Finding the Formula for An: A Linear Algebra Challenge

[roam]

Let $$A = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{array}\right)$$

Find a formula for Aⁿ

The Attempt at a Solution

This is a weird question, I asked a few tutors for help but none of them were able to help me. They said that this is a weird question.

Does anyone here know how to do that? My friends & I can’t figure it out. Therefore I decided to post it here to see if anyone can show us how this is done.

We haven’t studied eigenvalues & Eigenvectors or diagonalization etc… we have only studied basic Linear Algebra, just as a warning!


I have three guesses;

(I) $$A^n = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{array}\right)^n$$

(II) $$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1^n&2^n\\0&0&1^n\end{array}\right)$$

(III) $$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1^n&0\\0&0&1^n\end{array}\right)$$+$$\left(\begin{array}{ccc}0&0&0\\0&0&2\\0&0&0\end{array}\right)$$

Is any of them close to the correct answer?

 

[Defennder]

(II) comes pretty close to the correct answer. You should that the diagonal entries of A^n are pretty easy to figure out by trying out A^n for small N. The only problem is the a₂₃ entry. Again, just like above, try it out for small values of n. This follows an easy to identify pattern.

 

[roam]

I know that the a₂₃ entry is a problem & that's why (in my third guess) I suggested to separate it from the rest.

So you are suggesting that I substitute small values for n ie. n = 1,2... & then Aⁿ⁺¹ etc ? Could you please explain what can we exactly conclude from that?

$$A^1 = \left(\begin{array}{ccc}3&0&0\\0&1&2\\0&0&1\end{ar ray}\right)$$

$$A^2 = \left(\begin{array}{ccc}9&0&0\\0&1&4\\0&0&1\end{ar ray}\right)$$

$$A^3 = \left(\begin{array}{ccc}27&0&0\\0&1&8\\0&0&1\end{ar ray}\right)$$

EDIT: The only pattern I see is that a₁₁ & a₂₃ are the two entries that actually increase by every n.

 

[Defennder]

Your A^3 is incorrect. Recheck the entries. They are increasing, but how exactly are they increasing? I did say that your original answer was nearly correct.

 

[roam]

but how exactly are they increasing?

They are increasing in such pattern;

$$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1&2^n\\0&0&1\end{array}\right)$$

Your A^3 is incorrect. Recheck the entries.

$$A^2 = \left(\begin{array}{ccc}9&0&0\\0&1&4\\0&0&1\end{ar ray}\right)$$


$$A^{n+1} = A^{2+1} => A^3 = \left(\begin{array}{ccc}729&0&0\\0&1&64\\0&0&1\end{ar ray}\right)$$

 

[roam]

Is it something like:
If n = 2
A² = 9
A²⁺¹ = 9²⁺¹ = 729

Please tell me I'm so curious!

 

[statdad]

This is an exercise in 'block matrix' multiplication. let

$$A = \begin{bmatrix} 1 & \mathbf{0}'\\ \mathbf{0} & B\end{bmatrix}$$

where

$$\mathbf{0} = \begin{bmatrix} 0\\ 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 2\\0 & 1 \end{bmatrix}$$

Here

$$A^n = \begin{bmatrix} 1 & \mathbf{0}' \\ \mathbf{0} & B^n \end{bmatrix}$$

 

[HallsofIvy]

Using statdad's idea, let M= [3] and
$$N= \left[\begin{array}{cc}1 & 2 \\ 0 & 1\end{array}\right]$$
so that
$$A= \left[\begin{array}{cc} M & 0 \\ 0 & N\end{array}\right]$$
and
$$A^n= \left[\begin{array}{cc}M^n & 0 \\ 0 & N^n\end{array}\right]$$
Mⁿ is trivial and Nⁿ should be pretty easy to see after calculating a few low powers. (Which is always a good idea in problems like these!)

 

[HallsofIvy]

They are increasing in such pattern;

$$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1&2^n\\0&0&1\end{array}\right)$$




$$A^2 = \left(\begin{array}{ccc}9&0&0\\0&1&4\\0&0&1\end{ar ray}\right)$$


$$A^{n+1} = A^{2+1} => A^3 = \left(\begin{array}{ccc}729&0&0\\0&1&64\\0&0&1\end{array}\right)$$

No, that's not right at all! What is
$$\left(\begin{array}{ccc}9 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right)$$

Is it something like:
If n = 2
A² = 9
A²⁺¹ = 9²⁺¹ = 729

Please tell me I'm so curious!

No! I am seriously tempted to ask why you are trying to do matrix arithmetic if you can't do arithmetic with numbers! A³ is NOT 9³ because A is not 9. If A² is 9, then A is either 3 or -3. If A= 3, then A³= 3³= 27. If A= -3, A³= (-3)³= -27.

 

[roam]

No, that's not right at all! What is
$$\left(\begin{array}{ccc}9 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right)$$

Forgive my blur, I had a terrible headache along with a great hangover, and I appologize for the mistakes. Here's the corrected version of what I think:


...[3 0 0]
A¹ = [0 1 2]
...[0 0 1]

...[3² 0 0]
A² = [0 1 4 ]
...[0 0 1 ]

...[3³ 0 0]
A³ = [0 1 6 ]
...[0 0 1 ]

...[3⁴ 0 0]
A⁴ = [0 1 8 ]
...[0 0 1 ]
:
:
:
:
...[3ⁿ 0 0]
Aⁿ = [0 1 2n ]
...[0 0 1 ]


$$A^n = \left(\begin{array}{ccc}3^n&0&0\\0&1&2n\\0&0&1\end{ar ray}\right)$$

No! I am seriously tempted to ask why you are trying to do matrix arithmetic if you can't do arithmetic with numbers! A³ is NOT 9³ because A is not 9. If A² is 9, then A is either 3 or -3. If A= 3, then A³= 3³= 27. If A= -3, A³= (-3)³= -27

 

[Tobias Funke]

Here's another way to approach the problem and find the pattern. Have you learned that this matrix represents a certain linear transformation of R^3 to itself and what the columns of the matrix mean in that case? This transformation takes the vector (1,0,0) to (3,0,0), tripling its first coordinate. It takes (0,1,0) to (0,2,0), doubling its second coordinate, and it takes (0,0,1) to (0,2,1), adding 2 to its second coordinate.

Now if you know that matrix multiplication represents function composition, can you see what happens when you carry out this transformation n times? Looks like you do, but you said above that that's what you think, so if you're still unsure, can you prove it (by induction maybe)?

 

[Defennder]

That's complicating the problem unnecessarily. The OP hasn't learned eigvenvectos or diagonalisation, and he certainly doesn't even need to understand linear transformations to do this. He got it right with his last post, though.

 

[Tobias Funke]

Yeah I also misread the second column as (0,2,0). I don't know exactly what the OP knows, but you don't need to know anything about eigenvectors or diagonalization to know what I said. At least in my case, thinking in those terms actually makes the problem seem simpler and gives the pattern right away. But you're right of course-it's not necessary to solve the problem.