Question

Let {u,v,w} be a base for the vector space V. Determine if {u1,u2,u3} is a basis for V, where u1=u+v-3w; u2=u+3v-w and u3=v+w.

Answer 1

Given that any vector \(\vec{v}\in V\) can be written as \(\vec{v}=c_1u+c_2v+c_3w\), can \(\vec{v}\) be also written as \(\vec{v}=d_1u_1+d_2u_2+d_3u_3\)?

Answer 2

$$
Ax=b\\
\begin{bmatrix}
1&1 &-3 \\
1&3 &-1 \\
0& 1 & 1
\end{bmatrix}
\begin{bmatrix}
u\\
v\\
w
\end{bmatrix}
=
\begin{bmatrix}
u_1\\
u_2\\
u_3
\end{bmatrix}\\$$
Row-reducing A shows that \(u_1,u_2\) and \(u_3\) only fill a 2-dimensional space.
$$
\begin{bmatrix}
1&1 &-3 \\
0&1 &1 \\
0& 0 & 0
\end{bmatrix}
$$
Since {u,v,w} is a basis for V, they are independent and fill a 3-dimensional space.

Therefore, it is not possible for {u1,u2,u3} to be a basis for V.

Answer 3

I have that the matrix is [1 1 0|X]
[131|Y]
[-3-11|]
which I reduced and gave me 0=2x+1/2y+z
I just don't know if it is correct.

Answer 4

I'm not following exactly. What i tried to show above was that the two basis could not span the same space because the second spans a smaller space than V.

Answer 5

I have the matrix you showed, except that your rows are my columns and am trying to solve for a general vector (x,y,z) to what space it generates and then determine if it is a basis for V.

Answer 6

ok, I see that now.

Answer 7

I know what you don't understand, you see, you said \({u_1,u_2,u_3}\), it means you should expand them as column vectors. I mean