Question #f4e98

2 Answers
Mar 15, 2017

# tr( 2A^7+3A^5+4A^2+A+I ) = 6 #

Explanation:

If #A# is nilpotent with order #2#; then

# A^2 = 0 #

And more importantly;

# A^m = 0 AA m in NN, m ge 2#

From which we can deduce that:

# tr(A^m) = 0 AA m in NN, m ge 2#

And we can use the trace properties:

# tr(A+B) = tr(A)+tr(B) #
# tr(mA) \ \ \ \ = m \ tr(A) #
# tr(I_n) \ \ \ \ \ \ \ = n # where #I_n# is the #nxxn# identity matrix

And so:

# tr( 2A^7+3A^5+4A^2+A+I ) #
# " " = tr( 2A^7) +tr(3A^5)+tr(4A^2)+tr(A)+tr(I) #
# " " = 2tr( A^7) +3tr(A^5)+4tr(A^2)+tr(A)+tr(I) #
# " " = 0 +0+0+tr(A)+tr(I) #
# " " = tr(A)+tr(I) #
# " " = 3+3 #
# " " = 6 #

Mar 15, 2017

If #A^2=0# then

#B=2A^7+3A^5+4A^2+A+I=A+I#

and #"tr"(B)="tr"(A)+"tr"(I)=3+3=6#