Question

Let A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),show that(aI+bA)ⁿ=aⁿI+na^n-1bA,where I is the identity matrix of order 2 and n∈N

Answer 1

It is given that A= \(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) 

To show: (aI+bA)ⁿ=aⁿI+na^n-1bA 
We shall prove the result by using the principle of mathematical induction. 

For n = 1, we have: 
P(1):(aI+bA)=aI+ba⁰A=aI+bA 

Therefore, the result is true for n =1. 
Let the result be true for n = k. 
That is,
P(k):(aI+bA)^k=akI+ka^k-1bA 
Now, we prove that the result is true for n = k + 1. 
Consider:
(aI+bA)^k+1=(aI+bA)^k(aI+bA) 
=(a^kI+ka^k-1bA)(aI+bA) 
=a^k+1+ka^kbAI+a^kbIA+ka^k-1b²A² 
=a^k+1I+(k+1)a^kbA+ka^k-1b²A² ...(1) 

Now A²=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\)= \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) = O 

From (1), we have: 
(aI+bA)^k+1=a^k+1I+(k+1)a^kbA+O 
=a^k+1I+(k+1)a^kbA 

Therefore, the result is true for n = k + 1. 

Thus, by the principle of mathematical induction, we have: 

(aI+bA)ⁿ=aⁿI+na^n-1bA where A=\(\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\),n∈N