QuantumJG's first year uni maths revision thread.

[QuantumJG]

Show that if A is an m x n matrix and A(BA) is defined, then B is an n x m matrix.

Firstly let the size of B = y x z

if A(BA) is defined, then BA is defined => z = m

therefore the size of matrix BA is y x n, so for A(BA) to be defined => y = n

=> size of B is n x m. 

Is that an adequate proof? 

[jimmy999]

If you wanted to extend it you could say the number of columns of the first one must equal the number of rows of the second one. Something like that. I'm not exactly sure how detailed proofs need to be

[QuantumJG]

Let matrix , be a matrix that represents all 2 x 2 matrices

Then AB = BA when B is either

or

Show that matrix A which satisfies the above must be a scalar matrix.

[QuantumJG]

A scalar matrix is a matrix where all diagonal entries are equal, but I'm confused as to how to show A is a scalar matrix.

[Voltaire]

why are you reviewing the intro to matrixies stuff..?

you should do like change of basis, orthonormal set's, then diagonalization and conics.
you gotta understand how it all links up etc, dont be doing these isolated defition type Q's

[QuantumJG]

why are you reviewing the intro to matrixies stuff..?

you should do like change of basis, orthonormal set's, then diagonalization and conics.
you gotta understand how it all links up etc, dont be doing these isolated defition type Q's

why are you reviewing the intro to matrixies stuff..?

you should do like change of basis, orthonormal set's, then diagonalization and conics.
you gotta understand how it all links up etc, dont be doing these isolated defition type Q's

Lol. I have already done that.

I'm just going through some proof questions I didn't get during the semester.

[QuantumJG]

This question is really annoying me!

Let A be a square matrix satisfying A² = A and let matrix B be a square matrix the same size as A.

Show that (AB - ABA)² = 0 (obviously the 0 matrix)

The only thing I can think of is trying to prove AB = ABA, which would be easy if ABA = AAB = A²B = AB.

[zzdfa]

(AB-ABA)^2
=(AB-ABA)(AB-ABA)
=AB(AB-ABA)-(ABA)(AB-ABA)        matrix multiplication is distributive
=ABAB-ABABA-ABAAB+ABAABA         and again
=ABAB-ABABA-ABAB+ABABA          A^2=A
=0
sometimes you just have to get your hands dirty.

[zzdfa]

Let matrix , be a matrix that represents all 2 x 2 matrices

Then AB = BA when B is either

or

Show that matrix A which satisfies the above must be a scalar matrix.

Let be the first matrix and be the second.

then you have 2 equations:







Get those equations in terms of a,b,c,d (do the multiplication e.g. )

[QuantumJG]

(AB-ABA)^2
=(AB-ABA)(AB-ABA)
=AB(AB-ABA)-(ABA)(AB-ABA)        matrix multiplication is distributive
=ABAB-ABABA-ABAAB+ABAABA         and again
=ABAB-ABABA-ABAB+ABABA          A^2=A
=0
sometimes you just have to get your hands dirty.

Thanks.

I suck at matrix proofs.

[QuantumJG]

How can you do this distributivity proof?

(αa + βb) x c =αa x c + βb x c

the only thing I can think of is setting,

a = (a₁, a₂, a₃)

b = (b₁, b₂, b₃)

c = (c₁, c₂, c₃)

and showing that the RHS = LHS

is there a neater way? 

[humph]

Depends what you've already proved/can assume beforehand, but in general no, that's the most direct (albeit lengthy) way.

[/0]

Here is a proof for coplanar vectors, from Introduction to Electrodynamics - Griffiths. It also includes the source for the general case. (Which, unfortunately, I don't have)

[kamil9876]

I like that geometry

I agree with humph, was just about to say that it depends what you have proven earlier/take as the definition of cross product. Like for instance if you know the linearity (in a single row) property of determinants and know how determinants relate to it, this becomes trivial.