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If I have a homgenous matrix with one of the entries being $a$ and I need to determine which values of $a$ will give the matrix a space of solutions that has dimension $1$ (or dimension 2), how would I go about doing that?

For example ( just making this up from the top of my head):

$\begin{pmatrix} 1 &2&3&0\\4&5&6&0\\3&2&a&0 \end{pmatrix}$

I know I have to bring the matrix to RREF first, and I know which $a$ to choose to get infinitely many solutions or no solutions, but I'm not sure if there's a relation with dimension. Also my knowledge on linear algebra is just from an intro course on it.

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up vote 2 down vote accepted

Recall that the rank and the nullity of a matrix add up to the number of columns of the matrix (rank-nullity theorem). The nullity the dimension of the solution space. So, you can control the nullity by changing $a$ to adjust the rank of the matrix.

The rank of the matrix in your case is at most 3, since it is 3x4. The first two rows are linearly independent, so the rank is at least 2. This means that $a$ controls whether the rank of the matrix is 2 or 3, and therefore if the nullity is 2 or 1. If you choose $a$ such that its row is linearly independent of the above rows, the nullity will be 1. If it is linearly-dependent, the nullity will be 2.

Note however, that it is not guaranteed that you can choose $a$ to specify the dimension you want, because this is also controlled by the other numbers in the row.

For example, you can clearly see here that the rank will always be 3:

$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & a \end{pmatrix} $

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If I had a matrix A=$\begin{pmatrix}1&0&-1&0\\0&1&2&0\\0&0&a&0 \end {pmatrix}$ for example, how would you choose an $a$ such that the dimension would be, say 1? (if that's possible at all?) – Sujaan Kunalan Jun 20 '13 at 21:03
@SujaanKunalan if $a=0$ the rank of the matrix will be 2 so the nullity will be 2. Otherwise the matrix will be of rank 3 so the nullity will be 1. Do you understand why? – Bitwise Jun 20 '13 at 21:05
Well, actually since rank (A) + null(A)=n, we know n and rank (A) so you're just rearranging them. So I understand that part of it. – Sujaan Kunalan Jun 20 '13 at 21:13
Let's mark the ith row as $A_i$. If $a=0$, then $0A_1+0A_2=A_3$, so $A_3$ is a linear combination of $A_1$ and $A_2$. However, if $a$ is not zero, there clearly aren't any numbers $c,d$ such that $cA_1+dA_2=A_3$. So in this case the rows are linearly independent. – Bitwise Jun 21 '13 at 11:59

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