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I was asked whether the following two statements are correct or incorrect. Can anybody have a quick look to see if my reasoning is valid?

$S:R^3to R^5$, defined by $S(x,y,z)=(z,x+y,x+y+z,2x+2y+z,x+y-z)$

a) $text{Im},S=text{Sp}left{(1,0,1,1,-1),(1,1,2,3,0)right}$

b) $text{ker},S=text{Sp}left{(1,-1,0)right}$

I reasoned as follows:

I calculated the matrix required to convert $(x,y,z)$ into $(z,x+y,x+y+z,2x+2y+z,x+y-z)$:

$$begin{bmatrix}0&0&1\1&1&0\1&1&1\2&2&1\1&1&-1end{bmatrix}$$

Then I took the columns of the matrix as $text{Im},S$ (with one of the columns being redundant); the result implies that a) is incorrect (as $(1,1,2,3,0)$ is not linear combination of $(0,1,1,2,1)$).

However, multiplying the given matrix by $(1,-1,0)$ does give $(0,0,0,0,0)$, and so would any multiples of $(1,-1,0)$, so b is correct.

Thank you!

To be honest, I don't understand what you mean by "as $(1,1,2,3,0)$ is not linear combination of $(0,1,1,2,1)$" and why you think that it means that the first statement is wrong… In fact, both statements a) and b) are correct.

Let's start with a). If you convert this matrix into its reduced row echelon form, you will see that the first and thirds columns can be taken to form the basis for the image of this linear transformation:
$$operatorname{Im}S=operatorname{Span}{mathbf{v}_1,mathbf{v}_2}, text{ where } mathbf{v}_1=(0,1,1,2,1) text{ and } mathbf{v}_2=(1,0,1,1,-1).$$
What probably seems confusing is that $mathbf{v}_1$ is present in the given statement, but then there's some other vector instead of $mathbf{v}_2$. However, you may observe that $mathbf{v}'_2=(1,1,2,3,1)=mathbf{v}_1+mathbf{v}_2$, and it follows that in fact
$$operatorname{Im}S=operatorname{Span}{mathbf{v}_1,mathbf{v}_2}=operatorname{Span}{mathbf{v}_1,mathbf{v}'_2}.$$

Now, about the kernel. Your observation only implies that this vector $(1,-1,0)$ lies in the kernel, but not that it spans the kernel. That's why you can only address part b) after part a). Now that we know that the rank (dimension of the image) is $2$, from the rank-nullity theorem we determine that the dimension of the kernel is $1$, which together with your calculation means that statement b) is correct.

Your answer to part $(a)$ is incorrect. $(1,1,2,3,0)=(1,0,1,1,-1)+(0,1,1,2,1)$, so $text{Sp}{(1,1,2,3,0),(1,0,1,1,-1)}=text{Sp}{(0,1,1,2,1),(1,0,1,1,-1)}$, hence $(a)$ is correct. Your answer to $(b)$ is okay.

Since
$$begin{bmatrix}0&0&1\1&1&0\1&1&1\2&2&1\1&1&-1end{bmatrix}begin{bmatrix}x\y\zend{bmatrix}=(x+y)begin{bmatrix}0\1\1\2\1end{bmatrix}+zbegin{bmatrix}1\0\1\1\-1end{bmatrix}$$
It follows $text{Im},S=text{Sp}left{(0,1,1,2,1),(1,0,1,1,-1)right}$.

We know that $$Sp{v_1,v_2}=Sp{v_1,v_1+v_2}$$and since $$(0,1,1,2,1)+(1,0,1,1,-1)=(1,1,2,3,0)$$hence the result. Also your argument for b) is correct.