Help with exercise , linear transformation, prove it's linear and then find dimension of kernel and image I have this exercise and I have tried to solve it, can anyone check if I did it right?
Do I have to check that $f(u+v) = f(u)+f(v)$ if the matrix of the coefficients is already linearly independent? does it have any connection?
I'll post a picture because it is very long:

Thank you
 A: I start bringing back something it could be useful to remember:
Dot product:
Given two any vectors $\vec{v} = (x_v, y_v, z_v)$ and $\vec{w} = (x_w, y_w, z_w)$ you can define their dot product as $\vec{v}\cdot\vec{w} = \lvert \vec{v} \rvert \lvert \vec{w} \rvert \cos(\theta)$ where $\theta$ is the angle between the vectors. 
Imagine having two vectors on a plane, for instance $\vec{v} = (x_v, y_v)$ and $\vec{w} = (x_w, y_w)$ where $\vec{v}$ forms an angle $\alpha$ with the x-axis, and $\vec{w}$ forms an angle $\beta$ with it. Then $\theta = \beta - \alpha $ and $$\cos(\beta-\alpha) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$$ Replacing this in the above relation you find that
$$\vec{v}\cdot\vec{w} = \lvert \vec{v} \rvert \lvert \vec{w} \rvert( \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) ) = \lvert \vec{v} \rvert\cos(\alpha)\lvert \vec{w} \rvert\cos(\beta) +\lvert \vec{v} \rvert\sin(\alpha)\lvert \vec{w} \rvert\sin(\beta) $$
Where 
$$\begin{gather} 
 \lvert \vec{v} \rvert\cos(\alpha) = x_v \\
 \lvert \vec{w} \rvert\cos(\beta) = x_w \\
 \lvert \vec{v} \rvert\sin(\alpha) = y_v \\
 \lvert \vec{w} \rvert\sin(\beta) = y_w \\
\end{gather}$$
Therefore $$\vec{v}\cdot\vec{w} = x_vx_w + y_vy_w$$
And this can be generalized to higher dimension spaces, where you still calculate it by multiplying the corresponding coordinates and adding them together. 
Cross product
Given two any vector $\vec{v}$ and $\vec{w}$ you define their cross product $\vec{t} = \vec{v}\times\vec{w}$ as the new vector that satisfies


*

*$\vec{t}$ is perpendicular to both $\vec{v}$ and $\vec{w}$

*$\vec{t}$ is oriented in such a way that $\{\vec{v},\vec{w},\vec{t}\}$ is a positively oriented basis

