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Old 09-16-2014, 12:02 AM
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yaser yaser is offline
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Default Re: Multiplication matrix X and its transpose

Quote:
Originally Posted by tarquilu View Post
Hello,

I'm trying to compute the pseudo-inverse of matrix X with dimensions [d+1 x N], where N is the number of data points. Lecture 3, slide 16 shows that the dimensions "shrink" when multiplying (X^T)X from [d+1 x N][N x d+1] to [d+1 x d+1].

Although LFD on page 86 states that "In practice, (X^T)T is invertible in most of the cases since N is often much bigger than d+1, so there will likely be d+1 linearly independent vectors xn," I don't know if this statement is meant to explain the reason for the simplification to [d+1 x d+1] dimensions. Moreover, when creating X and computing (X^T)T, I obtain a matrix with [N x N] dimensions, not [d+1 x d+1] dimensions.

Could some please clarify this for me? Thank you.
1. X is in fact of dimensions N x d+1. The transpose is d+1 x N.

2. The remark about independent vectors is only meant to argue that the d+1 x d+1 matrix has full rank (d+1) since in principle it could be singular, with rank less than d+1 hence not invertible.

3. I am not clear on what (X^T)T is.
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