# Thread: Linear Algebra: dimension of image?

1. ## Linear 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

2. ## Re: Linear Algebra: dimension of image?

Yes. The rank of $T$ is the image of its image, which being a subspace of $\mathbb{R}^m$ is at most $m$.

3. ## Re: 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 Rn, and T maps into Rm, 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(Rn) ⊆ Rm, so any basis for im(T) can at most possess m elements.

the nifty thing about rank is we can choose any bases for Rn and Rm we like to calculate the matrix A for T. some bases are easier to work with than other ones.