I loved the way you chose a problem that would introduce some "awkward" cases, needing special consideration at the end. I must admit, I was waiting to see what advantage you would gain from writing v2 in a fractional form, where I would have chosen x=1 to avoid that.
The fraction would've re-appeared in P^-1 I believe since you'd divide by the determinant which wouldn't be one anymore, his choice of eigenvector seems to have just been to make the inverse (and maybe P^-1 itself) look nicer.
Since your matrices are upper triangular, you can avoid some steps. First, the eigenvalues of a triangular matrix are just the diagonal entries of the matrix (a fact worth pointing out), so they could be read off immediately if you want them. Second, you do not really need diagonalization at all to find the powers of the matrix, since it is easy to find and prove (by induction) a formula for the powers of a 2x2 triangular matrix: [[x,y],[0,z]]^k = [[x^k, y(x^k-z^k)/(x-z)],[0,z^k]] if x and z are distinct, and [[x,y],[0,x]]^k = [[x^k, ykx^(k-1)],[0,x^k]] if x=z.
There’s an alternative method for computing A^p for the matrix A = [[n, 1],[0,-2]] when n ≠ -2. The eigenvalues of A are n and -2, so we know that A^p = PD^pP^(-1) where D = [[n, 0],[0,-2]] but we need not compute P and P^(-1) explicitly. We just need to know that A^p = n^p B + (-2)^p C where B = P [[1, 0],[0,0]] P^(-1) and C = P [[0, 0],[0,1]] P^(-1). Setting p = 0 gives a first equation E0 : B + C = I (identity matrix) Setting p = 1 gives a second equation E1 : nB -2 C = A Combining the equations : 2E0 + E1 gives (n+2)B = 2I + A then nE0 - E1 gives (n+2)C =n I - A thus we have identified B and C without computing P and P^(-1).
There is another way to solve this, which is perhaps a bit faster, as long as 2n is a positive integer. If it is not, the meaning of the 2n th power is in any case not uniquely defined, since ((1)^{2n} is then already non-uniquely defined. For positive integer 2n, just split the matrix into the sum: {{n,1},(0, -2}}= D + N of a diagonal part D= {{n,0},(0, -2}} and a nilpotent part N:= {{0,1},(0, 0}}. Raising this to the (2n)th power consists of raising D to the (2n)th power, giving the diagonal matrix D^{(2n)} = {{n^{2}), (-2)^{2n}} and adding the sum, from j=0 to 2n -1, of the products D^j N D^{2n-j} = {{0, - 2^(2n-1) (n/2)^j}}, which is a finite geometric series, easily summed. The resulting matrix is the same as yours: {{n^{2n}, (n^{2n} - 4^n)/n+1, {0, 4^n}}. Setting it equal to {{a,b},{0, b- 5b}} gives the special solutions you found.
He has analogous understood in group theory in "abstract Algebra" coursework you'll have to take in your future to be complete in "group theory" that includes matrix groups. The 0 identity element in a set has to be 0^0 = 1 analogous to permutation of 0 choose 0. Contrary to false teachers saying 0^0 not being 1 have not included the fact that 0 is an "even integer" group element where now -1 is the smallest negative odd integer and 0 is the smallest even integer before negative integer sets.
The empty product is always equal to the multiplicative identity, just like an empty sum is always equal to 0. But in this case, the determinant is equal to -2n, so the only case where it's non-invertible is when n=0. In that case [[n,1],[0,-2]]^2n = [[0,1],[0,-2]]^0 is the empty product, and so is the identity.
I can tell this is meant for more advanced learners, but I'm wondering what connection exists between eigen values/vectors and diagonal matrices. When it can be diagonalised and such. It looks like a powerful tool to me.
No, it's basic linear algebra (at least in my country). The link between diagonal matrices and eigenvalues is the minimal polynomial: the matrix is diagonalizable if and only if its minimal polynomial is single-root split.
I think it doesn't really matter. The fraction only means you can't have n=-2. But at that point we already disallowed n=-2 in order to be able to diagonalise. We still have to check -2 manually, and we wouldn't get different solutions from using different eigenvectors (It would be wise to choose a simpler eigenvector however equally less educational for this video)
Something about raising a matrix with 0 determinant to the power 0 and calling it the identity doesn’t sit right with me. But I can’t quite put my finger on why that is.
He got that from understanding the limit as x--> 0 of x^x concepts forming 0^0 limit actually going to 1 proof analogy. Many TH-cam math teachers are inappropriately defining 0^0 not defined when it has an actual limit 1 all the time I've seen. It is solved in many other professors teaching L'Hopital's Rule and, yes, it can be applied by knowing "0" is an even integer to combine with limit of x^x goes to 1 as x approaches 0 from the negative xs and the positive xs directions ... An advanced numerical analysis acceptance (I'll leave such math abstract math concepts for your further studies)!!
