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If TU=I then show that rank of T is n.please ans is urgent.

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It is stated that TU = I, the identity matrix. This implies that T is invertible (non-singular) and the inverse of T is U. If T is non-singular, then the column vectors form a basis for the column-space of T (that is, all of the column vectors are linearly independent.) This can be shown by the fact that Tx = 0^n has only the trivial solution: x = 0^n from which follows from T being a non-singular matrix. The rank of T is defined to be the dimension of the column-space of T (or row-space, as the dimension of each are always identical.) Lastly, the dimension is defined to be the number of vectors forming a basis - in this case, n. Thus, the rank of T is n, or: rank(T) = n

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thank you very very much..indeeed it was very helpful solution.please tell me the name of any site/link etc where i can improve my linear algbera.
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I would watch the khan academy playlist on linear algebra. You can find this at YouTube. It is much help for me when I am having a hard time understanding a book or what my professor is teaching. There are over 100 videos in the playlist covering a large range. Good luck.
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