If T is a linear transformation from T: R^n-->R^m, must the rank of T be less than or equal to m?

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- Dec 12th 2012, 08:33 AMRachel123Linear Algebra: dimension of image?
If T is a linear transformation from T: R^n-->R^m, must the rank of T be less than or equal to m?

Thanks - Dec 12th 2012, 11:06 AMDrexel28Re: Linear Algebra: dimension of image?
Yes. The rank of $\displaystyle T$ is the image of its image, which being a subspace of $\displaystyle \mathbb{R}^m$ is at most $\displaystyle m$.

- Dec 12th 2012, 01:54 PMDevenoRe: Linear Algebra: dimension of image?
the rank of T (where T is a linear transformation of a finite-dimensional vector space into a finite-dimensional vector space) is defined to be the rank of any matrix A for T.

if the domain of T is R^{n}, and T maps into R^{m}, such a matrix will be an mxn matrix. this has only m rows, so only m (at most) of them can be non-zero when we row-reduce it.

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alternatively (and equivalently) if you define rank(T) = dim(im(T)) it should be clear im(T) = T(R^{n}) ⊆ R^{m}, so any basis for im(T) can at most possess m elements.

the nifty thing about rank is we can choose any bases for R^{n}and R^{m}we like to calculate the matrix A for T. some bases are easier to work with than other ones.