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Define T: M, n R by T(A) = a,, + a,, + + a (the trace of A). Prove that T is a linear transformation. Let A and B be to matrices in M. and let the entries of A, B be denoted a, and b respectively, for 1 si, jsn. Bỳ matrix addition, we know that the entries of the matrix A + Bare of the form-Select-v for 1si,jsn. By the definition of T and the commutative and associative property of addition, we have which of the following? T(A + B) = (a1 -b + + (ann bnn) = (a1 + + ann + (b,1 + + bn) = T(A) + T(B) T(A + B) = (a, b,) + --- + (a, b) = (a, + + a) + (b, + +b) = T(A) + T(B) %3D T(A + B) = (a, + b,) + + (a, + b) = T(A) + T(B) ("q + -- + 'q) + ("e + + e) = T(A + B) = (a1 +b) + + (an +ban) = (a, + + an) + (b + - + bn) = T(A) + T(B)

Define T: M, n R by T(A) = a,, + a,, + + a (the trace of A). Prove that T is a linear transformation. Let A and B be to matrices in M. and let the entries of A, B be denoted a, and b respectively, for 1 si, jsn. Bỳ matrix addition, we know that the entries of the matrix A + Bare of the form-Select-v for 1si,jsn. By the definition of T and the commutative and associative property of addition, we have which of the following? T(A + B) = (a1 -b + + (ann bnn) = (a1 + + ann + (b,1 + + bn) = T(A) + T(B) T(A + B) = (a, b,) + --- + (a, b) = (a, + + a) + (b, + +b) = T(A) + T(B) %3D T(A + B) = (a, + b,) + + (a, + b) = T(A) + T(B) ("q + -- + 'q) + ("e + + e) = T(A + B) = (a1 +b) + + (an +ban) = (a, + + an) + (b + - + bn) = T(A) + T(B)

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Define T: M, п, п →R by T(A) = a1 + a22 + -…- + a, nn (the trace of A). Prove that T is a linear transformation. Let A and B be to matrices in M "ח ,ה and let the entries of A, B be denoted and b ij respectively, for 1 < i, j< n. By matrix addition, we know that the entries of the matrix A +B are of the form -Select-v for 1 si,j< n. By the definition of T and the commutative and associative property of addition, we have which of the following? T(A + B) = (a, b,) + + (a bn) (a,, + + ap) + (b,1 + -- + b) = T(A) + T(B) ... %3D T(A + B) = (a, b;) + --- + (a, · b,) + a,) + (b, + + b,) %3D %3D = T(A) + T(B) ("q + - + 'q) + ("e + --- + 'e) = %3D ("q + "e) + .. + (*q+ Te) = (a + *** + b,) %3D %3D = T(A) + T(B) ("q + + q) + ("e + .. + Te) = %3D + (ann+ bnn T(A + B) = (a11 + b11) + . = (a1 %3D +.. + a nn+ (b,1 + * + bn) %3D T(A) + T(B) %3D
Transcribed Image Text:Define T: M, п, п →R by T(A) = a1 + a22 + -…- + a, nn (the trace of A). Prove that T is a linear transformation. Let A and B be to matrices in M "ח ,ה and let the entries of A, B be denoted and b ij respectively, for 1 < i, j< n. By matrix addition, we know that the entries of the matrix A +B are of the form -Select-v for 1 si,j< n. By the definition of T and the commutative and associative property of addition, we have which of the following? T(A + B) = (a, b,) + + (a bn) (a,, + + ap) + (b,1 + -- + b) = T(A) + T(B) ... %3D T(A + B) = (a, b;) + --- + (a, · b,) + a,) + (b, + + b,) %3D %3D = T(A) + T(B) ("q + - + 'q) + ("e + --- + 'e) = %3D ("q + "e) + .. + (*q+ Te) = (a + *** + b,) %3D %3D = T(A) + T(B) ("q + + q) + ("e + .. + Te) = %3D + (ann+ bnn T(A + B) = (a11 + b11) + . = (a1 %3D +.. + a nn+ (b,1 + * + bn) %3D T(A) + T(B) %3D
-Select-- 1<i,j<n. By the definition of T and the distributive property Now, let c be a scalar. By scalar multiplication, we know that the entries of the matrix CA are of the form of multiplication, we have which of the following? T(CA) = ca, can = c(a, + . + a) = CT(A) %3D %3D %3D T(CA) = ca, + … + + can %3D %3D = CT(A) ("e +. + e)ɔ %3D T(CA) = = c(a11 = CT(A) + + cann ca 11 %3D + .· + an) %3D %3D T(CA) = ca1 ca nn c(a 11 = CT(A) %3D %3D Therefore, T is a linear transformation. Need Help? Read It
Transcribed Image Text:-Select-- 1<i,j<n. By the definition of T and the distributive property Now, let c be a scalar. By scalar multiplication, we know that the entries of the matrix CA are of the form of multiplication, we have which of the following? T(CA) = ca, can = c(a, + . + a) = CT(A) %3D %3D %3D T(CA) = ca, + … + + can %3D %3D = CT(A) ("e +. + e)ɔ %3D T(CA) = = c(a11 = CT(A) + + cann ca 11 %3D + .· + an) %3D %3D T(CA) = ca1 ca nn c(a 11 = CT(A) %3D %3D Therefore, T is a linear transformation. Need Help? Read It
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