*His magnitude is $\lvert \vec{t}\rvert = \lvert \vec{v}\rvert \lvert \vec{w}\rvert \lvert\sin(\theta)\rvert$
And this particular operation, differently from the previous case, is only defined for three dimensional spaces. You can prove the same way as we did with the dot product that this can be written the following way if $\vec{v} = (x_v, y_v, z_v)$ and $\vec{w} = (x_w, y_w, z_w)$
$$\vec{t} = ( y_vz_w - z_vy_w ){\widehat{i}} + ( z_vx_w - x_vz_w ){\widehat{j}} + ( x_vy_w - y_vx_w ){\widehat{k}} = (y_vz_w - z_vy_w, z_vx_w - x_vz_w, x_vy_w - y_vx_w)  $$
You can find this remembering the following trigonometric formula, just in case you wanna try:
$$\sin(\beta - \alpha) = \sin(\beta)\cos(\alpha) - \cos(\beta)\sin(\alpha)$$
And you can check simply by computing this determinant, that you can find this vector as 
$$det\begin{bmatrix} 
\widehat{i} & \widehat{j} & {\widehat{k}} \\
x_v & y_v & z_v \\
x_w & y_w & z_w
 \end{bmatrix}$$
Triple product 
Given three any vector $\vec{v} = (x_v, y_v, z_v), \vec{w}=(x_w, y_w, z_w), \vec{t} = (x_t, y_t, z_t)$ of a three dimensional space you can define their triple product as follows
$$\vec{u}\times\vec{w}\cdot\vec{t} = (y_vz_w - z_vy_w, z_vx_w - x_vz_w, x_vy_w - y_vx_w)\cdot (x_t, y_t, z_t) = x_ty_vz_w - x_tz_vy_w + y_tz_vx_w - y_tx_vz_w + z_tx_vy_w - z_ty_vx_w$$
This can be rewritten as
$$\vec{u}\times\vec{w}\cdot\vec{t} = (x_ty_vz_w + y_tz_vx_w + z_tx_vy_w) - ( x_tz_vy_w + y_tx_vz_w + z_ty_vx_w)$$
And you can easily rewrite this one too in terms of a matrix determinant as 
$$det\begin{bmatrix} 
x_t & y_t & z_t \\
x_v & y_v & z_v \\
x_w & y_w & z_w \\
 \end{bmatrix}$$
For its geometrical interpretation take this as a reference, the image is really good for intuition.
Your exercise:
I assume you have already proved that the function is a linear application, this is not hard to do; you can write the associated matrix $A$ and interpretate its rows as vectors in a three dimensional space, so that it's evident that the absolute value of its determinant is the volume of the prism whose sides are the vectors $$\vec{t} = (1,-2,1), \vec{v} = (1,0,-3), \vec{w} = (1,k,k)$$ And the third one depends on a parameter $k$. When the determinant of this matrix is different from zero it means that the prism has a volume and so the three vectors do not lay on the same plane, therefore the three vectors are linearly independent (i.e. not coplanar). Instead if the determinant of this matrix is null then the volume of the prism is zero and this means that the prism was actually a parallelogram, which is a planar shape: all the three vectors lay on the same plane which means that they are linearly dependent. 
Let's now compute the triple product of $\vec{t}, \vec{v}, \vec{w}$: as you said 
$$det\begin{bmatrix} 
1 & -2 & 1 \\
1 & 0 & -3 \\
1 & k & k \\
 \end{bmatrix} = 6 + 6k = 6(k+1)$$
For every $k\neq -1$ the prism with sides $\vec{t}, \vec{v}, \vec{w}$ actually has a volume and so the three vectors are linearly independent: the only way you can obtain a null linear combination from them is by using all null coefficients: the kernel therefore is uniquely $\ker(f) = \{\vec{0}\}$ and his dimension is $0$. 
Instead when $k = -1$ the triple product is null and this means that the three vectors aren't linearly independent (they all lay on the same plane): there's not anymore an unique way to write every vector of the space and this means that you can obtain the null vector with a non trivial linear combination, so the kernel must be at least a line and have a dimension such that $dim(ker(f)) > 0$.
If you want to find out the exact dimension you have to compute the gaussian elimination after substituting $k = -1$. I have the impression that if you do you could write $Row_2 = Row_1 + 2Row_3$ (i.e. you could delete the second row when solving the system) but the remaining are non parallel, so $dim(ker(f)) = 1$
A: As I said in the comments, it's necessary to verify a map is linear before considering a matrix for it. Fortunately, your working for showing linearity is fine, except you basically stopped short of saying "Therefore, $f(\vec{u} + \vec{v}) = f(\vec{u}) + f(\vec{v})$."
In terms of row-reducing the matrix, I think you've been skipping too many steps. To determine the kernel, you do need to row reduce the augmented matrix:
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 1 & 0 & -3 & 0 \\ 1 & k & k & 0 \end{array}\right).$$
You can row reduce this like any other matrix, but you should be aware when dividing a row by a multiple of $k$ that you might be dividing by $0$. In such cases, you should treat the $0$ case separately (in this case, this should happen when $k = 0$).
The first step, in any case, is to subtract the top row from the second and third rows:
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 2 & -4 & 0 \\ 0 & k + 2 & k - 1 & 0 \end{array}\right).$$
We can then divide the second row by $2$:
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & k + 2 & k - 1 & 0 \end{array}\right)$$
and subtract $k + 2$ times the second row from the third (it doesn't matter if $k = 0$ here; this just turns out to be a nothing operation!):
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 3k + 3 & 0 \end{array}\right).$$
This is where we want to divide by $3k + 3$. The problem is when $3k + 3 = 0$, i.e. when $k = -1$. In this case, the matrix becomes
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right),$$
which has a kernel of dimension $1$ (one column without a pivot, so one free parameter) and thus a rank of $2$. Otherwise, $3k + 3 \neq 0$, and we can divide safely, yielding
$$\left(\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right),$$
a matrix with $3$ pivots, and a trivial kernel.
A: You have no idea how much i appreciate this, now everything is clear with the 6+6k.
I calculated RREF in matlab to do it faster and i didn't consider K=-1.
I guess that was my problem.
So from now on every time i have dynamic parameters like K,S etc i will do it manually.
I found out that if K=-1 then Det=0, and it has a Kernel of 1 dimension as you said, and the basis for the null space is [3 2 1].
So my answer for the exercise should be that it is linear, and the dimension of image and kernel depend on k, if k = -1 we got a plane, so R2, (isn't it a projection from R3->R2 ? ) and we got a kernel of dimension 1, and an image of dimension 2.
Otherwise, if K =/ -1 then we have a prism and it's volume depends on the parameter K, we have a kernel of 0 and an image of dimension 3 ? Right ? 