Plugging your own paper in a youtube comment is pretty low. Having it on viXra doesn't make you look a lot better, and straight up writing what an LLM told you in there casts an especially big shadow on you
Your eigenvector and eigenvalue and PDP^-1 and [0]^0 = I knowledge is astounding as a math professor. I see you understand "group theory" and abstract Algebra courses along with Linear Algebra advanced topics to know I have attended a lecture from a senior level mathematics major, if not graduate level, professor. From Abstract Algebra permutations theory in sets of "nothing choose nothing being equal to the identity element in sets" I can see your brilliance in the zero matrix to the power of zero equal to the identity matrix I people will have to learn (if they haven't 😂)! By the way, from abstract Algebra coursework: f(x) = x^x does satisfy 0^0 = 1 to maintain e^(0) = 1 + all x multiplied 0s in series expansion math. Teachers are inappropriately forgetting that 0 is in the set of "even integers" when combining with the set of "odd integers" in group theory to think 0^0 is anything other than the identity of multiplication element "1" in integers mathematics (that is in additive identities forgetful mathematicians trying to say 0^0 = 1 😂)!!
@@theblinkingbrownie4654 ... Stay in class student! Stop being a bully and belittle people who have major university academics! I wasn't here for English coursework! I'm a 3.5+ graduate with getting Cs in all my English classes. So don't make another major graduate feel unsuccessful in life doing English not so well but math and science work with As and Bs to build up a University GPA again feeling good! It's like belittling an outstanding medical doctor writing prescriptions poorly he assigns that work to secretaries so that English knowledged professions (and nurses) don't have to be belittled for how really they aren't otherwise needed in a doctor's place of occupation. 🤯👎
![[LIVE] : ONE ลุมพินี 84 วันนี้!! คู่เอก "ก้องศึก vs เมืองไทย"](http://i.ytimg.com/vi/3kuzsAIDf1c/mqdefault.jpg)
I loved the way you chose a problem that would introduce some "awkward" cases, needing special consideration at the end. I must admit, I was waiting to see what advantage you would gain from writing v2 in a fractional form, where I would have chosen x=1 to avoid that.
The fraction would've re-appeared in P^-1 I believe since you'd divide by the determinant which wouldn't be one anymore, his choice of eigenvector seems to have just been to make the inverse (and maybe P^-1 itself) look nicer.
@@raddish4440 Good point! I admit I just watched the video and didn't actually work through the alternative approach.
This is such a good mastery question for a linear algebra course. You give a really wonderful presentation of technique.
Since your matrices are upper triangular, you can avoid some steps. First, the eigenvalues of a triangular matrix are just the diagonal entries of the matrix (a fact worth pointing out), so they could be read off immediately if you want them. Second, you do not really need diagonalization at all to find the powers of the matrix, since it is easy to find and prove (by induction) a formula for the powers of a 2x2 triangular matrix:
[[x,y],[0,z]]^k = [[x^k, y(x^k-z^k)/(x-z)],[0,z^k]] if x and z are distinct, and [[x,y],[0,x]]^k = [[x^k, ykx^(k-1)],[0,x^k]] if x=z.
De que putas hablas ue
It's been a few years since I studied linear algebra and this was a really nice refresher. Thanks!
There’s an alternative method for computing A^p for the matrix A = [[n, 1],[0,-2]] when n ≠ -2. The eigenvalues of A are n and -2, so we know that A^p = PD^pP^(-1) where D = [[n, 0],[0,-2]] but we need not compute P and P^(-1) explicitly. We just need to know that A^p = n^p B + (-2)^p C where B = P [[1, 0],[0,0]] P^(-1) and C = P [[0, 0],[0,1]] P^(-1). Setting p = 0 gives a first equation E0 :
B + C = I (identity matrix)
Setting p = 1 gives a second equation E1 :
nB -2 C = A
Combining the equations : 2E0 + E1 gives (n+2)B = 2I + A then nE0 - E1 gives (n+2)C =n I - A thus we have identified B and C without computing P and P^(-1).
Very nice video, featuring many areas of Linear Algebra. I'd love to see more of this kind.
There is another way to solve this, which is perhaps a bit faster, as long as 2n is a positive integer.
If it is not, the meaning of the 2n th power is in any case not uniquely defined, since ((1)^{2n}
is then already non-uniquely defined.
For positive integer 2n, just split the matrix into the sum: {{n,1},(0, -2}}= D + N
of a diagonal part D= {{n,0},(0, -2}} and a nilpotent part N:= {{0,1},(0, 0}}.
Raising this to the (2n)th power consists of raising D to the (2n)th power, giving the diagonal
matrix D^{(2n)} = {{n^{2}), (-2)^{2n}} and adding the sum, from j=0 to 2n -1, of the products
D^j N D^{2n-j} = {{0, - 2^(2n-1) (n/2)^j}}, which is a finite geometric series, easily summed.
