Navigating Linear Algebra Ep. 8: Similar Matrices

We start our investigation of similar matrices, as well as study how to find the inverse of an invertible matrix.

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[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.)

[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.

[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!

[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 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!

[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.

[yassinezaoui]

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 p<>0 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.

[yassinezaoui]

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.

[yassinezaoui]

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<---R2 - 2R1

A1= [ 2, 4 ] I1= [ 1, 0 ]
[ 1, 0 ] [ -2, 1 ]

R1<--->R2

A2= [ 1, 0 ] I2= [ -2, 1 ]
[ 2, 4 ] [ 1, 0 ]

R2<---R2-2R1

A3= [ 1, 0 ] I3= [ -2, 1 ]
[ 0, 4 ] [ 5, -2 ]

R2<---1/4 R2

A4= [ 1, 0 ] =I I4= [ -2 , 1 ]=A^-1
[ 0, 1 ] [ 5/4, -1/2 ]

Since det(A)=2*8-5*4=-4 and thanks to the note given, we can verify that A^-1 is indeed what we have found

[axemenace]

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

[ishansingh]

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