A linear map is always 1-to-1 if the null space has dimension zero.

The converse of this statement is also true. A linear map is 1-to-1 if the null space has dimension 0.

The linear map, \(\displaystyle f,\) is not 1-to-1 because more than one vector goes to the zero vector. In fact, all vectors go to the same vector for the zero mapping!

For example, the vector \(\displaystyle (1,1)\) and \(\displaystyle (10,4)\) both go to the same vector. Thus \(\displaystyle f\) is not 1-to-1

This linear mapping is not 1-to-1. We know this because the null space is spanned by two vectors. Therefore the null space has dimension \(\displaystyle 2\). A linear map is 1-to-1 only if it has a null space with dimension zero. This linear map doesn't have a null space of dimension zero, therefore it is not 1-to-1.

The answer is no because the dimension of the null space is not zero. This comes from the equation

\(\displaystyle d=n+r\)

We know that the domain is \(\displaystyle R^4\) which has dimension of \(\displaystyle 4\). Therefore

\(\displaystyle d=4\)

Also from the problem statement \(\displaystyle r=2\).

Plugging these into the equation gives

\(\displaystyle d=n+r\)

\(\displaystyle 4=n+2\)

\(\displaystyle n=2\)

Since the dimension of the null space is \(\displaystyle 2\) and not \(\displaystyle 0\), then the function can't be 1-to-1.

If the dimension of the null space is zero then the linear map is 1-to-1.

For this problem, we are told the null space is only the zero vector. Therefore the null space has dimension \(\displaystyle 0\). Since the null space has dimension \(\displaystyle 0\), then \(\displaystyle f\) is 1-to-1.

The answer is yes because the dimension of the domain and the rank are equal. This implies that the dimension of the null space is zero.

This comes from the equation

\(\displaystyle d=n+r\)

We know that the domain is \(\displaystyle R^4\) which has dimension of \(\displaystyle 4\). Therefore

\(\displaystyle d=4\)

Also from the problem statement \(\displaystyle r=4\).

Plugging these into the equation gives

\(\displaystyle d=n+r\)

\(\displaystyle 4=n+4\)

\(\displaystyle n=0\)

Since the dimension of the null space is \(\displaystyle 0\) then the function is 1-to-1.

Notice that whenever \(\displaystyle d=r\), then \(\displaystyle n\) is always zero. Thus whenever \(\displaystyle d=r\), the linear mapping is 1-to-1.

Any vector in the null space satisfies \(\displaystyle f(x,y) = (0,0,0)\).

Therefore we get the following equation:

 \(\displaystyle (5x+5y,x+y,0) = (0,0,0)\)

Thus \(\displaystyle x=-y\). Hence the null space is any vector in form \(\displaystyle (a,-a)\) where \(\displaystyle a\) is any real number. Therefore, any point on the line \(\displaystyle y=x\) gets mapped to the zero vector in \(\displaystyle R^3\) 

The zero map takes all vectors to the zero vector. Therefore, the entire domain of the map is the null space. The domain of this map is \(\displaystyle R^2\). Thus \(\displaystyle R^2\) is the null space.

The image of the the zero map is the zero vector. A single vector has dimension \(\displaystyle 0\). Therefore the dimension of the image is zero. Hence the rank is zero. 

For \(\displaystyle T\) to be a linear transformation, it must hold that 

\(\displaystyle T(f+g)= T(f)+ T(g)\)

and

\(\displaystyle T(cf) = cT(f)\)

for all \(\displaystyle f,g\) in the domain of \(\displaystyle T\) and and for all scalar \(\displaystyle c\).

Let \(\displaystyle f, g \in C_{ \mathbb{R}}\).

\(\displaystyle T(f) = \sum_{j=1}^{100} f'(j)\) and \(\displaystyle T(g) = \sum_{j=1}^{100} g'(j)\), so

\(\displaystyle T(f)+ T(g)= \sum_{j=1}^{100} f'(j)+ \sum_{j=1}^{100} g'(j)\)

By the sum rule for finite sequences,

\(\displaystyle T(f)+ T(g)= \sum_{j=1}^{100}[ f'(j) + g'(j)]\)

By the derivative sum rule,

\(\displaystyle T(f)+ T(g)= \sum_{j=1}^{100}( f '+g')(j)\)

\(\displaystyle T(f)+ T(g)= \sum_{j=1}^{100}( f +g)'(j)\)

\(\displaystyle T(f)+ T(g)= T(f+g)\)

The first condition is met.

Let \(\displaystyle f \in C_{ \mathbb{R}}\) and \(\displaystyle c\) be a scalar.

\(\displaystyle T(f) = \sum_{j=1}^{100} f'(j)\)

\(\displaystyle c T(f) = c\sum_{j=1}^{100} f'(j)\)

By the scalar product rule for finite sequences,

\(\displaystyle c T(f) = \sum_{j=1}^{100}c f'(j)\)

By the scalar product rule for derivatives,

\(\displaystyle c T(f) = \sum_{j=1}^{100}(c f)'(j)\)

\(\displaystyle c T(f) =T(c f)\)

The second condition is met.

\(\displaystyle T\) is a linear transformation.