As an _empty product,_ which _by convention_ is equal to the corresponding multiplicative identity (a neutral element of multiplication) in every algebraic structure. So you do not really "prove" it, since it's a _convention_ (of an empty product), but it's a fairly reasonable convention.
Similar to why zero factorial is also *0! = 1* (for the same reason, as an empty product, by convention), so you don't have to "prove" it either - it is simply _defined_ that way.
It works within the algebra of matrices as well: *M⁰ = (N−N)⁰ = O⁰ = I₂* where *I₂ = [1 0; 0 1]* is the identity *2×2* matrix (a multiplicative identity of the matrix ring of *2×2* matrices), and *O = [0 0; 0 0]* is a zero *2×2* matrix.
Poor matrix *_N_* - no one cares about her at all. Let's make a question that involves her in a game. We already know what *_M²⁰⁰⁵_* is, but what is *_Mᴺ_* though?🤔
Matrix to the power of another matrix? 😳 (MN)^2005 is also really easy to compute actually, as MN is diagonal; MN = [2^2 0 ; 0 5^2] ==> (MN)^2005 = [2^4010 0 ; 0 5^4010] Now M doesn’t have to be ignored ^ ^
But really, why not? We already know how to evaluate matrix powers *_Aⁿ_* for _integer_ exponents *n ∈ ℤ.* What can stop us from generalizing it to _whatever_ exponents (including matrix ones) in the same way as we did for our beloved _real_ numbers?
@@allozovsky I've seen all kinds of weird and wonderful generalisations, but never ever a matrix to the power of a matrix. Maybe it's something to invent
@@allozovsky Okay I found it: M^N = exp(Nlog(M)); we compute log(M) using the Taylor series expansion for log(I+(M-I)), which will give another matrix. Then multiply by N, and use Taylor series for exponential. There is a problem in this, which is Nlog(M) does not equal log(M)N, so there are 2 different possible answers which is quite frustrating. Also, the series for log(M) only converges if |a| < 1, where |a| is the component of M with the largest absolute value
Yes, u can generalise this to actually find the power of any matrix. Assuming the eigenvalue decomposition exists, X = UDU^(-1) => X^n = UDU^(-1)UDU^(-1)…UDU^(-1) And all the U’s cancel with the U^(-1) in the middle, leaving: X^n = U D^n U^(-1) , And D^n is easy to compute as it is just each diagonal element raised to power n
@@josephlorizzo8997 No worries, I didn't explain it greatly but the D is a diagonal matrix with eigenvalues on the leading diagonal, and U is a matrix with the corresponding eigenvectors as its columns. The assumption it exists falls under the matrix X being full rank, and so doesn't have any eigenvalues equal to 0. Another cool fact: for any analytic f(x) which admits a power series f(x) = a0 + a1x + a2 x^2 + ... , we get the result: f(X) = U f(D) U^(-1), where f(D) is f applied to all diagonal elements. f(x) = x^n gives us the result with the power, f(x) = e^x = 1 + x + x^2/2 + ... enables us to calculate the exponential e^X of a matrix easily. We can compute all sorts such as sin(X), arsinh(X), erf(X), X! (the factorial of a matrix using the Pi function Taylor Series (for X positive semi-definite)), log(X) == log(I + (X-I)) (with care of convergence, only for || X - I || < 1) and the list goes on... this is so cool! and in fact, setting f(x) = x^(-1) also works and this gives another formula to invert X: X^(-1) = U D^(-1) U^(-1) so yeah, your method can be generalised to almost anything to give such amazing results - this is why I love math so much, its full of surprises!
@@adw1z In this case, D = diag(sqrt(10), -sqrt(10)) (diagonal matrix with diagonal elements sqrt(10) and -sqrt(10)) because the solutions for det(M-aI) = 0 are a1 = sqrt(10) and a2 = -sqrt(10), so D^2005 = diag(sqrt(10)^2005, (-sqrt(10))^2005) = (10^1002) * D, therefore M^2005 = (10^1002) * M. For this particular problem, you don't even need to know U.
Option B barely misses to be the second solution, because M² = N², thus their even powers are equal.
I love your Bible verse at the end of each session. God Bless.
Never stop learning ❤
Excellent explanation Sir. Thanks 👍
love these videos.
Can you prove / show how 0 to the power of 0 = 1?
As an _empty product,_ which _by convention_ is equal to the corresponding multiplicative identity (a neutral element of multiplication) in every algebraic structure. So you do not really "prove" it, since it's a _convention_ (of an empty product), but it's a fairly reasonable convention.
