The key insight needed to solve this problem is the

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The key insight needed to solve this problem is the following: a 11 a 12 a 21 a 22 1 0 = a 11 a 21 and a 11 a 12 a 21 a 22 0 1 = a 12 a 22 . That is, a matrix multiplying a vector of the form ( 1 0 ) or ( 0 1 ) (a “standard basis vector”) equals the respective column of the matrix. So, a matrix A representing T can be found by observing that T 1 0 = First column of A and T 0 1 = Second column of A . Since T is linear, T ((1 , 0)) and T ((0 , 1)) can be found. Note that T - 1 2 and T 2 - 3 are already known and 0 1 = 2 - 3 + 2 - 1 2 . Therefore, T 2 - 3 + 2 T - 1 2 = T 2 - 3 + 2 - 1 2 ( T is linear) = T 2 - 2 - 3 + 4 = T 0 1 . So, T 0 1 = T 2 - 3 + 2 T - 1 2 = 2 - 13 + 2 - 1 8 = 0 3 . 50
Now, it has been found that T - 1 2 = - 1 8 , T 2 - 3 = 2 - 13 , and T 0 1 = 0 3 . To determine T ((1 , 0)) , note that 1 0 = - - 1 2 + 2 0 1 , so, T 1 0 = T - - 1 2 + 2 0 1 = - T - 1 2 + 2 T 0 1 = - - 1 8 + 2 0 3 = 1 - 2 . Therefore, A = T 1 0 T 0 1 = - 1 0 2 3 is the matrix representing A . At the beginning of the section it was proven in theorem 3.1.6 that matrix multiplication is a linear transformation. Now, it will be proven that the only linear transformation is matrix multiplication. That is, given any linear map T : R m R n , there is an n × m matrix A satisfying T ( x ) = Ax . The key idea was illustrated in the previous problem: the matrix representing a linear map is determined by the linear map acting on the “standard basis vectors.” Definition 3.1.26. The i th standard basis vector in R n is the vector e i = 0 . . . 0 1 0 . . . 0 where e i has a 1 as the i th component and all other components are 0. 51
Example 3.1.27. The standard basis vectors for R 3 are e 1 = 1 0 0 , e 2 = 0 1 0 , and e 3 = 0 0 1 . Theorem 3.1.28. Let T : R m R n be a linear map. Then, T ( x ) = Ax where A in the n × m matrix given by A = ( T ( e 1 ) T ( e 2 ) · · · T ( e m ) ) , i.e. A is the matrix with T ( e i ) as its i th column. Proof. Let x R m . Then, x 1 x 2 . . . x m = x 1 e 1 + x 2 e 2 + · · · + x m e m . So, since T is linear, T ( x ) = T x 1 x 2 . . . x m = T ( x 1 e 1 + x 2 e 2 + · · · + x m e m ) (Expanding x in terms standard basis vectors) = x 1 T ( e 1 ) + x 2 T ( e 2 ) + · · · + x m T ( e m ) (Linearity of T ) = ( T ( e 1 ) T ( e 2 ) · · · T ( e m ) ) x 1 x 2 . . . x m . Therefore, T is represented by A = ( T ( e 1 ) T ( e 2 ) · · · T ( e m ) ) . 52
3.2 Matrices 3.2.1 Matrix Algebra Definition 3.2.1 (Matrix addition and scalar multiplication) . Let A and B be n × m matrices and let c be a scalar. The sum A + B is defined componentwise: A + B = a 11 · · · a 1 m . . . . . . . . . a n 1 · · · a nm + b 11 · · · b 1 m . . . . . . . . . b n 1 · · · b nm = ( a 11 + b 11 ) · · · ( a 1 m + b 1 m ) . . . . . . . . . ( a 1 n + b n 1 ) · · · ( a nm + b nm ) . The product cA is defined componentwise: cA = c a 11 · · · a 1 m .

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