# Thread: Determinant/Dimension problem

1. ## Determinant/Dimension problem

This is from my text. My instructor listed this as one for us to think about to help us prep for the exam on Tuesday. I really am stuck though, which is scaring me.

Here it goes:

Consider the matrix $A=\left[ \begin{array}{ccc}
x_{1} & a_{1} & b_{1} \\
x_{2} & a_{2} & b_{2} \\
x_{3} & a_{3} & b_{3} \end{array} \right]
\in M_{3}(k)
$

where the $a$'s and $b$'s are fixed and the $x$'s may vary. Show that a set of all $(x_{1} , x_{2}, x_{3})$ such that $det(A)=0$ is a subspace of $k^{3}$ of dimension at least 2.

I know that the determinant is a linear transformation in each column, but I don't really know how that helps me. I asked a classmate who said she thought Rank-Nullity theorem might help.. but I am still clueless. Any and all explanation on how to prove this would be great. Thanks!

2. Can you show it's a subspace?

Assuming that you can, if you want to show that the dimension is at least 2, consider the following two cases :

(1) $\vec a = (a_1, a_2, a_3)$ and $\vec b =(b_1, b_2, b_3)$ are linearly independent : then the subspace contains the span of $\vec a, \vec b$, which is two dimensional (in fact, try to show that it's equal to the span of $\vec a, \vec b$);

(2) $\vec a, \vec b$ are linearly dependent : then the space is all of $k^3$! (Can you see why?)

Can you generalize to higher dimensions?

3. I feel really bad, but I don't see how to do any of that. How do I show a column of a matrix is a subspace, and of what?

1. I understand that the span of {a,b} is two dimensional as then it would be a basis (smallest spanning set, largest linearly independent set). Since I am not sure on what the subspace is, I dont really know how it would be equal.

2. Maybe once I get the first parts cleared up this will become more apparent. I dont see how this is true, and how {x1, x2, x3} or the determinant help me.

4. DISREGARD: My computer posted multiple times

5. DISREGARD: My computer posted multiple times

6. Originally Posted by Bruno J.
Can you show it's a subspace?

Assuming that you can, if you want to show that the dimension is at least 2, consider the following two cases :

(1) $\vec a = (a_1, a_2, a_3)$ and $\vec b =(b_1, b_2, b_3)$ are linearly independent : then the subspace contains the span of $\vec a, \vec b$, which is two dimensional (in fact, try to show that it's equal to the span of $\vec a, \vec b$);

(2) $\vec a, \vec b$ are linearly dependent : then the space is all of $k^3$! (Can you see why?)

Can you generalize to higher dimensions?
I feel really bad, but I don't see how to do any of that. How do I show a column of a matrix is a subspace, and of what?

1. I understand that the span of {a,b} is two dimensional as then it would be a basis (smallest spanning set, largest linearly independent set). Since I am not sure on what the subspace is, I dont really know how it would be equal.

2. Maybe once I get the first parts cleared up this will become more apparent. I dont see how this is true, and how {x1, x2, x3} or the determinant help me.

7. Figured it out... don't need any help.