The fact is almost usual in mathematical theory of elasticity where the double dot product of second order tensors is the work of stress tensor for the strain tensor. In such a case, it does not matter what the matrices rank is, when the product is zero it means that the matrices are orthogonal 'with respect to the energy' i.e. the stress (say the first matrix) makes no work for the strain (the second one). The product is zero when the first matrix is symmetric and the second is skew-symmetric, for example, but not only. You can have a diagonal matrix for a pure shear one and the product is zero as well
The inner product (dot product or scalar product) of two matrices (think vectors in this case) can be visualized as the 'projection' of one matrix onto another (in vector terms: multiply the length of the first vector times the length of the second vector times the cosine of the angle between the two vectors). When the angle between the vectors is 90 degrees (i.e., the vectors are orthogonal to each other), the cosine of 90 degrees is zero, so the resulting product (vector 1 X vector 2 X cos of angle between vector 1 and vector 2) is also zero. Just think of a matrix as a n-dimensional set of vectors, where the number of rows in the matrix is n and the columns of the matrix represent the components of the vector (e.g. x, y, z, ...)
Taking the inner product of two matrices (or vectors) results in a numeric (scalar, number) value. In the case that this inner product is zero, you know that the two objects (matices, vectors) are orthogonal (mutually perpendicular) to each other. In the case where the inner product is zero, the matrices (vectors) are linearly independent and form a basis set which 'spans' the space, meaning that every vector can be expressed as a linear combination of the basis set (the vectors/matrices that you started with.
Here is a link that might help.
http://mathworld.wolfram.com/DotProduct.html
Another link that explains things.
http://www.millersville.edu/~bikenaga/linear-algebra/basis/basis.html
@Mehdi: You should doublecheck whether you have actually to consider what you write here. What you write is that you transform matrices to vectors by simply concatenating their row-vectors and then cosider the normal dot product of these long vectors. I know of no problem which requires this queer handling of matrices. So my guess is that you misunderstood something. If I'm wrong here, you are in a good position: you then have only to understand what it means for two vectors that their dot product vanishes. I guess you know this already. If not, Eric's explanation may guide you.
@Mehdi, the way you define dot product is telling us that it is not necessary to take matrix A=[[a1,a2],[a3,a4]] (byrow). Instead you could simply take the vector va=[a1,a2,a3,a4] and proceed. As for the PDE which kind of conversion do you use?
@Mehdi. Sorry that a part of what I wrote was nonsense. Actually, to consider your kind of dot product is common in discussing n*n matrices as models of tensor products of n dimensional vectors. Normally your scalar product then is written
as Trace(a^t *b) where * means the matrix product and ^t means transposition.
For complex spaces, matrices, vectors, ... the transposition is replaced by Hermitean conjugation and the n*n matrices then form a n*n-dimensional Hilbert space. With respect to your PDE example I would ask the same question as Demetris.
Thanks to all friends.
But, my problem is still unresolved. However, I have noticed very good points here.
My main problem is a geometric interpretation.
Think about the columns of the two matrices as vectors in R^n: v_1,...,v_n for the first matrix A and w_1,...,w_n for the second matrix B. What you are computing is then the sum of dot products
v_1 \cdot w_1 + ... + v_n \cdot w_n
When all the vectors are of unit length then you are computing the sum of the cosines of the angles between the vectors. Thus
cos(a_1) + ... + cos(a_n)
where a_i is the angle between the vectors v_i and w_i.
That is the geometric interpretation
For the multiplication of matrices (inner product of matrices) the following conditions should be taken into account:
1- The number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.
2- The result will have the same number of rows as the 1st matrix and the same number of columns as the 2nd matrix.
If a matrix where all elements are zero is obtained by multiplying two matrices, you have then obtained the "null matrix". If the number of rows is 1 then you have obtained the "null vector" which is the geometrical representation of the number "0" on the plane (if you have two columns), or the null vector in the space (if you have three columns) or in the space of dimension "n" (if it has n columns). What these matrices and vectors represents, will depend on the type of problem that you want to resolve.
@Jorge.
