This question comes from page 142 of the book "Pattern Recognition and Machine Learning" by Christopher M. Bishop. Since it takes pages to arrive at the result, I excerpt the major settings as follows.

The goal is to derive the maximum likelihood solution for parameters $\bf w$ in a deterministic linear model with additive Gaussian noise $$t={\bf w}^T{\bf\phi}({\bf x})+\epsilon$$ where ${\bf w}=(w_0,\ldots,w_{M-1})^T$ and ${\bf\phi}$ is a vector of basis functions $(\phi_0,\ldots,\phi_{M-1})^T$ and hence should be in bold typeface.

Now consider a data set of inputs ${\bf X}=\{{\bf x}_1,\ldots,{\bf x}_N\}$ with corresponding target values ${\bf t}=(t_1,\ldots,t_N)^T$. Taking the (transpose of the) gradient of the log likelihood function and setting this gradient to zero gives enter image description here

The book proceeds with solving for $\bf w$ to obtain $${\bf w}_{\mathrm{ML}}=({\bf\Phi}^T{\bf\Phi})^{-1}{\bf\Phi}^T{\bf t}.$$

Here $\bf\Phi$ is an $N\times M$ matrix, called the design matrix:

enter image description here.

My questions is, matrix ${\bf\Phi}^T{\bf\Phi}$ is invertible only if $\bf\Phi$ has linearly independent columns. If we denote columns of $\bf\Phi$ by ${\bf\varphi}_j, j=1,\ldots,M$ (again should be bold-faced), why are ${\bf\varphi}_j$'s linearly independent? In general, if $M>N$, these columns are necessarily linear dependent in $\mathbb R^N$, so how could ${\bf\Phi}^T{\bf\Phi}$ be invertible to form the Moore-Penrose pseudo-inverse? Such a linear independence is also needed in subsequent geometrical interpretation in which $M<N$ and the subspace spanned by these column vectors is claimed to have dimensionality $M$. I checked the book but see nowhere make this claim. Neither can I figure out the independence from general definition of basis functions in the book. I'll appreciate it if you help me understand such a linear independence between ${\bf\varphi}_j$'s.

  • $\begingroup$ Hi: If $M$ greater than $N$, then that means that you have more coefficients to estimate than you have observations, This is is not a good thing and will lead to lack of invertibility. The lack of invertibility can occur in other ways even when $M$ is not greater than $N$ the case but $M$ being greater than $N$ will insure it. $\endgroup$
    – mlofton
    Jul 24, 2022 at 3:44

1 Answer 1


You are right, $\boldsymbol{\Phi}^T\boldsymbol{\Phi}$ is only invertible if all the columns of $\boldsymbol{\Phi}$ are independent, which is impossible if $M>N$ and should be presumed for $M\le N$. And if Bishop did not mention this explicitly, it is unfortunate.

However, recall that the observations $\mathbf x_i$ could usually be considered as samples from some continuous probability distribution, and thus the columns of $\mathbf \Phi$ could be considered as samples from a continuous distribution, too. Now, with respect to this probability measure, the set of matrices $\mathbf\Phi$ with $N\ge M$ that are not of maximal rank is of measure zero, i.e. those "singular" matrices $\mathbf\Phi$ occur with probability zero. Thus, from this point of view, it is justified to presume maximal rank of $\mathbf \Phi$ for $N\ge M$. But, again, you are right, and the singular situation should have at least been mentioned.

  • 1
    $\begingroup$ It's cool to think of full-rank almost surely. For those who are wondering, think of the probability of a function, say determinant, of a continuous r.v. to be exactly 0. Full-rank of $\bf\Phi$ with $M\le N$ means that all its $M\times M$ sub-determinants are zero. $\endgroup$
    – zzzhhh
    Jul 24, 2022 at 7:31

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