This one has been bothering me for days, supposedly if A is identity matrix, det(A) has to be 1, does this mean if the determinant of a matrix equals 1, that matrix has to be necessarily the identity matrix?
I don't even know where to start with this... Any ideas?
>>7719569
row 2 and 3 are flipped then row 2' is multiplied by δ and row 3' by γ so det = 1 => det = -δγ
>>7719569
What happens to the determinant if you multiply a row by a constant?
>>7719569
>if the determinant of a matrix equals 1, that matrix has to be necessarily the identity matrix?
det(
[cos(ϕ) sin(ϕ) ]
[-sin(ϕ) cos(ϕ)]
)=1 for all ϕ
>>7719569
Of course not
E.g.:
(-1 0)
(2 -1)
>>7719583
When you co-factor expand that row, all the coefficients now have that constant as a multiple
>>7719581
didn't even see that relation, was just trying to find the values and solve the determinant normally but i only had two variables for one equation, thanks man, forreal.
>>7719590
So the determinant is multiplied by the constant? What about when you swap adjacent rows?
>>7719594
if you multiply any line of a matrix by a constant you also have to multiply the determinant by that constant.
Also, what's the most efficient way to calculate determinants of matrices bigger than 3x3? Like this one.
>>7719594
Oh, and when you swap any lines you have to multiply the determinant by -1 each time you do it.
>>7719602
I think I figured this one out:
Basically the steps to simplify the matrix are
L2-L1
L3-L1
L4-L1
L5-L1
L6-L1
L3-L2
L4-L2
L5-L2
L6-L2
L4-L3
L5-L3
L6-L3
L5-L4
L6-L4
L6-L5
and you finally end up with a diagonal matrix so you can multiply every element of the diagonal and end up with lambda to the power of 6.
Does anyone here know of an operation one can perform on a matrix (like taking the determinant, or any sort of norm) which will return a unique number for all unique matrices (e.g. no multiplicity)? The determinant obviously won't do this, nor will the condition number and all the various matrix norms and induced matrix norms I've tried. Anyone?
What is the good read on matrixes&related stuff?
One that doesn't skip proofs nor come up with too much random shit
>>7719622
Well, [math]\mathbb R^n[/math] is of the same cardinality as [math]\mathbb R[/math], so you can find a bijection between those. I highly doubt there is a well-behaved algebraically useful bijection, though.
>>7719622
>Does anyone here know of an operation one can perform on a matrix (like taking the determinant, or any sort of norm) which will return a unique number for all unique matrices (e.g. no multiplicity)
for a n x m integer matrix A
2^n*3^m*5^[A(1)]*7^[A(2)]*11^[A(3)]*...
for any real n x m matrix
the size n in binary
concatenated with 2
concatenated with m in binary
concatenated with 2
concatenated with sign(A(i)) for i =1 to n*m
concatenated with 2
concatenated with each entry's digits interlaced (padded with zeros as needed) in binary (with no repeating .1111... forever)
>>7719636
And matrices with an infinite number of entries? :D
that's a smart use of multiplicity and primes my friend. Thank you!
>>7719636
mind if I ask for an example? I'd very much appreciate it just to make sure I don't mess it up...
>>7719639
[0 12357]
[3.14159265... -52]
let this resulting number be in base 11 and use A instead of 2 with binary as a separator (I'm fucking lazy)
2 A 2 A 0001 A 0100 0200 0300 0505 0732 . 0010 0040 0010 0050 0090 0020 0060 0050....
>>7719648
You know, that actually really helped me quite a bit. So thank you very very much! Usually I spend like... days, searching for answers to my stupid questions. Weeks if I have to actually figure it out myself. Very much appreciate it, and appreciate your understanding. Thanks!! :D
>>7719638
>And matrices with an infinite number of entries
You need more numbers/elements than are in the set of real numbers to encode all functionals
>>7719657
Not to mention if integers a formula for encoding the primes. ;) That was still a pretty neat use of the fundamental theorem of arithmetic I think... I don't quite need it in the infinite case, but would be an interesting challenge to find a useful bijection (perhaps outside the reals). Functionals, wow haven't heard that term used in a while - I must be out of practice. Awesome. :D
>>7719569
how are you supposed to calculate the determinant of this?
>>7719694
Well there's certainly a pattern to all submatrices, so I'd use a formula like Penrose's. Check out all the formulae on the wiki for determinants.
>>7719694
>determinant
since the matrix is infinite and the entries are constant, the determinant is either 0 or infinite with a probability of (100-epsilon)%
tryinig it out for sizes 1, 2 and 3 you see that its simply lamba^N
>>7719733
The columns and rows both terminate. It's clearly not infinite.
>(100-epsilon)%
>>7719694
Use column operations (subtracting a column doesn't change the det). It's just [math]\lambda^n[/math].
>>7719694
This is one of those "See the trick?" questions.
Subtract lambda off the diagonal, get a matrix with rank 2 (or 1?) So lambda is an eigenvalue of rank N-2 or N-1. You have the trace being the sum of the eigenvalues too. You need one more eigenvalue, then take the product to get the determinant.
If the rank is 1 after subtracting lambda, you are done, but I don't quite see it. I think you can row and column reduce after subtracting, get something... not sure.
>>7719760
I see, but the first column will always stay (λ,1,...,1) you can't triangulate like that. Do you mind to explain more thoroughly?
>>7719768
It becomes a lower triangular matrix. It can be shown by expansion that the determinant of a triangular matrix is just the product of its diagonals.
Hope that helps.
>>7719772
Ah I see, so basically you need one of the halves of the matrix (divided by the diagonal) to only have zeros as entries right?
Yeah man, I think I got it, thanks.
>>7719569
A =
3 4
2 3
Det A = 1
A =/= I
QED
>>7719569
No. There are plenty of examples ITT of non-identity matrices with determinant one, so I'll just give some thoughts.
If det(A) = 1 implies A = I, it would follow that for any positive number R there is a unique matrix M with det(M) = R. But by dividing the top row of M by R and multiplying the bottom row of M by R, we could obtain a new matrix M' with det(M') = R, contrary to the uniqueness of M.
A good way to think about this is to think of a circle of radius 1. Every vector has length one, but that doesn't mean there's only one vector.
Any isometry, such as reflections or rotations, will have determinant equal to 1.
It is however fact possible to build a quotient group where the only element with determinant equal to 1 is the identity.
