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The rank of a linear transformation T is the dimension of its image, which is also called the column space. The rank of a matrix representing the linear transformation can be found by reducing the matrix to its row echelon form and counting the non-zero rows. This is the same as the number of linearly independent columns in the matrix. The rank is an important property that helps us understand the behaviour of the transformation and the structure of its image.

The transpose of a linear transformation T, denoted by \(T^t\), is a mapping between the dual spaces of the original transformation's domain and codomain. In other words, if T is a linear transformation from V to U, then \(T^t\) is a linear transformation from \(U^*\) to \(V^*\), where \(V^*\) and \(U^*\) are the dual spaces of V and U, respectively. The dual space of a vector space is the set of all linear functionals on that space. In terms of matrices, the transpose of a linear transformation T is represented by the transpose of the matrix representing T. This is obtained by interchanging the rows and columns of the matrix.

Let's denote the bases of V and U by \(B_V\) and \(B_U\), and their dimensions by n and m, respectively. Also, let's represent the matrix of the linear transformation T with respect to these bases by A. Then, our goal is to show that the rank of A is equal to the rank of its transpose.

Let's calculate the image of the linear transformation T and its transpose, denoted by Im(T) and Im(\(T^t\)), respectively. Im(T) is a subspace of U and is spanned by the column vectors of A. Im(\(T^t\)) is a subspace of \(V^*\) and is spanned by the row vectors of A.

To show that the rank of T is equal to the rank of its transpose, we need to prove that the number of linearly independent column vectors in A (which forms a basis for Im(T)) is equal to the number of linearly independent row vectors in A (which forms a basis for Im(\(T^t\))). Let's perform Gaussian elimination on the matrix A to obtain its row echelon form. Suppose we have r non-zero rows after this process. We know that the number of non-zero rows corresponds to the number of linearly independent column vectors in A, which is equal to the rank of T. Now, consider the transpose of A, denoted by \(A^t\). Notice that the number of non-zero rows in \(A^t\) is equal to the number of non-zero columns in the original matrix A, which is r. Thus, the number of linearly independent row vectors in A is equal to the number of linearly independent column vectors in A. Therefore, we conclude that the rank of T is equal to the rank of its transpose, as desired: \(\operatorname{rank}(T)=\operatorname{rank}\left(T^{t}\right)\).