User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Extra Credit
I didn't get the extra credit on my linear algebra exam, can someone help me with it? If I remember correctly, it was something like:

Find the values of $a$ and $b$ such that the rank of the following matrix is $2$.


 * $\begin{bmatrix} 1 & 0 & 0 \\ 0 & a - 2 & b \\ 0 & 2 & a + 2 \\ 0 & 0 & 3 \end{bmatrix}$

--GFauxPas 15:12, 4 April 2012 (EDT)


 * If that really is the matrix, the solution is the empty set; the first, third and fourth row constitute a $3\times3$ matrix with nonzero determinant. --Lord_Farin 15:28, 4 April 2012 (EDT)


 * Can you maybe help me figure out a suitable version where there is a solution? In any event, we have not gotten to determinants in class other than (briefly) relationship between 0 determinant and being singular. --GFauxPas 15:37, 4 April 2012 (EDT)

Consider


 * $\begin{bmatrix} 1 & 0 & 0 \\ 0 & a - 2 & b \\ 0 & 2 & a + 2 \\ 0 & 1 & 3 \end{bmatrix}$

I strongly suspect that there are cases where this does not have maximal rank (while it is evident that it will have rank at least two, taking the first and second column).

A useful characterization of rank is 'dimension of the range' (taken as a linear space); this corresponds directly to the number of linearly independent columns. So in our present case, the rank will be $3$ unless the latter two columns are linearly dependent. I am sure you can work out $a$ and $b$ from there. --Lord_Farin 16:57, 4 April 2012 (EDT)

$a$ has to be $4$ or else all three columns will have a pivot:


 * $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 3 \\ 0 & 2 & b \\ 0 & 0 & 0 \end{bmatrix}$

$a = 4, b=6$? --GFauxPas 17:31, 4 April 2012 (EDT)


 * Quite sure that is correct. --Lord_Farin 17:38, 4 April 2012 (EDT)


 * Hooray thank you LF --GFauxPas 17:39, 4 April 2012 (EDT)

Just thought of something else.

We have $\rho\left({\mathbf A}\right) + \nu\left({\mathbf A}\right) = \text{number of columns} = 3$, which we might not have a page for.

From Null Space Contains Only Zero Vector iff Columns are Independent, $\nu\left({\mathbf A}\right) = 1$ iff the column vectors are independent.

etc.

Actually, I should add a corollary to that page: nullity is one iff column vectors are independent... --GFauxPas 23:32, 4 April 2012 (EDT)


 * Well, we do have Rank Plus Nullity Theorem, if that's what you mean. By the way, did you mean $\nu\left({\mathbf A}\right) = 0$? --abcxyz 23:47, 4 April 2012 (EDT)

Algorithm
Let $T: \R^n \to \R^m$ be a linear transformation.

Let $\mathbf A_T$ be the matrix representation of $T$, where $\mathbf A_T$ is $m \times n$.