## ksaimouli one year ago Linear algebra question

1. ksaimouli

$\left[\begin{matrix}1 & 0&-4 \\0 & 3&-2 \\ -2 & 6& 3\end{matrix}\right]$

2. ksaimouli

Let ^ be A, b= $\left(\begin{matrix}4 \\ 1\\ -4\end{matrix}\right)$

3. ksaimouli

columns of A is a1,a2,a3 and W= span{a1,a2,a3}

4. ksaimouli

Is b in W? how many vectors are in {a1,a2,a3}

5. ksaimouli

If W is in span of A, means C1a1+C2a2+C3a3=W?

6. anonymous

Looks like you're just checking to see if $$b$$ can be written as a linear combination of $$a_1,a_2,a_3$$, which amounts to solving for $$c_1,c_2,c_3$$ such that $c_1a_1+c_2a_2+c_3a_3=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$ which means you'll be solving a system of 3 equations with 3 unknowns.

7. anonymous

In other words, $\begin{cases} c_1-2c_3=4\\ 3c_2+6c_3=1\\ -4c_1-2c_2+3c_3=-4 \end{cases}$

8. ksaimouli

True, I found it to be unique. Which means b is in the span of W. How many vectors are in W?

9. anonymous

Well, I'd say there's an infinite number of vectors. $$a_1,a_2,a_3$$ are linearly independent, so $$W$$ forms a basis of $$\mathbb{R}^3$$, which contains an infinite number of vectors.

10. ksaimouli

I see, how would be interpret if the question was Is W in b?

11. anonymous

I don't think that question would make sense. $$W$$ is a set/space of vectors, while $$b$$ is just a vector. A set can't belong to an element.

12. ksaimouli

Okay, you said "number of vectors. a1,a2,a3 are linearly independent" is it because they are not scalar multiplies of each other?

13. anonymous

Yes, the only way to get $$k_1a_1+k_2a_2+k_3a_3=\begin{pmatrix}0\\0\\0\end{pmatrix}$$ is if all the $$k=0$$. It's more accurate to say "linear combination" in place of "scalar multiples" here. (The second term is a special case of the first term.)

14. ksaimouli

thank you so much :-)