The resulting matrix is the same as yours: {{n^{2n}, (n^{2n} - 4^n)/n+1, {0, 4^n}}. Setting it equal to
{{a,b},{0, b- 5b}} gives the special solutions you found.
I think raising matrix M to power k ≤ 0 isn't well defined when M is non-invertible, I'd prefer defining M⁰ = I only for invertible mattices
He has analogous understood in group theory in "abstract Algebra" coursework you'll have to take in your future to be complete in "group theory" that includes matrix groups. The 0 identity element in a set has to be 0^0 = 1 analogous to permutation of 0 choose 0. Contrary to false teachers saying 0^0 not being 1 have not included the fact that 0 is an "even integer" group element where now -1 is the smallest negative odd integer and 0 is the smallest even integer before negative integer sets.
The empty product is always equal to the multiplicative identity, just like an empty sum is always equal to 0.
But in this case, the determinant is equal to -2n, so the only case where it's non-invertible is when n=0.
In that case [[n,1],[0,-2]]^2n = [[0,1],[0,-2]]^0 is the empty product, and so is the identity.
I can tell this is meant for more advanced learners, but I'm wondering what connection exists between eigen values/vectors and diagonal matrices. When it can be diagonalised and such. It looks like a powerful tool to me.
No, it's basic linear algebra (at least in my country).
The link between diagonal matrices and eigenvalues is the minimal polynomial: the matrix is diagonalizable if and only if its minimal polynomial is single-root split.
Surprisingly, entertained.
Is there a reason to not go for v_2 = [ -1, n+2 ] to avoid the fractions?
I think it doesn't really matter. The fraction only means you can't have n=-2. But at that point we already disallowed n=-2 in order to be able to diagonalise.
We still have to check -2 manually, and we wouldn't get different solutions from using different eigenvectors
(It would be wise to choose a simpler eigenvector however equally less educational for this video)
@@cedrichartz390 A good question, well answered by @raddish4440 in response to me above.
Something about raising a matrix with 0 determinant to the power 0 and calling it the identity doesn’t sit right with me. But I can’t quite put my finger on why that is.
He got that from understanding the limit as x--> 0 of x^x concepts forming 0^0 limit actually going to 1 proof analogy. Many TH-cam math teachers are inappropriately defining 0^0 not defined when it has an actual limit 1 all the time I've seen. It is solved in many other professors teaching L'Hopital's Rule and, yes, it can be applied by knowing "0" is an even integer to combine with limit of x^x goes to 1 as x approaches 0 from the negative xs and the positive xs directions ... An advanced numerical analysis acceptance (I'll leave such math abstract math concepts for your further studies)!!
nice video! you are using in an elegant wah the tools provided by Liner Algebra
Thanks very much👍
Meanwhile, I think this is an important news for any mathematician:
Division by Zero 1/0 = 0/0 = 0 is applied in many fields:
viXra:2402.0068 submitted on 2024-02-14 21:47:20 ,
Division by Zero 1/0 = 0/0 = 0 and Computers Real.div: New Information and Many Applications
Plugging your own paper in a youtube comment is pretty low. Having it on viXra doesn't make you look a lot better, and straight up writing what an LLM told you in there casts an especially big shadow on you
Where were you 15 years ago when I was struggling with this lol
Your eigenvector and eigenvalue and PDP^-1 and [0]^0 = I knowledge is astounding as a math professor. I see you understand "group theory" and abstract Algebra courses along with Linear Algebra advanced topics to know I have attended a lecture from a senior level mathematics major, if not graduate level, professor. From Abstract Algebra permutations theory in sets of "nothing choose nothing being equal to the identity element in sets" I can see your brilliance in the zero matrix to the power of zero equal to the identity matrix I people will have to learn (if they haven't 😂)!
By the way, from abstract Algebra coursework: f(x) = x^x does satisfy 0^0 = 1 to maintain e^(0) = 1 + all x multiplied 0s in series expansion math. Teachers are inappropriately forgetting that 0 is in the set of "even integers" when combining with the set of "odd integers" in group theory to think 0^0 is anything other than the identity of multiplication element "1" in integers mathematics (that is in additive identities forgetful mathematicians trying to say 0^0 = 1 😂)!!
I try to avoid correcting grammar but in this case I am having a lot of trouble trying to understand what you are saying, so just letting you know
@@theblinkingbrownie4654 ... Stay in class student! Stop being a bully and belittle people who have major university academics! I wasn't here for English coursework! I'm a 3.5+ graduate with getting Cs in all my English classes. So don't make another major graduate feel unsuccessful in life doing English not so well but math and science work with As and Bs to build up a University GPA again feeling good! It's like belittling an outstanding medical doctor writing prescriptions poorly he assigns that work to secretaries so that English knowledged professions (and nurses) don't have to be belittled for how really they aren't otherwise needed in a doctor's place of occupation. 🤯👎
UR IQ is extremely high...