proof related to matrices

[logistic_guy]

Let \(\displaystyle \mathcal{A}\) be the set of \(\displaystyle 2 \times 2\) matrices with real number entries. Recall that matrix multiplication is defined by

\(\displaystyle \begin{bmatrix}a & b \\c & d \end{bmatrix} \begin{bmatrix}p & q \\r & s \end{bmatrix} = \begin{bmatrix}ap + br & aq + bs \\cp + dr & cq + ds \end{bmatrix}\)

Let \(\displaystyle M = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\)

and let \(\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}\)

Prove that if \(\displaystyle P, Q \in \mathcal{B}\), then \(\displaystyle P + Q \in \mathcal{B}\) (where \(\displaystyle +\) denotes the usual sum of two matrices).

 

[khansaheb]

Let \(\displaystyle \mathcal{A}\) be the set of \(\displaystyle 2 \times 2\) matrices with real number entries. Recall that matrix multiplication is defined by

\(\displaystyle \begin{bmatrix}a & b \\c & d \end{bmatrix} \begin{bmatrix}p & q \\r & s \end{bmatrix} = \begin{bmatrix}ap + br & aq + bs \\cp + dr & cq + ds \end{bmatrix}\)

Let \(\displaystyle M = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\)

and let \(\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}\)

Prove that if \(\displaystyle P, Q \in \mathcal{B}\), then \(\displaystyle P + Q \in \mathcal{B}\) (where \(\displaystyle +\) denotes the usual sum of two matrices).

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

This is how I will attack this problem. I will assume that \(\displaystyle P, Q \in \mathcal{B}\). And I will use the distributive property of matrix multiplication.

\(\displaystyle M(P + Q) = MP + MQ\)

Since \(\displaystyle P\) and \(\displaystyle Q\) are in \(\displaystyle \mathcal{B}\), I can switch their position.

\(\displaystyle MP + MQ = PM + QM = (P + Q)M\)

Since \(\displaystyle M(P + Q) = (P + Q)M\), then \(\displaystyle P + Q \in \mathcal{B}\).