Doubt about the kronecker delta

[jonjacson]

Well I don't understand this equality:

A^{j}_{i}*A_{j}^{k}=\delta_{i}^{k}

It is true because it is the result of a calculation. But assuming it is true ¿What happens if one of the A_{i}^{j} is zero?.

Then it does not matter wich is the value of the A_{j}^{k} the equality will be false because 0*any number= 0, but supposedly it should be 1, due to delta having i=k.

So the point is ¿What happens if the coefficients are zero?¿Is it really true?.

[wisvuze]

What is A i,j ? Are these entries of a matrix? What is the matrix?

[jedishrfu]

i think there are some constraints on the Aij in that Aij = - 1/Aji or something like that right?

Anyway I found this reference:

http://books.google.com/books?id=O2zuQNwRIHsC&pg=PA44&lpg=PA44&dq=Aji*Akj%3D%CE%B4ki&source=bl&ots=6Lekugs6Hh&sig=VJGUjDRxnFalcMVUF08z8FwOs8w&hl=en&sa=X&ei=p-MFT-fcBfKCsAKWstGRCg&sqi=2&ved=0CB0Q6AEwAA#v=onepage&q=Aji*Akj%3D%CE%B4ki&f=false

which indicates that the Aij are not the same matrix as in your formula instead they are inverses.

As an aside: I cut and pasted your formula (markup and all) into google and it found this reference - just amazing.

[jonjacson]

I was making a post but you found my original source XD. So you see the expression, A is simply the coefficients that change one base to another.

¿What would happen if one of that coefficients would be zero?¿Would be true the equality with the kronecker delta?.

[jedishrfu]

speed of light issues my response to wisuze got in first somehow

[wisvuze]

The A i,j' matrix cannot give you a zero vector if you input a basis element, since it represents a map that changes one basis into another. And the bases have the same dimension ( bases for the same space )

[jedishrfu]

The A i,j' matrix cannot give you a zero vector if you input a basis element, since it represents a map that changes one basis into another. And the bases have the same dimension ( bases for the same space )

they could be zero if it was only a scaling transform right? (ie Aij = 0 where i not = j)

[jonjacson]

I was making an example, extraordinarily simple:

e1=(1,0) ; e2=(0,1)

e1'=(2,0); e2'=(0,2)

I see that e' is not orthonormal so the e1'*e1'≠1 or 0 so the expression is not the same as a delta kronecker.

I am going to work with other numerical example and then I will post the result.

*Remark: I am tired of studying this topic without numerical examples to fix ideas, it's really frustrating :( .

[wisvuze]

they could be zero if it was only a scaling transform right? (ie Aij = 0 where i not = j)

By scaling transform, I assume you mean a diagonal matrix? Anyway, these are specific matrices ( ones that take you from the { e_j' } basis to the { e_i } basis ), the matrix is determined by the relations e_j' = a1e1 + ... + anen ( and the ai's then form a column of the matrix ). Whatever this relation is, it could be via scaling , the resulting matrix images of the e_j' 's must be linearly independent and spanning, since the result is another basis for the vector space V. We cannot have a 0 vector in this image ( so a scaling matrix with a 0 factor on the diagonal couldn't be a change of basis matrix from V to V )

[Fredrik]

Recall that the definition of matrix multiplication is $(AB)^i_j=A^i_k B^k_j$ (when the row indices are written upstairs, and the column indices downstairs). So the equality $A^j_i A^k_j=\delta^j_i$ is just saying that the matrix A satisfies $A^2=I$. If this isn't immediately obvious, first note that the left-hand side can also be written as $A^k_j A^j_i$, which according to the definition of matrix multiplication is equal to $(A^2)^k_i$.

There are lots of examples of matrices that satisfy this condition, e.g. the 2×2 matrix
\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}

[Fredrik]

In the book's notation, $A^i_{j'}$ and $A^{i'}_j$ are the row i, column j components of two different matrices. (Zoom in if you find the primes hard to see). I would prefer a notation that uses different symbols instead of just primes. For example, suppose that $\{e_i\}$ and $\{f_i\}$ are two orthonormal bases for a vector space V. Since $\{e_i\}$ is a basis, every $f_i$ is a linear combination of the $e_i$. $$f_i=A^j_i \,e_j.\qquad\text{(3.14)}$$ Since $\{f_i\}$ is a basis, every $e_i$ is a linear combination of the $e_i$. $$e_i=B^j_i\, f_j\qquad\text{(3.15)}$$ Now use (3.14) in (3.15). $$e_i=B^j_i\, f_j=B^j_i\, A^k_j e_k.\qquad\text{(3.16)}$$ Since $\{e_i\}$ is linearly independent, this implies that $$B^j_i\, A^k_j=\delta^k_i.\qquad\text{(3.17)}$$ If we let A denote the matrix with $A^k_j$ on row k, column j, for all k and j, and B the matrix with $B^j_i$ on row j, column i, for all j and i, then (3.17) says that AB=I, or equivalently, that $B=A^{-1}$.