MATH311ansHW8

# MATH311ansHW8 - Math 311 Homework 8 partial answers and...

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Math 311, Homework 8 partial answers and solutions 3.5C.1. (a) Any matrix A with entries a ij is a linear combination of the matrices E ij : A = m i =1 n j =1 a ij E ij , so these matrices span all of M m,n . Conversely, any linear combination m i =1 n j =1 a ij E ij is the matrix with entries a ij . So if this linear combination is the zero matrix, then all its entries, and so all the coefficients in the linear combination, have to be zero. This means that the matrices { E ij } are linearly independent. As a result, they form a basis, and the dimension of the space of m × n matrices is mn . (b) A similar argument shows that a basis for the space of diagonal n × n matrices are the matrices { E ii : i = 1 , 2 , . . . , n } . As a result, this space has dimension n . 3.5C.3. Suppose a 1 x 1 + a 2 x 2 + . . . + a k x k = 0 . Then using the linearity of the function f , a 1 f ( x 1 ) + a 2 f ( x 2 ) + . . . + a k f ( x k ) = f ( a 1 x 1 + a 2 x 2 + . . . + a k x k ) = f ( 0 ) = 0 . In other words, if the vectors { x 1 , x 2 , . . . , x k } are linearly dependent, then so are the vectors { f ( x 1 ) , f ( x 2 ) , . . . , f ( x k ) } . In other words, if the vectors { f ( x 1 ) , f ( x 2 ) , . . . , f

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