Similar to why zero factorial is also *0! = 1* (for the same reason, as an empty product, by convention), so you don't have to "prove" it either - it is simply _defined_ that way.
It works within the algebra of matrices as well:
*M⁰ = (N−N)⁰ = O⁰ = I₂*
where *I₂ = [1 0; 0 1]* is the identity *2×2* matrix (a multiplicative identity of the matrix ring of *2×2* matrices), and *O = [0 0; 0 0]* is a zero *2×2* matrix.
very educational, nice!
Let's Go!!
that was fun to watch :)
Awesome problem
Poor matrix *_N_* - no one cares about her at all.
Let's make a question that involves her in a game.
We already know what *_M²⁰⁰⁵_* is, but what is *_Mᴺ_* though?🤔
Matrix to the power of another matrix? 😳
(MN)^2005 is also really easy to compute actually, as MN is diagonal;
MN = [2^2 0 ; 0 5^2]
==> (MN)^2005
= [2^4010 0 ; 0 5^4010]
Now M doesn’t have to be ignored ^ ^
@@adw1z Why not? We, math lovers, can generalize any operation to whatever domain of its operands.
But really, why not? We already know how to evaluate matrix powers *_Aⁿ_* for _integer_ exponents *n ∈ ℤ.* What can stop us from generalizing it to _whatever_ exponents (including matrix ones) in the same way as we did for our beloved _real_ numbers?
@@allozovsky I've seen all kinds of weird and wonderful generalisations, but never ever a matrix to the power of a matrix. Maybe it's something to invent
@@allozovsky Okay I found it:
M^N = exp(Nlog(M));
we compute log(M) using the Taylor series expansion for log(I+(M-I)), which will give another matrix. Then multiply by N, and use Taylor series for exponential.
There is a problem in this, which is Nlog(M) does not equal log(M)N, so there are 2 different possible answers which is quite frustrating. Also, the series for log(M) only converges if |a| < 1, where |a| is the component of M with the largest absolute value
I managed to solve it before I watch the full solution! I guess I'm not as rusty as I thought I was.
Cool.
Wow
Hi sir
can't It be done with diagolanization? i mean in general, this Is a very special case
Yes, u can generalise this to actually find the power of any matrix. Assuming the eigenvalue decomposition exists,
X = UDU^(-1)
=> X^n = UDU^(-1)UDU^(-1)…UDU^(-1)
And all the U’s cancel with the U^(-1) in the middle, leaving:
X^n = U D^n U^(-1) ,
And D^n is easy to compute as it is just each diagonal element raised to power n
@@adw1z perfect, then what i thought was actually true, thank you very much
@@josephlorizzo8997 No worries, I didn't explain it greatly but the D is a diagonal matrix with eigenvalues on the leading diagonal, and U is a matrix with the corresponding eigenvectors as its columns. The assumption it exists falls under the matrix X being full rank, and so doesn't have any eigenvalues equal to 0.
Another cool fact: for any analytic f(x) which admits a power series f(x) = a0 + a1x + a2 x^2 + ... , we get the result:
f(X) = U f(D) U^(-1),
where f(D) is f applied to all diagonal elements. f(x) = x^n gives us the result with the power, f(x) = e^x = 1 + x + x^2/2 + ... enables us to calculate the exponential e^X of a matrix easily. We can compute all sorts such as sin(X), arsinh(X), erf(X), X! (the factorial of a matrix using the Pi function Taylor Series (for X positive semi-definite)), log(X) == log(I + (X-I)) (with care of convergence, only for || X - I || < 1) and the list goes on... this is so cool!
and in fact, setting f(x) = x^(-1) also works and this gives another formula to invert X:
X^(-1) = U D^(-1) U^(-1)
so yeah, your method can be generalised to almost anything to give such amazing results - this is why I love math so much, its full of surprises!
@@adw1z In this case, D = diag(sqrt(10), -sqrt(10)) (diagonal matrix with diagonal elements sqrt(10) and -sqrt(10)) because the solutions for det(M-aI) = 0 are a1 = sqrt(10) and a2 = -sqrt(10), so D^2005 = diag(sqrt(10)^2005, (-sqrt(10))^2005) = (10^1002) * D, therefore M^2005 = (10^1002) * M. For this particular problem, you don't even need to know U.
@@jlmassiryes that’s a very nice spot, how D^2005 brings back a multiple of itself, and so we get back a UDU^-1 term which is just M!