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During my research I have come across matrices this type

$$C=B\left(B^T B\right)^{-1}B^T\ ,$$

where $B$ is an $m\times n$ real matrix. If $B^TB$ is not invertible, then $\left(B^T B\right)^{-1}$ should be interpreted as the Moore-Penrose pseudoinverse. the I am interested in the diagonal elements of $C$.

Clearly, if $m=n$ and $B$ is invertible, then $C$ is the identity matrix $I_{m\times m}$. It is pretty easy to show that this is still the case when $m<n$ and the rows of $B$ are linearly independent.

However, when the rows of $B$ are not independent, there is not much I can say. From the physical context where this came up, I have a very strong suspicion (supported by numerical experiments) that the diagonal elements of $C$ are bounded from above by $1$ (and clearly from below by $0$). My question is whether someone could give me a hint as to show this, or where to look for in the literature.

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    $\begingroup$ en.wikipedia.org/wiki/Projection_%28linear_algebra%29 $\endgroup$ – Federico Poloni Feb 26 '15 at 12:51
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    $\begingroup$ (in particular, you'll need the formula and the part where it says that $\|Pv\|\leq \|v\|$, which you can use for the elements of the canonical basis $v=e_i$ to get what you need). $\endgroup$ – Federico Poloni Feb 26 '15 at 13:06
  • $\begingroup$ After years of experience, I'm still shocked from my own ignorance. Thanks. If you write it as an answer I will accept it. $\endgroup$ – yohbs Feb 26 '15 at 17:32
  • $\begingroup$ Don't feel that way -- This is a direction of linear algebra which, although elementary, is not usually explored in the "classical" courses; it is not shocking that a mathematician working in another field has never seen this in detail. $\endgroup$ – Federico Poloni Feb 26 '15 at 17:38
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$C$ is an idempotent matrix, so its eigenvalues are either 1 or zero. $C$ is also Hermitian so Schur's Theorem says the sum of the $k$ largest eigenvalues is greater then the sum of the $k$ largest diagonal elements of $C$ for $k=1,2,...,m$. No diagonal element of $C$ can therefore exceed unity.

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  • $\begingroup$ I accepted your answer, but I think @FedericoPoloni's answer is somewhat simpler. The crux is $C$ is idempotent $\Rightarrow$ its eigenvalues are $\in [0,1]$ $\Rightarrow$ foe any $v$ $||Pv||\le||v||$, and specifically for elemets of the cannonical basis $\endgroup$ – yohbs Mar 1 '15 at 9:05

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