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Math Help - difference between dimension and rank of a matrix

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    Member Jskid's Avatar
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    difference between dimension and rank of a matrix

    I'm having trouble seeing the difference between the dimensions and the rank of a matrix. Could someone please give me an example where the two are differnt?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jskid View Post
    I'm having trouble seeing the difference between the dimensions and the rank of a matrix. Could someone please give me an example where the two are differnt?
    Since you didn't give that much pertinent background I'll assume, for simplicities sake, that you're speaking of some A\in\text{Mat}_n\left(\mathbb{R}\right). Then, if one views A as linear homomorphism \mathbb{R}^n\to\mathbb{R}^n then \text{rk}(A)=\dim\text{im}(A).
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    Quote Originally Posted by Jskid View Post
    I'm having trouble seeing the difference between the dimensions and the rank of a matrix. Could someone please give me an example where the two are differnt?
    Strictly speaking, the term "dimension" does not apply to matrices- it applies to vector spaces. However, a matrix is a linear transformation from one vector space to another. The "dimension" of the matrix can be thought of as the dimensions of those two vectors spaces- if a matric has, say, m rows and n columns, it is a linear transformation from a vector space, U, with dimension n to a vector space, V, of dimension m. Some texts will refer to "n by m" as the "dimension" of the matrix. On the other hand, we can think of the collection of all "n by m" matrices as a vector space in its own right- it has dimension mn.

    If a matrix is "one-to-one", then it would map the vector space of dimension n to to a subspace of dimension n. In that case, the "rank" of the matrix is n. If it is not one-to-one, then it maps U, of dimension n, to a subspace of V of dimension less than n. (If m< n, then the matrix can't be one-to-one.) In either case, the rank of the matrix is the dimension of that subspace.
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