Question

Let A=[{0 1}{0 0}] show that (aI + bA)^n =a^nI+na^n-1bA Where I is the identity matrix of order 2

Answer 1

\[A=\left[\begin{matrix}0 & 1 \\ 0 & 0\end{matrix}\right]\]
Show that \( (aI + bA)^n = a^n I + na^{n-1} bA \)

Answer 2

of order 2?

and as far as I can tell, an exponent on a matrix would imply a diagonalization of eugene values .... maybe

Answer 3

Yeah 2x2

Answer 4

and i didnt understand the last part

Answer 5

\begin{pmatrix}a&0\\0&a\end{pmatrix}+\begin{pmatrix}0&b\\0&0\end{pmatrix}

Answer 6

that didnt format like I wanted :)

Answer 7

Lol

Answer 8

\begin{pmatrix}a&b\\0&a\end{pmatrix}^n

is what I hope I see correctly so far

Answer 9

I think i have to prove it using Mathematical Induction

Answer 10

ugh! thats a bit above my abilities at the moment; just starting the linearA stuff this term.

Answer 11

Oh :(

Answer 12

show that its true for some n; then work it out for n+1 is what I can recall from math induction

Answer 13

Yeah ,For n then k & k+1

Answer 14

its the extra variables, a and b thatll throw me for a loop

Answer 15

good luck with it :)

Answer 16

Haha , :D Thanks !!

Answer 17

does it mean this
\[\begin{pmatrix}a&b\\0&a\end{pmatrix}^n=\begin{pmatrix}a^n&na^{n-1}b\\0&a^n\end{pmatrix}\]??

Answer 18

\[\begin{pmatrix}a&b\\0&a\end{pmatrix}\times \begin{pmatrix}a&b\\0&a\end{pmatrix}=\begin{pmatrix}a^2&2ab\\0&a^2\end{pmatrix}\] by computation, so it seems good

Answer 19

yeah, i get that when i substitute the matrix in (aI + bA)^n =a^nI+na^n-1bA

Answer 20

you could also use this step as the case n = 2 to start your induction proof, or you could use n = 1 which is trivial, too trivial really, so it will be a proof by induction right?

Answer 21

say we know it is true for all
\[k\leq n\] and we want to prove it is true for
\[k=n+1\]

Answer 22

then write
\[\begin{pmatrix}a&b\\0&a\end{pmatrix}^{n+1}=\begin{pmatrix}a&b\\0&a\end{pmatrix}^n\times \begin{pmatrix}a&b\\0&a\end{pmatrix}\]

Answer 23

"by induction" we know it is true for
\[\begin{pmatrix}a&b\\0&a\end{pmatrix}^n\] i.e. we know that
\[\begin{pmatrix}a&b\\0&a\end{pmatrix}=\begin{pmatrix}a^n&na^{n-1}b\\0&a^n\end{pmatrix}\]

Answer 24

so finish by multiplying
\[\begin{pmatrix}a&b\\0&a\end{pmatrix}^{n+1}=\begin{pmatrix}a&b\\0&a\end{pmatrix}^n\times \begin{pmatrix}a&b\\0&a\end{pmatrix}\]
\[=\begin{pmatrix}a^n&na^{n-1}b\\0&a^n\end{pmatrix}\times \begin{pmatrix}a&b\\0&a\end{pmatrix}\]
and the last step is matrix multiplication, which should give you the necessary answer

Answer 25

namely
\[=\begin{pmatrix}a^n&na^{n-1}b\\0&a^n\end{pmatrix}\times \begin{pmatrix}a&b\\0&a\end{pmatrix}=
\begin{pmatrix}a^{n+1}&(n+1)a^nb\\0&a^{n+1}\end{pmatrix}\]

Answer 26

which it does!

Answer 27

Oh Alright :D
Thankyou so Much :)

Answer 28

yw, hope it is clear.

Answer 29

Yes Very clear :)