Navigating Linear Algebra Ep. 8: Similar Matrices

[@LetsSolveMathProblems]

The problems for this video can be found in this link: github.com/LetsSolveMathProblems/Navigating-Linear-Algebra/blob/main/Episode%208.pdf. Please free to post solutions to these problems in the comment section, as well as peer-review other people's proofs. :)

[@buxeessingh2571]

I am glad you are back.

[@cantcommute]

For problem 2: A can be written as RE(A) + iIM(A) for the real and complex entries respectively. Suppose a real inverse exists, then A⁻¹A=A⁻¹RE(A) + iA⁻¹IM(A)=I. So we must have A⁻¹IM(A)=0. But rankA⁻¹=n and rankIM(A)>0 hence rankA⁻¹IM(A)>0 (row echelon form of A⁻¹ is I and the product of the echelon forms of matrices C,D has the same rank as their product I think). So rankA⁻¹IM(A)!=0, a contradiction.

[@cantcommute]

For problem 1: We do, in sequence, the operations: scale first row by 1/2, at negative 5 of first row to second row, add second row to first row and finally multiply second row by -1/2. This gives the matrix [[-2,5/4],[1,1/2]] which can be checked to be correct!

[@LetsSolveMathProblems]

For problem 1, this is almost correct: You should have -1/2, not 1/2. Also, you may wish to specify that you are writing "columns first" (i.e., [-2, 5/4] is the first column, not the first row, as is usually done).

[@kiit8337]

Is this help ful for jee

[@legendarynoob6732]

Thank you for this :) Very helpful.

[@cantcommute]

For problem 5: A diagonal matrix with diagonal elements a,b,c... nth's power has diagonal elements aⁿ,bⁿ,cⁿ... and zeros everywhere else. This makes computation nice cause instead of multiplying B n times you get Aⁿ for free and just have to multiply S and S⁻¹ (which is also made a lot easier! AS is just multiplying all rows of S by corresponding diagonal entry of A.)

[@LetsSolveMathProblems]

For problem 5, this is correct.

[@cantcommute]

For problem 6: If it's rank 1 and nilpotent then is must have the form [[0,0],[a,0]] or [[0,b],[0,0]]. We just need to show that there exists a transform S that takes us between these forms and vice versa. To this end, you can do a long computation to then find that the transform [[0,1],[a/b,0]] works when going from [[0,0],[a,0]] to [[0,b],[0,0]], the transform [[1,0],[0,a/b]] works when going from [[0,0],[a,0]] to [[0,0],[b,0]] and the transform [[1,0],[0,b/a]] works when going from [[0,a],[0,0]] to [[0,b],[0,0]]. These should be all the cases.

[@LetsSolveMathProblems]

For problem 6, I did not explicitly check your work, but this is likely correct. :)

[@yassinezaoui4555]

Pb5: This is too helpful to calculate powers of B if B is similar to a diagnoal matrix A since we can write B^m = SA^mS^-1 where S is invertible and powers of a diagnoal matrix are the easiest to compute : if A = diag(a1,...,an) then A^m = diag(a1^m,..,an^m) (this can be proven by induction if needed) then we will need to multiply it by S^-1 to the left and S to the right to end up with B^m.

[@LetsSolveMathProblems]

For problem 5, this is correct.

[@yassinezaoui4555]

Pb3: (a) We can guarantee that A a 2x2 invertible matrix is not similar to a^-1 its inverse if we choose |det(A)|1 since if A is similar to A^-1 => det(A)=det(A^-1) but we know as well det(A^-1)=1/det(A) so det²(A)=1 -det(A)|=1. We can take A form the pb1 since det(A)=-4 (not +/- 1) or we can take other more sipmle exemples such as diagonal matrices : diag(2,1), diag(5,1/4), ... diag(a,b) where |ab|1 (b) Yes , absolutely, we can take any non-null projection p from R² to R² and take its matrix with respect to the usual basis for exemple A=diag(1,0) ( e1|--->e1 and e2|--->0) or diag (0,1) (e1|--->0 and e2|--->e2) Since p0 and p is a projection so p²=p so p^n= id (n=0) , p (n>=1) hence A its matrix can't be nilpotent and can't be invertible as well since the only invertible projection is the identity and for our exemple we can see as well taht det(A)=0.

[@LetsSolveMathProblems]

For problem 3, this is correct.

[@cantcommute]

For problem 3: (a) the elementary matrix [[1/2,0],[0,1]] has inverse [[2,0],[0,1]] but have different traces so they are not similar. (b) the matrix [[1,0],[0,0]] works.

[@LetsSolveMathProblems]

For problem 3, this is correct.

[@cantcommute]

For problem 2: (a) we can write this matrix as A=RE(A)+iIM(A) where we're considering the real and imaginary components respectively. Then if we take an in inverse we have A⁻¹A=A⁻¹RE(A)+iA⁻¹IM(A)=I. This mandates that A⁻¹IM(A)=0, which is impossible (for real matricies) as rankA⁻¹=n while rankIM(A)>0 so rankA⁻¹IM(A)>0 hence A⁻¹IM(A)!=0. (b) Consider S= [[i,1],[0,1]] with inverse S⁻¹= [[-i,1],[0,1]] and A=[[1,1],[0,1]] then B=S⁻¹AS =[[1,0],[-i,1]] so it is possible!

[@LetsSolveMathProblems]

For (a), this is correct. For (b), given that you are writing "columns first", I think your S^{-1} should be [[-i,i],[0,1]]; given this, I think your (b) still works! :)

[@yassinezaoui4555]

Pb 1: let's apply the same elementary operations that lead to I form A to I and we wil end up with A^-1. A= [ 2 , 4 ] I= [ 1 , 0 ] [ 5 , 8 ] [ 0 , 1 ] R2

[@LetsSolveMathProblems]

For problem 1, this is correct.

[@axemenace6637]

At this point, do you have a clearer idea of how many videos in the series are left before you plan to stop?

[@LetsSolveMathProblems]

Yes! I think we will have two more episodes (one on eigenvectors and the other on generalized eigenvectors). Even though many important topics (like inner products) will be left out, I hope these 10 episodes are insightful enough for viewers who are new to abstract linear algebra.

[@ishansingh2391]

Please speak in your normal accent as it seems too much made up accent and the video's quality degrades only and only due to fake accent