When the number of rows is the same as the number of colums i.e. $m=n$, the matrix is said to be square and of order $n$ (or $m$ since they are the same).
\nIn an $n \\times n$ square matrix $A$, the leading diagonal (or principal diagonal) is the top-left to bottom right collection of elements $a_{11}, a_{22}, a_{33}, . . . ,a_{nn}$.
\nThe sum of the elements in the leading diagonal of $A$ is called the trace of the matrix and we write it as $tr(A)$.
\nIf $ A= \\left( \\begin{array}{ccc} a_{11} & a_{12} & ... & a_{1n}\\\\ a_{21} & a_{22} & ... & a_{2n}\\\\ \\vdots & \\vdots & \\vdots & \\vdots\\\\ a_{n1} & a_{n2} & ... & a_{nn} \\end{array} \\right) $ then $ tr(A)= a_{11}+a_{22}+...+a_{nn}$
", "advice": "Remembering that, for a square matrix, the trace is the sum of the elements in the leading diagonal.
\nBegin at the top left element and work down to the bottom right element, adding as you go.
\n$A=\\var{A}$
\n$tr(A)=\\var{A[0][0]}+\\var{A[1][1]} +\\var{A[2][2]}=\\var{trA} $
\n\n
$B=\\var{B}$
\n$tr(B)=\\var{B[0][0]}+\\var{B[1][1]}=\\var{trB} $
\n\n
$C=\\var{C}$
\n$tr(C)=\\var{C[0][0]}+\\var{C[1][1]} +\\var{C[2][2]}+\\var{C[3][3]} +\\var{C[4][4]}=\\var{trC} $
\n\n
$D=\\var{D}$
\n$tr(D)=\\var{D[0][0]}+\\var{D[1][1]} +\\var{D[2][2]}+\\var{D[3][3]}=\\var{trD} $
\n\n
$E=\\var{EE}$
\n$tr(E)=\\var{EE[0][0]}+\\var{EE[1][1]} +\\var{EE[2][2]}=\\var{trE} $
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\n$tr(A)=$ [[0]]
\n\n
$B=\\var{B}$
\n$tr(B)=$ [[1]]
\n\n
$C=\\var{C}$
\n$tr(C)=$ [[2]]
\n\n
$D=\\var{D}$
\n$tr(D)=$ [[3]]
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$E=\\var{EE}$
\n$tr(E)=$ [[4]]
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