Proving Properties of 2x2 Matrices

In summary, In part 1, A is a matrix of order 2x2 whose main diagonal's entries' sum is zero. In part 2, given: A^2 is a scalar matrix and the sum of the entries of the main diagonal of AB-BA is zero. In part 3, given: C is a matrix of order 2x2. Prove that the sum of the entries of the main diagonal of AB-BA is equal to C (AB-BA).f
  • #1
1. A is a matrix of order 2x2 whose main diagonal's entries' sum is zero. Prove that A^2 is a scalar matrix.

2. Given: A and B are two matrices of order 2x2. Prove that the sum of the entries of the main diagonal of AB-BA is zero.

3. A, B and C are three matrices of order 2x2. Given: A^2 is a scalar matrix and the sum of the entries of the main diagonal of AB-BA is zero. Prove that C (AB-BA) ^2= (AB-BA) ^2*C?
 
  • #2
Hi and welcome to MHB! What have you tried for each of these? Which one would you like to start with?
 
  • #3
1. A is a matrix of order 2x2 whose main diagonal's entries' sum is zero. Prove that A^2 is a scalar matrix.
A is of the form $\begin{bmatrix}a & b \\ c & -a\end{bmatrix}$ What do you get when you square that?


2. Given: A and B are two matrices of order 2x2. Prove that the sum of the entries of the main diagonal of AB-BA is zero.
Let $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and let $B= \begin{bmatrix}p & q \\ r & s\end{bmatrix}$. What is AB- BA?

3. A, B and C are three matrices of order 2x2. Given: A^2 is a scalar matrix and the sum of the entries of the main diagonal of AB-BA is zero. Prove that C (AB-BA) ^2= (AB-BA) ^2*C?
If $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ then $A^2= \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}$. Since $A^2$ is a "scalar matrix" $A^2= \begin{bmatrix}P & 0 \\ 0 & P^2\end{bmatrix}$ so $a^2+ bc= bc+ d^2= P$, $ab+ bd= ac+ cd= 0$.
 
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  • #4
A is of the form $\begin{bmatrix}a & b \\ c & -a\end{bmatrix}$ What do you get when you square that?



Let $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ and let $B= \begin{bmatrix}p & q \\ r & s\end{bmatrix}$. What is AB- BA?


If $A= \begin{bmatrix}a & b \\ c & d\end{bmatrix}$ then $A^2= \begin{bmatrix}a^2+ bc & ab+ bd \\ ac+ cd & bc+ d^2\end{bmatrix}$. Since $A^2$ is a "scalar matrix" $A^2= \begin{bmatrix}P & 0 \\ 0 & P^2\end{bmatrix}$ so $a^2+ bc= bc+ d^2= P$, $ab+ bd= ac+ cd= 0$.

Thank you very much for the response, sir.
In regard to #2 -- I calculated AB-BA.
However, I did not see how the sum of the two entries in the diagonal equal zero.

I obtained the following entries:

a11= rb-cp
a22=qc-dp

How should I proceed?
 
  • #5
Thank you very much for the response, sir.
In regard to #2 -- I calculated AB-BA.
However, I did not see how the sum of the two entries in the diagonal equal zero.

I obtained the following entries:

a11= rb-cp
a22=qc-dp

How should I proceed?
You need to look at that again.
I get \(\displaystyle (AB - BA)_{11} = (ap + br) - (ap + cq) = br - cq\) and \(\displaystyle (AB - BA)_{22} = (cq + ds) - (br + ds) = cq - br\)

-Dan
 
  • #6
You need to look at that again.
I get \(\displaystyle (AB - BA)_{11} = (ap + br) - (ap + cq) = br - cq\) and \(\displaystyle (AB - BA)_{22} = (cq + ds) - (br + ds) = cq - br\)

-Dan
I obtained the same result.
Thank you very much :)

Also, my bad. The "given" part in the 3rd question is not given.
The question is written as follows:

A, B and C are three matrices of order 2x2. Prove that C (AB-BA) ^2= (AB-BA) ^2*C.
You can use the results obtained in parts 1 and 2 in order to solve part 3.
Does this mean that A and B in part 3 are the same matrices used in parts 1 and 2 (as in, matrix's A diagonal's entries' sum is zero)?
 

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