Question 5 Practice Midterm
ShengyueHuo
I got [-10,1,8] , is that right???
aleksanderjanczewski: May 5, 2015, 9:41 p.m.
You should get that image is spanned by the following vectors: $v_1= \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$ $v_2= \begin{bmatrix} 2 \\ -2 \\ -6 \end{bmatrix}$ Whereas kernel is spanned by any multiple of the following vector: $v_3= \begin{bmatrix} -10 \\ -1 \\ -8 \end{bmatrix}$
aleksanderjanczewski: May 5, 2015, 9:42 p.m.
CORRECTION: You should get that image is spanned by the following vectors: $v_1= \begin{bmatrix} 1 \\ 3 \\ 1 \end{bmatrix}$ $v_2= \begin{bmatrix} 2 \\ -2 \\ -6 \end{bmatrix}$ Whereas kernel is spanned by any multiple of the following vector: $v_3= \begin{bmatrix} -10 \\ 1 \\ 8 \end{bmatrix}$
ShengyueHuo: May 5, 2015, 9:48 p.m.
My answer for the kernel is the same. I think the image is spanned by [1 3 1] , [2 -2 -6] and [1 4 2]. Why you do not include [1 4 2]?
aleksanderjanczewski: May 5, 2015, 9:56 p.m.
Because vectors $v_a= \begin{bmatrix} 1 \\ 4 \\ 2 \end{bmatrix}$ is a not independent and can be expressed in terms of $v_1$ and $v_2$. Also think about it this way. You are in R3, so dimension of image + dimension of kernel must equal to 3. In your case dimension of image is 3, what would imply that kernel is only a zero vector, and as you said you found a vector spanning kernel, this tells you that image will not be spanned by 3 vectors.
ShengyueHuo: May 5, 2015, 10:12 p.m.
Thank you so much! It is so helpful!!!