Matrix Multiplication

[ksdhart2]

Matrix Multiplication; Prove that [tex]S_{\theta }^n=S_{n\theta}[/tex]

Hi all,

I'm having some difficulties with my Linear Algebra course. I've solved part (i), and part (iii) seems fairly straightforward if I can solve part (ii).

2) For any angle \(\displaystyle \theta\), let \(\displaystyle S_{\theta }=\begin{pmatrix}cos\left(\theta \right)&-sin\left(\theta \right)\\ sin\left(\theta \right)&cos\left(\theta \right)\end{pmatrix}\)

(i) Prove that \(\displaystyle S_{\alpha }\times \:S_{\beta }=S_{\alpha +\beta }\) for any angles \(\displaystyle \alpha \text{ and } \beta\).

(ii) Prove that \(\displaystyle S_{\theta }^n=S_{n\theta }\) for any natural number n.

(iii) If \(\displaystyle N=\begin{pmatrix}\frac{1}{2}&-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}\), find N², N³, N⁴, N⁵, and N²³¹, using your results from part (ii)

Part (i) was really straightforward, just a matter of using the known trig identities and showing that they were equal to the dot products.

Now, for part (ii) I can see that it holds true for n=2 and n=3, just by doing by the matrix multiplication, but how can I prove it works for any natural n? The only way I can think of to do that would be to use eigenvalues and/or eigenvectors, but we haven't learned about those yet, so I highly doubt that's the intended method... any advice on what I might be missing?

 

[Harry_the_cat]

Hi all,

I'm having some difficulties with my Linear Algebra course. I've solved part (i), and part (iii) seems fairly straightforward if I can solve part (ii).



Part (i) was really straightforward, just a matter of using the known trig identities and showing that they were equal to the dot products.

Now, for part (ii) I can see that it holds true for n=2 and n=3, just by doing by the matrix multiplication, but how can I prove it works for any natural n? The only way I can think of to do that would be to use eigenvalues and/or eigenvectors, but we haven't learned about those yet, so I highly doubt that's the intended method... any advice on what I might be missing?

It can be done using Proof by Induction - have you learnt that method of proof yet?

 

[ksdhart2]

Oh, wow. That makes so much more sense than what I was trying. That hint, plus looking at the problem again with refreshed eyes and a refreshed brain helped me see exactly what I'd been missing. Thanks so much!

Using the results from part (i), letting \(\displaystyle \alpha=\beta=\theta\):

\(\displaystyle S_{\theta }\times S_{\theta }=S_{\theta +\theta} \Rightarrow S^2_{\theta }=S_{2\theta }\)

Assume it works for n=k:

\(\displaystyle S^k_{\theta }=S_{k \theta }\)

Prove it for n = k + 1:

\(\displaystyle S_{\theta }^{k+1}=S_{\left(k+1\right)\theta }\)

\(\displaystyle S_{\theta }^k\times S_{\theta }=S_{k \theta +\theta }\)

Returning to part (i) again:

\(\displaystyle S_{k \theta +\theta }=S_{k\theta }\times S_{\theta }\)

The left and right sides are equal, so that's the end of my proof.

 

[Harry_the_cat]

Oh, wow. That makes so much more sense than what I was trying. That hint, plus looking at the problem again with refreshed eyes and a refreshed brain helped me see exactly what I'd been missing. Thanks so much!

Using the results from part (i), letting \(\displaystyle \alpha=\beta=\theta\):

\(\displaystyle S_{\theta }\times S_{\theta }=S_{\theta +\theta} \Rightarrow S^2_{\theta }=S_{2\theta }\)

Assume it works for n=k:

\(\displaystyle S^k_{\theta }=S_{k \theta }\) Yes, this is your assumption.

Prove it for n = k + 1:

\(\displaystyle S_{\theta }^{k+1}=S_{\left(k+1\right)\theta }\) This is what you need to prove. You need to prove this statement based on your assumption above. It will require going back to the original matrix.

\(\displaystyle S_{\theta }^k\times S_{\theta }=S_{k \theta +\theta }\)

Returning to part (i) again:

\(\displaystyle S_{k \theta +\theta }=S_{k\theta }\times S_{\theta }\)

The left and right sides are equal, so that's the end of my proof.

see comment in red

 

[pythagorean314159]

Oh, wow. That makes so much more sense than what I was trying. That hint, plus looking at the problem again with refreshed eyes and a refreshed brain helped me see exactly what I'd been missing. Thanks so much!

Using the results from part (i), letting \(\displaystyle \alpha=\beta=\theta\):

\(\displaystyle S_{\theta }\times S_{\theta }=S_{\theta +\theta} \Rightarrow S^2_{\theta }=S_{2\theta }\)

Assume it works for n=k:

\(\displaystyle S^k_{\theta }=S_{k \theta }\)

Prove it for n = k + 1:

\(\displaystyle S_{\theta }^{k+1}=S_{\left(k+1\right)\theta }\)

\(\displaystyle S_{\theta }^k\times S_{\theta }=S_{k \theta +\theta }\)

Returning to part (i) again:

\(\displaystyle S_{k \theta +\theta }=S_{k\theta }\times S_{\theta }\)

The left and right sides are equal, so that's the end of my proof.

Very good! Now you can use the result for part iii.