

The trace of the matrix is denoted by Tr(A) or tr.A.

$
\operatorname{Tr}(\mathrm{A})=\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{ii}}
$

Let us consider the square matrix of order $3 \times 3$ as shown below. The elements of the matrix are $a_{11}, a_{12}, a_{13}$ $\qquad$ , $a_{33}$. The principal diagonal elements are $a_{11}, a_{22}, a_{33}$. So trace of the matrix is the sum of all principal diagonal elements.
$
\begin{aligned}
& A=\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right] \\
& \operatorname{Tr}(A)=a_{11}+a_{22}+a_{33}
\end{aligned}
$

Eg.
For a given matrix $\mathrm{A}$,
$
A=\left[\begin{array}{ccc}
-2 & 4 & 7 \\
8 & 3 & -1 \\
5 & -6 & 9
\end{array}\right], \operatorname{Tr}(\mathrm{A})=-2+3+9=10
$

Properties of a trace of the matrix:

Given below are some properties of the trace of a matrix. Let us consider two square matrices A and B

$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}} ; \mathrm{B}=\left[\mathrm{b}_{\mathrm{ij}}\right]_{\mathrm{n} \times \mathrm{n}}$ and $\mathrm{k}$ be a scalar, then

i) If A is a square matrix of order ‘n’ and k is a scalar quantity then the trace of multiplication of k and A is equal to the Multiplication of k into the trace of A.

Tr(kA)=k·Tr(A)

ii)If A is the square matrix of order ‘n x m’ and B be square matric of order ‘m x n then, the trace of the sum( or subtraction) of Matrix A and B is equal to the sum (or subtraction) of Trace of A and Trace of B.

Tr(A ± B) = Tr(A) ± Tr(B)

iii) If A is the square matrix of order ‘n x m’ and B be square matric of order 'm x n' then, the Trace of Matrix AB is equal to the Trace of matrices B A.

Tr(AB) = Tr(BA)

iv) If A is square matrices of order ‘n’ then, the Trace of Matrix A is equal to the Trace of the transpose of Matrix A.

Tr(A) = Tr(A’)

v)If A is the square matrix of order ‘n x m’ and B be square matric of order ‘m x n’ then, the trace of Matrix AB is not equal to the Trace of Matrix A multiplied by the Trace of Matrix B.

Tr(AB) ≠ Tr(A).Tr(B)

vi) if I is the Identity matrix of order ‘n’ then the trace of the identity matrix is equal to n.

Tr(I) =n

vii) Trace of zero or null matrix is always zero.

Tr(0)=0S

Summary

The features of the trace of a matrix make it an adaptable tool for analyzing and comprehending matrices and their transformations. It is a useful topic with applications in many fields of mathematics. The trace of a matrix provides a valuable measure of its diagonal entries' sum, offering insights into matrix properties and transformations.

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Solved Examples Based on Trace of Matrices

Example 1: Let and $2 A-B=\left[\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]$. If $\operatorname{Tr}(\mathrm{A})$ denotes the sum of all diagonal elements of the matrix $A$, then $\operatorname{Tr}(\mathrm{A})-\operatorname{Tr}(\mathrm{B})$ has a value equal to [JEE MAIN 2021]

Solution:

$\begin{aligned} & A+2 B=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right) \\ & 2 A-B=\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right) \\ & \Rightarrow 4 A-2 B=\left(\begin{array}{ccc}4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4\end{array}\right)\end{aligned}$

Add (1) and (2), we get

$\begin{equation}
\begin{aligned}
& \Rightarrow 5 \mathrm{~A}=\left(\begin{array}{ccc}
5 & 0 & 10 \\
10 & -5 & 15 \\
-5 & 5 & 5
\end{array}\right) \\
& A=\left(\begin{array}{ccc}
1 & 0 & 2 \\
2 & -1 & 3 \\
-1 & 1 & 1
\end{array}\right) \text { and } 2 A=\left(\begin{array}{ccc}
2 & 0 & 4 \\
4 & -2 & 6 \\
-2 & 2 & 2
\end{array}\right) \\
& \therefore \mathrm{B}=\left(\begin{array}{ccc}
2 & 0 & 4 \\
4 & -2 & 6 \\
-2 & 2 & 2
\end{array}\right)-\left(\begin{array}{ccc}
2 & -1 & 5 \\
2 & -1 & 6 \\
0 & 1 & 2
\end{array}\right) \\
& \mathrm{B}=\left(\begin{array}{ccc}
0 & 1 & -1 \\
2 & -1 & 0 \\
-2 & 1 & 0
\end{array}\right) \\
& \operatorname{tr}(\mathrm{A})=1-1+1=1 \\
& \operatorname{tr}(\mathrm{B})=-1 \\
& \operatorname{tr}(\mathrm{A})=1 \text { and } \operatorname{tr}(\mathrm{B})=-1 \\
& \therefore \operatorname{tr}(\mathrm{A})-\operatorname{tr}(\mathrm{B})=2
\end{aligned}
\end{equation}$

