for which $a$ values will $U$ and $V$ span the same space

Question

having trouble with this question... if anyone could give me a hint on where to begin that would be lovely.

Let the set of vectors $U= \{u_1,u_2\}$ $\space$ when $u_1 = (1,a,2)\space $and$\space u_2=(a,1,0) $

and also $V= \{v_1,v_2\}$ $\space$ when $v_1 = (-3,0,2)\space $and$\space v_2=(3,3,a) $

for which $a$ values will $U$ and $V$ span the same space?

Answer 1

Hints:

It is enough and necessary that $\;u_i\in\text{Span}\{v_1,v_2\}\;,\;\;v_i\in\text{Span}\{u_1,u_2\}\;$ (and you can save yourself some work if both sets are linearly independent. Can you see why?).

Anyway:

$$(1,a,2)=\alpha(-3,0,2)+\beta(3,3,a)\iff\begin{cases}&-3\alpha+&3\beta=1\\&&3\beta=a\\&2\alpha+&a\beta=2\end{cases}$$

From the first two equations we get

$$\beta=\frac a3\;,\;\;\alpha=\frac{a-1}3$$

and then 3rd equation gives

$$\frac23a-\frac23+\frac{a^2}3=2\implies a^2+2a-8=0\iff (a+4)(a-2)=0$$

Take it from here.

Answer 2

Hint/suggestion:

Try taking the following matrices to reduced row-echelon form:

$$\left(\begin{array}{ccc}1&a&2\\a&1&0 \end{array} \right)$$

and

$$\left(\begin{array}{ccc}-3&0&2\\3&3&a \end{array} \right)$$

If the reduced row-echelon forms are the same then the matrices have the same row space (and conversely).

You will have to deal with some specific cases to avoid dividing by $0$ in the process.

Answer 3

The subspaces $U$ and $V$ have dimension $2$; you know that you must have $\dim(U+V)=2$. This means that the matrix having the four vectors as columns has rank $2$: \begin{align} \begin{bmatrix} -3 & 3 & 1 & a \\ 0 & 3 & a & 1 \\ 2 & a & 2 & 0 \end{bmatrix} &\to \begin{bmatrix} 1 & -1 & -1/3 & -a/3 \\ 0 & 3 & a & 1 \\ 2 & a & 2 & 0 \end{bmatrix} \\[10px]&\to \begin{bmatrix} 1 & -1 & -1/3 & -a/3 \\ 0 & 3 & a & 1 \\ 0 & a+2 & 8/3 & 2a/3 \end{bmatrix} \\[10px]&\to \begin{bmatrix} 1 & -1 & -1/3 & -a/3 \\ 0 & 1 & a/3 & 1/3 \\ 0 & a+2 & 8/3 & 2a/3 \end{bmatrix} \\[10px]&\to \begin{bmatrix} 1 & -1 & -1/3 & -a/3 \\ 0 & 1 & a/3 & 1/3 \\ 0 & 0 & (8-a^2-2a)/3 & (a-2)/3 \end{bmatrix} \end{align} Thus we must have $a=2$ or the rank would be $3$. This is a necessary condition.

Now, considering that both subspaces have dimension $2$, conclude that, for $a=2$ they're indeed equal.