P roof let φ and ψ be the maps represented by a and

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P ROOF . Let φ and ψ be the maps represented by A and B , respec- tively. Recall that AB corresponds to map composition φ ψ . Obviously, Im( φ ψ ) Im φ . Hence rank( AB ) = dimIm( φ ψ ) dimIm( φ ) = rank( A ). Similarly, Im( φ ψ ) is spanned by φ ( S ) where S is any basis of Im( ψ ). Hence rank( AB ) = dimIm( φ ψ ) dimIm( ψ ) = rank( B ). Thus the result follows. Our notion of rank in Definition 6.18 is sometimes also referred to as column rank of matrix A . One may also define the row rank of A as the maximal number of linearly independent rows of A . However, column rank and row rank always coincide. For any matrix A , Theorem 6.24 rank( A ) = rank( A 0 ). For the proof of this theorem we first show the following result. Let A be a m × n matrix. Then rank( A 0 A ) = rank( A ). Lemma 6.25 P ROOF . We show that ker( A 0 A ) = ker( A ). Obviously, x ker( A ) im- plies A 0 Ax = A 0 0 = 0 and thus ker( A ) ker( A 0 A ). Now assume that x ker( A 0 A ). Then A 0 Ax = 0 and we find 0 = x 0 A 0 Ax = ( Ax ) 0 ( Ax ) which implies that Ax = 0 so that x ker( A ). Hence ker( A 0 A ) ker( A ) and, consequently, ker( A 0 A ) = ker( A ). Now notice that A 0 A is an n × n matrix. Theorem 6.22 then implies rank( A 0 A ) - rank( A ) = ( n - nullity( A 0 A ) ) - ( n - nullity( A )) = nullity( A ) - nullity( A 0 A ) = 0 and thus rank( A 0 A ) = rank( A ), as claimed. P ROOF OF T HEOREM 6.24 . By Theorem 6.23 and Lemma 6.25 we find rank( A 0 ) rank( A 0 A ) = rank( A ) . As this statement remains true if we replace A by its transpose A 0 we have rank( A ) rank( A 0 ) and thus the statement follows.
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6.4 S IMILAR M ATRICES 45 Let A be an m × n matrix. Then Corollary 6.26 rank( A ) min { m , n } . Finally, we give a necessary and sufficient condition for invertibility of a square matrix. An n × n matrix A is called regular if it has full rank , i.e., if rank( A ) = n . Definition 6.27 A square matrix A is invertible if and only if it is regular. Theorem 6.28 P ROOF . By Theorem 6.13 a matrix is invertible if and only if nullity( A ) = 0 (i.e., ker( A ) = { 0 } ). Then rank( A ) = n - nullity( A ) = n by Theorem 6.22 . 6.4 Similar Matrices In Section 5.3 we have seen that every vector x V in some vector space V of dimension dim V = n can be uniformly represented by a coordinate vector c ( x ) R n . However, for this purpose we first have to choose an arbitrary but fixed basis for V . In this sense every finitely generated vector space is “equivalent” (i.e., isomorphic) to the R n . However, we also have seen that there is no such thing as the basis of a vector space and that coordinate vector c ( x ) changes when we change the underlying basis of V . Of course vector x then remains the same. In Section 6.2 above we have seen that matrices are the representa- tions of linear maps between R m and R n . Thus if φ : V W is a linear map, then there is a matrix A that represents the linear map between the coordinate vectors of vectors in V and those in W . Obviously matrix A depends on the chosen bases for V and W .
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