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Using linear algebra to explain, can someone show the intuition? I can show that the ranks are the same by using properties of rank but can't get my head around the whole projection thing more than just thinking hazily.

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Let the number of observations be $n$, let $p$ count the parameters, and let $r$ designate the rank of the $n\times p$ design matrix $X$ (which, by definition, is the dimension of the image of $X$).

The SVD

A Singular Value Decomposition expresses $X$ as a product

$$X = U\Sigma V^\prime$$

where the matrices $U$ (dimensions $n\times r$) and $V$ (dimensions $p \times r$) are orthogonal and $\Sigma$ is an $r\times r$ diagonal matrix with no zeros. A nonzero $X$ always has an SVD.

(Here's one proof: the columns of $V$ must be the eigenvectors of $X^\prime X$ corresponding to nonzero eigenvalues while the columns of $U$ must be the eigenvectors of $XX^\prime$ corresponding to nonzero eigenvalues. Those eigenvectors and eigenvalues exist because both $X^\prime X$ and $XX^\prime$ are nonzero real symmetric matrices: this is part of the Spectral Theorem. Although in the SVD it is arranged that all elements of $\Sigma$ be nonnegative, we won't need that here.)

Interpreting the SVD

One way to view the SVD is that it expresses the columns of $X$ as linear combinations of the columns of $U$: the coefficients are the columns of $\Sigma V^\prime$. You may therefore think of $U$ as being an orthonormal frame for the image of $X$, which is an $r$-dimensional subspace $\mathbb W\subset \mathbb{R}^n$. ("Orthonormal" means "orthogonal" and of unit length; "orthogonal" means mutually perpendicular, which is a crucial simplification.) Indeed, it is appealing to consider this geometrically: upon choosing bases for all the vector spaces in question, for $\beta\in\mathbb{R}^p$, $X$ determines a linear transformation from $\mathbb{R}^p$ into $\mathbb{R}^n$ in three steps:

  1. $\beta \to V^\prime \beta$ is a vector in $\mathbb{R}^r$.

  2. $\Sigma$ rescales each of the $r$ basis vectors of $\mathbb{R}^r$.

  3. The resulting $r$ coefficients determine a linear combination of the columns of $U$: that is, a unique vector in $\mathbb W$. (Equivalently, the original $r$ coefficients $V^\prime \beta$ specify linear combinations of the orthogonal columns of $U\Sigma$.)

The image of $\beta$ in step (1) consists of all vectors spanned by the $r$ rows of $V$, and therefore has dimension $r$. Because the diagonal elements of $\Sigma$ are nonzero, the rescaling in (2) does not change that dimension. Thus the dimension of the space generated in (3) is also $r$. Consequently, the rank of $X$ is $r$.

In statistical language, $V$ finds identifiable linear combinations of the parameters $\beta$ and the diagonal elements of $\Sigma$ establish scale factors in the space $\mathbb W$ spanned by the columns of $X$, which is the space of all possible vectors $y$ that can be exactly represented as linear combinations of those columns.

More About Projections

Here's a related algebraic argument.

Any orthonormal frame $U$ determines a projection matrix $UU^\prime$. Specifically, left multiplying any vector $y\in\mathbb{R}^n$ by $U^\prime$ computes the coefficients of $y$ for each of the columns of $U$. Obviously this has rank $r$: since the columns of $U$ each get projected to themselves, the image of the linear transformation $UU^\prime$ is precisely $\mathbb W$.

You probably know of a different looking formula for the "projection matrix": namely, $P=X(X^\prime X)^{-} X^\prime$ where $(X^\prime X)^{-}$ is a generalized inverse of $X^\prime X$. Using the SVD we may simplify this:

$$P = (U\Sigma V^\prime)((U\Sigma V^\prime)^\prime\, (U\Sigma V^\prime))^{-} (U\Sigma V^\prime)^\prime = UU^\prime.$$

This is because terms of the form $V^\prime V=I_r=U^\prime U$ are identity matrices, which disappear in the multiplications, and the generalized inverse of $\Sigma^2$ is just $\Sigma ^{-2}$.

It is now obvious that $P$ has rank $r$.

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  • $\begingroup$ This is great stuff! Can I think about it as each entry in the dependent variable needs to be modified by the projection matrix by each on of the vectors on a basis of the column space of the model matrix for the final projection to inhabit the vector space of the model matrix - hence the cardinality of the column space of any basis of the MM and Prjt. matrices have to agree? $\endgroup$ – Antoni Parellada Feb 7 '17 at 20:26
  • $\begingroup$ @Antoni I believe it's simpler than that. When you have an orthonormal frame for a vector subspace, the projection of any vector $y$ (the "dependent variable") into that subspace can be found by separately finding the component of $y$ on each element of the frame. This is the underlying idea behind orthogonal polynomials and various experimental design matrices, too: when all independent variables are uncorrelated (and thereby can serve as the columns of $U\Sigma$), the coefficient of each can be found without reference to any of the others. $\endgroup$ – whuber Feb 7 '17 at 20:44
  • $\begingroup$ That is not a proof of the existence of the SVD. $\endgroup$ – Federico Poloni Feb 7 '17 at 22:17
  • $\begingroup$ @Federico Although it's no proof, it's a sketch of one. Where do you think it falls short or is invalid? $\endgroup$ – whuber Feb 7 '17 at 22:35
  • $\begingroup$ @whuber There may be multiple choices for $U$ and $V$. For instance, if $X$ is orthogonal, $X'X=XX'=I$, so any choice of orthogonal $U$ and $V$ works. Not all these choices, though, produce a diagonal $U'XV$, which you need to complete the proof. $\endgroup$ – Federico Poloni Feb 8 '17 at 7:06

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