# Linear algebra help

1. May 6, 2005

### 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 = $$\left(\begin{array}{c}\frac{2}{7}\\\frac{13}{84}\\0\end{array}\right) + t \left(\begin{array}{c}\frac{-1}{7}\\\frac{5}{7}\\1\end{array}\right)$$
$$\left(\begin{array}{ccc}3&1&1\\2&-4&10\\-1&3&-7\end{array}\right) \left(\begin{array}{c}x_{1}\\x_{2}\\x_{3}\end{array}\right) = \left(\begin{array}{c}2\\-2\\1\end{array}\right)$$

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 :)

2. May 6, 2005

### HallsofIvy

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

Did you notice that there is no "n" in that? Are you claiming that xn 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.

3. May 6, 2005

### shan

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

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?

4. May 6, 2005

### shan

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