As is clear from a post of the owner of this question, the 'inner product' under consideration here is not the matrix product, but the trace of it. So your thoughts don't refer to the question under consideration.
Dear Ulrich
May I said that at the end of my reply I mentioned "What these matrices and vectors represents, will depend on the type of problem that you want to resolve". This means that if the result of the inner product of two matrices is 0 you can not say anything from the scientific point of view unless you know the problem is trying to solve or under consideration. This is the main part of my reply.
@Eric Hall: I'm sorry, but I don't agree that AB = 0 means that A and B (as matrices) are orthogonal. This I think holds only for vectors, but not for n x n-matrices or n x m-matrices (if n > 1 in the first case and n and/or m > 1 in the second case).
the reason is the following: AB=0 indicates that:
either A or B = 0 (so a matrix with only "0" as entries)
or that A and B have no inverse A(^-1) or B(^-1), because then it would hold:
A(^-1)AB = 1B = 0 (so B can only be made up of "0" due to the definition of the identity-matrix 1)
the same holds for ABB(^-1) = A1 = 0
if the matrices would be orthogonal then, per definition, it has to hold that the transposed matrix = the inverse matrix. But this directly contradics my short proof above. Hence they cannot be orthogonal.
I hope, I didn't overlook something here - if I did, please let me know. But if I am right then you can at least say: "it means that the matrices are not orthogonal ones".
@Johannes Gruenwald: you are right. but i wrote in my question where Rank(A)=Rank(B)
Dear Mehdi,
sorry, but what do you mean by "not full rank"? - do you mean that there is one line or row in you matrix which has no entries? - then your problem is trivial.
Furthermore, your definition of the inner product is unconventional: usually it is defined (for n x n-matrices, for n > 1) as: C = AB = c(ij) = sum(k)[a(ik)b(kl)] which means "multiply each entry of the first line of A with each entry of the first ROW of B), this gives you a new matrix
but you had: AB = a11b11 + a22b22 + ... which gives you a number, but using matrix multiplication should give you another matrix which should again be diagonal. This cannot have finite entries as you want to have AB = 0.
So with your matrices A, B definitions (not zero) this problem is not solvable with the usually defined matrix multiplikation.
If you use your definition (which I have never seen before, to be honest) - it works only for diagonal matrices and you will get a linear combination of your diagonal elements. This linear combination can only be 0, if all coefficients vanish (i.e. your entries a(ik) or your b(kl)), per definition, which again means that the entries of your second matrix are linear independent.
BUT: AB also vanishes, if you can express some a(ik)b(kl) by linear combination of your other terms (which is also possible), for example:
diagA=(1, 1, 1) and diagB=(1,1,-2) --> AB (according to your definition) = 1 + 1 - 2 = 0
and this would be the definition of linear dependence.
Hence, in order to solve your puzzle, AB and your definition have to be linear dependent and linear independent at the same time, which is impossilbe. Therfore, I would say, your problem is not solvable...
Dear Johannes
The point is that: the problem is solvable.(Attachment)
inner product in two vector is a number, why inner product of two matrices is a matrix?
What do you think about Vec(A)?
There are various definitions of the product of matrices involved here that differ from the usual matrix product:
One is the Hadamard product, see
http://en.wikipedia.org/wiki/Hadamard_product_%28matrices%29
The other is the Frobenius inner product, see
http://en.wikipedia.org/wiki/Frobenius_inner_product#Frobenius_product
This is related to the "regarding the matrices as vectors and take the usual inner product of the vectors" and maybe represented by taking a suitable trace, both of which have been mentioned above. It may also be regarded as a sum over all entries in the Hadamard product. With respect to this inner product, one can define orthogonality in the usual way. This inner product of matrices is a number, not a matrix. And, of course, it is possible to define the "length" of a matrix A by a Frobenius norm, and the "angle" between to matrices via cos alpha = (A,B) / sqrt( (A,A) (B,B) ) where (A,B) denotes the Frobenius product of A and B.
In order to see more clearly the application to PDE as stated by Mehdi, it would be helpful to know what are the definitions of the matrices A_4x4 and B_4x4.
