Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across this post lying dormant on some online forum. I am putting it here verbatim, it seems to me worth a lot.


By Prof. S. D. Agashe, IIT Bombay

(Source: Vector Calculus, by Durgaprasanna Bhattacharyya, University Studies Series,Griffith Prize Thesis, 1918, published by the University of Calcutta, India, 1920, 90 pp)

Chapter IV: The Linear Vector Function, article 15, p.24:

"The most general vector expression linear in $r$ can contain terms only of three possible types, $r$, $(a.r)b$ and $c\times r$, $a$, $b$, $c$ being constant unit vectors. Since $r$, $(a.r)b$ and $c\times r$ are in general non-coplanar,it follows from the theorem of the parallelepiped of vectors that the most general linear vector expression can be written in the form $\lambda . r + \mu (a.r)b + \nu (c\times r)$, where $\lambda, \mu, \nu$ are scalar constants".

Bhattacharyya does not prove this. Has anyone seen a similar result and its proof?

Bhattacharyya uses this to show that the divergence of the linear function is ($3 \lambda + a.b$), that the curl is ($a \times b + 2c$). He goes on to define div and curl of a differentiable function as the div and curl of the (linear) derivative function. The div and curl of a linear function are defined in terms of certain surface integrals.

I am excited about this result because it seems to provide an excellent route to div and curl, as Bhattacharyya himself remarks.

Sorry for a rather long and "technical" communication.

share|improve this question
2  
This is specific to 3 dimensions. Otherwise $c \times r$ is not so simple (wrong number of components) and you need more than one $a$ and $b$ –  Ross Millikan Feb 28 '11 at 4:11
    
@Ross, Is it not applicable for n dimension?? –  Dilawar Mar 1 '11 at 13:57
1  
A cross product is more properly viewed as an antisymmetric matrix. An antisymmetric $n \ times n$ matrix has $\frac{n(n-1)}{2}$ components. In 3 dimensions, this is 3 and we can call it a (pseudo)vector, but in 4 dimensions it is 6. The general linear transformation in 3 dimensions needs 9 components, which you have, but in 4 dimensions you need 16. With 4 each from $r$ and $b$ and 6 from the "cross product" you are still 2 short. There is some discussion at en.wikipedia.org/wiki/Cross_product –  Ross Millikan Mar 1 '11 at 14:12
add comment

2 Answers 2

up vote 6 down vote accepted

The claim is true. Any $3\times3$ matrix can be expressed as $$ A= \lambda I+ a b^T + B $$ where $\lambda$ is real, $a$ and $b$ are 3-vectors and $B$ is skew (so that $Bx=c\times x$ for some vector $c$).

To prove this, choose an orthogonal matrix $Q$ to diagonalize the symmetric part of $A$. Then $Q^TAQ=D+K$ where $D$ is diagonal and $K$ is skew. If the diagonal entries of $D$ are not all distinct then it is easy to write $D=\lambda I+\hat a \hat b^T$ and we finish as below. If the entries are all distinct, we can suppose that $Q$ was chosen so that the largest eigenvalue of $D$ is first, the smallest second and the middle last. Then for some positive $\mu$ and $\nu$, the matrix $D$ can be written $$ D = \lambda I + \mu \begin{pmatrix} 1 & 0 & 0\cr 0 &-\nu^2&0\cr 0&0&0 \end{pmatrix} =\lambda I + \hat a\hat b^T+\hat K, $$ with $$ \hat a= \mu\begin{pmatrix}1\cr \nu\cr0\end{pmatrix}, \quad \hat b= \begin{pmatrix} 1 \cr -\nu\cr 0 \end{pmatrix}, \quad \hat K=\mu\begin{pmatrix} 0&\nu&0\cr -\nu & 0&0\cr 0&0&0 \end{pmatrix} . $$ Let $a=Q\hat a$, $b=Q\hat b$, and $B=Q(\hat K+K)Q^T$, and you're done.

share|improve this answer
    
Any 3x3 matrix can be expressed as $A = \lambda I + ab^T + B$. Is it a general result? Theorem? Can you give me link for its proof? –  Dilawar Mar 1 '11 at 14:01
    
I think I got the proof of it. Still, any further ref is much appreciated. –  Dilawar Mar 2 '11 at 2:59
add comment

I'm not sure if it is such a good approach. The most general vector linear in $r$ is $M r$ where $M$ is a 3x3 matrix. The number of unfixed constants in your formula is 12, while you only need 9 in general. If you set $b=c$ it seems more sensible, assuming $r\times c\neq0$. Then it's essentially just the expansion of a vector in the basis $r$, $c$, $r\times c$.

The matrix $M$ can be divided into 3 parts, the trace part, a traceless symmetric part and a antisymmetric part: $$ M = \frac13\mathrm{tr}(M)\,I + \frac12(M+M^t-\frac23\mathrm{tr}(M)\,I) + \frac12(M-M^t) $$ where $I$ is the identity matrix and ${}^t$ denotes transpose.

So, the terms in $M\cdot r$ correspond to the terms $$ x r + y (r\cdot a)c + z (r\times c)\ .$$ You can see exactly how by just taking the gradient of both sides to get $$ M = x\, I + y\, c a^t + z\, \varepsilon\cdot c$$ where $\varepsilon$ is the totally antisymmetric tensor and $c a^t$ can be chosen to be traceless, ie $a \cdot c=0$. In terms of components, this reads $$ M_{ij} = x\, \delta_{ij} + y\, c_i a_j + z\, \sum_k\varepsilon_{ijk} c_k \ ,$$ where $\delta$ is the Kronecker delta symbol.

share|improve this answer
    
There's something not quite right about this.... –  Simon Feb 28 '11 at 4:46
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.