Sum of the main diagonal of a matrix

[diogomgf]

Hello every one,

I'm stuck in a math problem that goes as follow:

If the sum of the elements composing the main diagonal of a matrix \(\displaystyle A_{2*2} \) is 0, then \(\displaystyle A^2 \) is a scalar matrix.

With this information, show that for any \(\displaystyle B,C,F \in M_{2*2}\) :

\(\displaystyle (B \cdot C - C \cdot B)^2 \cdot F = F \cdot (B \cdot C - C \cdot B)^2 \).

 

[ksdhart2]

Well, considering that you've made 38 posts to this forum (thus you've definitely read The Rules, right? ), and you've shown no work of your own, I'm forced to assume you have none to shown and are stuck at the very beginning. That's okay, I can work with that. Let's begin by defining the matrices \(B\), \(C\), and \(F\) in the most generic terms possible:

\(\displaystyle B = \begin{bmatrix}
b_{11} & b_{12} \\ b_{21} & b_{22}
\end{bmatrix}, \:

C = \begin{bmatrix}
c_{11} & c_{12} \\ c_{21} & c_{22}
\end{bmatrix}, \:

F = \begin{bmatrix}
f_{11} & f_{12} \\ f_{21} & f_{22}
\end{bmatrix}\)

What does that make the matrix \(BC\)? Bearing in mind that matrix multiplication is (generally) not commutative, what does that make \(CB\)? And what do you get when you do the subtraction \(BC - CB\)? Specifically, how does this result relate to the given fact about scalar matrices? The final step is to review the definition of scalar matrix, to show why matrix multiplication is commutative if one of the matrices involved is a scalar matrix.

 

[diogomgf]

Well, considering that you've made 38 posts to this forum (thus you've definitely read The Rules, right? ), and you've shown no work of your own, I'm forced to assume you have none to shown and are stuck at the very beginning. That's okay, I can work with that. Let's begin by defining the matrices \(B\), \(C\), and \(F\) in the most generic terms possible:

\(\displaystyle B = \begin{bmatrix}
b_{11} & b_{12} \\ b_{21} & b_{22}
\end{bmatrix}, \:

C = \begin{bmatrix}
c_{11} & c_{12} \\ c_{21} & c_{22}
\end{bmatrix}, \:

F = \begin{bmatrix}
f_{11} & f_{12} \\ f_{21} & f_{22}
\end{bmatrix}\)

What does that make the matrix \(BC\)? Bearing in mind that matrix multiplication is (generally) not commutative, what does that make \(CB\)? And what do you get when you do the subtraction \(BC - CB\)? Specifically, how does this result relate to the given fact about scalar matrices? The final step is to review the definition of scalar matrix, to show why matrix multiplication is commutative if one of the matrices involved is a scalar matrix.

\(\displaystyle

BC - CB = \begin{bmatrix} b_{11} \cdot c_{11} + b_{12} \cdot c_{21} & b_{11} \cdot c_{12} + b_{12} \cdot c_{22} \\ b_{21} \cdot c_{11} + b_{22} \cdot c_{21} & b_{21} \cdot c_{12} + b_{22} \cdot c_{22} \end{bmatrix} - \begin{bmatrix} c_{11} \cdot b_{11} + c_{12} \cdot b_{21} & c_{11} \cdot b_{12} + c_{12} \cdot b_{22} \\ c_{21} \cdot b_{11} + c_{22} \cdot b_{21} & c_{21} \cdot b_{12} + c_{22} \cdot b_{22} \end{bmatrix},

\)

\(\displaystyle

BC - CB = \begin{bmatrix} b_{11} \cdot c_{11} + b_{12} \cdot c_{21} - c_{11} \cdot b_{11} - c_{12} \cdot b_{21} & b_{11} \cdot c_{12} + b_{12} \cdot c_{22} - c_{11} \cdot b_{12} - c_{12} \cdot b_{22} \\ b_{21} \cdot c_{11} + b_{22} \cdot c_{21} - c_{21} \cdot b_{11} - c_{22} \cdot b_{21} & b_{21} \cdot c_{12} + b_{22} \cdot c_{22} - c_{21} \cdot b_{12} - c_{22} \cdot b_{22} \end{bmatrix}

\)

It is indeed a matrix where:

\(\displaystyle ( BC - CB )_{11} + ( BC - CB )_{22} = 0 \), so squared it is a scalar matrix.

Thanks for showing me the way.

I got lost because I started solving \(\displaystyle (BC - CB )^2 \cdot F\) without decomposing it into an abstract matrix.