Help with Linear Algebra: Find Solutions & Verify

[shan]

This first one, I just want to verify that I've understood what the question is asking.

It says: If $$x_{n+1} = Ax_{n}$$, write an expression for $$x_{n}$$. The matrix A = $$\left(\begin{array}{ccc}2&0&0\\1&3&0\\-3&5&4\end{array}\right)$$

From what I understand, this question wants me to do an eigenvector decomposition so this is what I came up with (after finding the eigenvectors and eigenvalues):
$$x_{n} = c_{1}(4 * \left(\begin{array}{c}0\\0\\1\end{array}\right)) + c_{2}(3 * \left(\begin{array}{c}0\\\frac{-1}{5}\\1\end{array}\right)) + c_{3}(2 * \left(\begin{array}{c}\frac{1}{4}\\\frac{-1}{4}\\1\end{array}\right))$$
Is that what the question was asking for??

The other question (or help I need) is this one: Verify that there are infinitely many least squares solutions which are given by x = $$\left(\begin{array}{c}\frac{2}{7}\\\frac{13}{84}\\0\end{array}\right) + t \left(\begin{array}{c}\frac{-1}{7}\\\frac{5}{7}\\1\end{array}\right)$$
It's talking about this system...
$$\left(\begin{array}{ccc}3&1&1\\2&-4&10\\-1&3&-7\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right) = \left(\begin{array}{c}2\\-2\\1\end{array}\right)$$

I multiplied both sides by the inverse of the matrix and got this system:
$$\left(\begin{array}{ccc}14&-8&30\\-8&26&-60\\30&-60&150\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right) = \left(\begin{array}{c}-1\\13\\-25\end{array}\right)$$
and I then row reduced it to..
$$\left(\begin{array}{cccc}14&-8&30&1\\0&\frac{150}{7}&\frac{-300}{7}&\frac{95}{7}\\0&0&0&0\end{array}\right)$$
I thought I was going on the right track since I can see that this has infinitely many solutions but when I tried to find x...
$$x_{3} = t$$
$$x_{2} = \frac{19}{30} + 2t$$
$$x_{1} = \frac{13}{30} - t$$
which definitely does not verify the above question. Can someone tell me where I went wrong? Thanks :)

 

[HallsofIvy]

$$x_{n} = c_{1}(4 * \left(\begin{array}{c}0\\0\\1\end{array}\right)) + c_{2}(3 * \left(\begin{array}{c}0\\\frac{-1}{5}\\1\end{array}\right)) + c_{3}(2 * \left(\begin{array}{c}\frac{1}{4}\\\frac{-1}{4}\\1\end{array}\right))$$

Did you notice that there is no "n" in that? Are you claiming that x_n is constant and does not depend on n? There should be some nth powers in there.

In (2) you've completely neglected the "least squares" requirement.

 

[shan]

$$x_{n} = c_{1}(4 * \left(\begin{array}{c}0\\0\\1\end{array}\right)) + c_{2}(3 * \left(\begin{array}{c}0\\\frac{-1}{5}\\1\end{array}\right)) + c_{3}(2 * \left(\begin{array}{c}\frac{1}{4}\\\frac{-1}{4}\\1\end{array}\right))$$

Did you notice that there is no "n" in that? Are you claiming that x_n is constant and does not depend on n? There should be some nth powers in there.

Ah, I think I see what you mean... so should it have been more like..
$$x_{n} = 4^(n+1) c_{1} \left(\begin{array}{c}0\\0\\1\end{array}\right) + 3^(n+1) c_{2} \left(\begin{array}{c}0\\\frac{-1}{5}\\1\end{array}\right) + c^(n+1) c_{3} \left(\begin{array}{c}\frac{1}{4}\\\frac{-1}{4}\\1\end{array}\right)$$

In (2) you've completely neglected the "least squares" requirement.

What do you mean by that? I thought that to solve a least squares problem, you use $$A^T Ax = A^T b$$ where x represents the unknowns, in this case, x1, x2 and x3?

 

[shan]

Oops, nevermind about the second question, I figured it out :)