Dear Mehdi,
because there are only two multiplication for matrices defined (as far as I know):
1) matrix multiplication which I explained earlier
2) scalar multiplication - but here a number is multiplied by a matrix
an inner vector product is a special case where you can map two vectors onto a number. The most easy geometrical interpretation is this:
if you make the inner product of two vectors, you multiply their values with the cos of the angle between them - this is clearly a number.
if you multiply a matrix with a vector you basicaly rotate the vector and hence you end up with a vector
if you multiply two matrices you make two separate rotations around two seperate axis, but this can also (often) be done be one rotation around a third axis - so you end up with a matrix again
ok, Herbert Homeier was faster than me (he posted while I was typing my anser) :)
I didn't know that, but if this is a Frobenius product then it indicates that the tr(AB) = 0
that again would mean that you can express one of the components of the diagonal elements by the two others and that should indicate linear dependence, shouldn't it?
Dear Mehdi
What I try to explain is that from the mathematical point of view of, the result of the inner product of two matrices is another matrix which can be the “null matrix”. If the matrices are of the type V = (v1, v2,... vn) and U = (u1, u2,... un), the inner product of these two matrices (in this particular case vectors) will be V.U = v1.u1 + v2 u2 +... vn. un which is a number (scalar). If you get V. U = 0 in spaces of two and three dimensions that means that vectors are perpendicular to each other. For higher dimensions of the space the concept of perpendicularity lose its geometric interpretation. The interpretation of the result of the inner product of these two vectors will depend on the specific problem you want to solve.
Finally, if the inner product of two matrices is another matrix what representing this matrix will also depend on the problem that you try to resolve. If this matrix is the “null matrix”, then from the point of view of mathematics the two matrices are orthogonal. This is an abstract concept without a geometric representation in a space of more than three dimensions. In real live what the “null matrix” obtained as result of the inner product means will dependent of the specific problem you want to resolve.
Mehdi,
you were asked by three of the contributors what the connection between partial differential equations and matrices in your example is. Don't you think that the effort of the contributors should not go unhonored in this respect?
Dear Ulrich
I am very thankful from contributors.
I'm Sorry, I do not understand what you mean. Do i disrespected to someone?
Dear Mehdi,
don't worry, there seems to be a little language problem. Please recall the first sentence of my previous message. It says what the matter is about. My second sentence probably is formulated a bit too pathetical (a weakness of my style which I'm aware of) and can be ignored. In more dry words is asks you to try to give an answer to the question of these 3 gentlemen (which include me). We then would be happy and would not feel disrespected in any way.
The fact is almost usual in mathematical theory of elasticity where the double dot product of second order tensors is the work of stress tensor for the strain tensor. In such a case, it does not matter what the matrices rank is, when the product is zero it means that the matrices are orthogonal 'with respect to the energy' i.e. the stress (say the first matrix) makes no work for the strain (the second one). The product is zero when the first matrix is symmetric and the second is skew-symmetric, for example, but not only. You can have a diagonal matrix for a pure shear one and the product is zero as well
The orthogonality defined by A:B=0 is the same orthogonality of vectors say the 'component' of the one on the direction of the other is null. The orthogonality in term of Transpose = Inverse refers, in my opinion, to orthogonal transformation of the matrix operators with respect to the chenges of basis.
@Vincenzo,
you are right - also with your observation that 'orthogonality in terms of Transpose = Inverse' which Johannes brought into the discussion, has nothing to do with the problem under consideration.
Dear Lajos
my definition and your definition are the same.
and, matrices A and B are not positive definite and not inversable and not zero.
Now, Do is the geometric interpretation?
Dear Mehdi, in an abstract sense, there is a geometric interpretation, as several contributors including me have already pointed out at length.
If you want to obtain more specific information regarding your PDE-related problem, the following question is still open:
"In order to see more clearly the application to PDE as stated by Mehdi, it would be helpful to know what are the definitions of the matrices A_4x4 and B_4x4."
- Badges
- Science topic
- Similar topics
- Mathematics