Basis/Spans

**Zocken** · November 4th 2009, 04:36 PM

Let $v_1=\begin{bmatrix}1 \\ 1 \\ a\end{bmatrix},v_2=\begin{bmatrix}1 \\ a \\ 1\end{bmatrix},$ and $v_3=\begin{bmatrix}a \\ 1 \\ 1\end{bmatrix}$

For which values of the parameter a is $v_1$ , $v_2$ and $v_3$ a basis for $R^3$ ?

**Bruno J.** · November 4th 2009, 04:39 PM

Hint : evaluate the determinant of the matrix whose columns are those three vectors. Solve for the values of $a$ for which this determinant is nonzero. (Do you see why?)

**Zocken** · November 4th 2009, 04:52 PM

Bruno, I found the determinant of the matrix to be $a^3-3a+2$ and the values of a that make it nonzero are $a=/-2$ and $a=/1$

Is this right? And could you explain why this works (I am used to row reducing from class)? Thanks

**Bruno J.** · November 4th 2009, 05:04 PM

How can you possibly get $a^3$ in there? You should get a second degree polynomial in $a$ .

This works, because for the three vectors to be a basis of $\mathbb{R}^3$ they should form the sides of a parallelepiped having nonzero volume, and the determinant is that volume. Or, more abstractly put, if $\beta=\{v_1,v_2,v_3\}$ is a basis, then the change of basis transformation from $\beta$ to the standard basis is invertible; its matrix representation is the one whose determinant you evaluated (with a small mistake).

**math2009** · November 4th 2009, 05:13 PM

Determinant is easy method. But when dimension is high, it's difficult to solve equation.

I suggest processing by rref

$rref(A)=rref[\vec{v}_1\ \vec{v}_2\ \vec{v}_3]=\begin{bmatrix} 1&1&a\\0&a-1&1-a\\0&0&2-a-a^2 \end{bmatrix}$

if three vectors span a basis, then rank(A)=3, $a-1\neq 0 \& 2-a-a^2\neq 0$

**Zocken** · November 4th 2009, 11:22 PM

thank you

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