Hence, the answer is the option 2.Example 2

\begin{equation}
\text { : The number of all } 3 \times 3 \text { matrices A, with entries from the set }\{-1,0,1\} \text { such that the sum of the diagonal elements of } A A^T \text { is } 3 \text {, is }
\end{equation}

Solution

Let matrix $\mathrm{A}$ be
$
\begin{aligned}
& A=\left[\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right] \\
& A^T=\left[\begin{array}{lll}
a & d & g \\
b & e & h \\
c & f & i
\end{array}\right] \\
& \operatorname{trace}\left(A A^T\right)=a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3
\end{aligned}
$

So out of 9 elements, 3 elements must be equal to 1 or −1, and the rest elements must be 0.

Possible causes

$\begin{equation}
\begin{array}{cl}
0,0,0,0,0,0,1,1,1 & \rightarrow \text { Total possibilities }={ }^9 C_6 \\
0,0,0,0,0,0,-1,-1,-1 & \rightarrow \text { Total possibilities }={ }^9 C_6 \\
0,0,0,0,0,0,1,1,-1 & \rightarrow \text { Total possibilities }={ }^9 C_6 \times 3 \\
0,0,0,0,0,0,-1,1,-1 & \rightarrow \text { Total possibilities }={ }^9 C_6 \times 3 \\
\text { Total number of cases }={ }^9 C_6 \times 8=672
\end{array}
\end{equation}$

Hence, the answer is the option 1

Example 3 Let $\mathrm{A}$, other than $l$ or $-l$, be a $2 \times 2$ real matrix such that $a^2=l, l_{\text {being the unit matrix. Let }} \operatorname{Tr}(A)$ be the sum of diagonal elements of $\mathrm{A}$. [JEE MAIN 2013]

Statement 1: $\operatorname{Tr}(A)=0$
Statement 2: $\operatorname{det}(A)=-1$

1)statement 1 is true; statement 2 is false.

2) statement 1 is true; statement 2 is true; statement 2 is not the correct explanation for statement 1

3)statement 1 is true; statement 2 is true; statement 2 is the correct explanation for statement 1

4)statement 1 is false; statement 2 is true;

Solution:

$
\begin{aligned}
& {\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& {\left[\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& b(a+d)=0, b=0 \text { or } a=-d \\
& c(a+d)=0, c=0 \text { or } a=-d \\
& a^2+b c=1, b c+d^2=1 \\
&
\end{aligned}
$
' $a$ ' and 'd' are the diagonal element, $a+d=0$
Statement 1 is true
Now, $\operatorname{det}(\mathrm{A})=\mathrm{ad}-\mathrm{bc}$
Now, from (3) $a^2+b c=1$ and $d^2+b c=1$
So, $a^2-d^2=0$
Adding $a^2+d^2+2 b c=2$
$
=(a+d)^2-2 a d+2 b c=2
$
or $0-2(a d-b c)=2$
So, $a d-b c=1 \Rightarrow \operatorname{det}(A)=-1$
So, statement -2 is also true.
But statement -2 is not the correct explanation of statement-I

Hence, the answer is option 2

\begin{equation}
\text { Example 4: If the element of a matrix A is defined by } a_{i j}=i^2-j^2 \text { and } \mathrm{A} \text { is a square matrix of order } 3 \times 3 \text {. Then } \operatorname{tr}(A)=
\end{equation}

1) 14

Example 5: Let A be a $2 \times 2$ real matrix with entries from $\{0,1\}$ and $|A| \neq 0$ Consider the following two statements :
(P) If $\mathrm{A} \neq \mathrm{I}_2$, then $|\mathrm{A}|=-1$
(Q) If $|\mathrm{A}|=1$, then $\operatorname{tr}(\mathrm{A})=2$,

where denotes 2 × 2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then:

1) (P) is false and (Q) is true

2) Both (P) and (Q) are false

3) (P) is true and (Q) is false

4) Both (P) and (Q) are true

Solution:

$
\begin{aligned}
& |A| \neq 0 \\
& \text { For }(\mathrm{P}): \mathrm{A} \neq \mathrm{I}_2
\end{aligned}
$

So, $\mathrm{A}=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right]$ or $\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right]$ or $\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]$ or $\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ or $\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$ $|A|$ can be -1 or 1

So (P) is false
For $(\mathrm{Q}):|\mathrm{A}|=1$
$
\begin{aligned}
& \mathrm{A}=\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right] \text { or }\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \text { or }\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& \Rightarrow \operatorname{tr}(\mathrm{A})=2 \\
& \Rightarrow \mathrm{Q} \text { is true }
\end{aligned}
$

Hence, the answer is the option 1