# Thread: Linear Independence and bases help

1. ## Linear Independence and bases help

Hi:
I was reading over my linear algebra book, and trying to study for a midterm, but they didn't provide these conceptual answers in the back of the book.
So I was wondering if anyone could show me how the following three questions are done.

Thanks.

2. bump

3. A. $suppose\ \vec{x}\in V, \mathfrak{B}=(\vec{v}_1,\vec{v}_2);then\ \vec{x}=span(\vec{v}_1,\vec{v}_2)$
If $(\vec{v}_1,c\vec{v}_1+\vec{v}_2)$ linear dependence, then $k\vec{v}_1=c\vec{v}_1+\vec{v}_2\rightarrow (k-c)\vec{v}_1=\vec{v}_2$, it conflict with $(\vec{v}_1,\vec{v}_2)$ linear independence, $\therefore (\vec{v}_1,c\vec{v}_1+\vec{v}_2)$ linear independence.
$\vec{x}=k_1\vec{v}_1+k_2\vec{v}_2=(k_1-k_2 c)\vec{v}_1+k_2(c\vec{v}_1+\vec{v}_2)=span(\vec{v} _1,c\vec{v}_1+\vec{v}_2)$, so $(\vec{v}_1,c\vec{v}_1+\vec{v}_2)$ is basis of V.
It's the same to rest proof.

B. $L:R^4\rightarrow R^5, A\in R^{5\times 4},L(\vec{x})=A\vec{x}.$
$c_1L(\vec{x}_1)+c_2L(\vec{x}_2)+c_3L(\vec{x}_3)=0\ rightarrow c_1A\vec{x}_1+c_2A\vec{x}_2+c_3A\vec{x}_3=0$ $\rightarrow A(c_1\vec{x}_1+c_2\vec{x}_2+c_3\vec{x}_3)=0$
$\because ker(L)={\vec{0}},\therefore c_1\vec{x}_1+c_2\vec{x}_2+c_3\vec{x}_3=0$
$\because (\vec{x}_1,\vec{x}_2,\vec{x}_3)\ is\ linearly\ independent,\therefore c_1=c_2=c_3=0$
$\therefore \{L(\vec{x}_1),L(\vec{x}_2),L(\vec{x}_3)\}\ is\ linearly\ independent$

C.
$\mathfrak{B}_a=(\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&1&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&0&0\\0&0&1 \end{bmatrix})$

$\mathfrak{B}_b=(\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&1&0\\0&0&0 \end{bmatrix},\begin{bmatrix} 0&0&0\\0&0&0\\0&0&1 \end{bmatrix},\begin{bmatrix} 0&1&0\\0&0&0\\0&0&0 \end{bmatrix}\begin{bmatrix} 0&0&1\\0&0&0\\0&0&0 \end{bmatrix}\begin{bmatrix} 0&0&0\\0&0&1\\0&0&0 \end{bmatrix})$

4. Cheers Math2009!