1. ## linear transformations

consider the linear transformation T:R4--> R4 defined by
[w] [w+z]
T [x] = [y]
[y] [x-y]
[z] [w]

for each i have to justify the answer
a. is T injective?

b. is T surjective?

c.is T invertible?

2. Hi,
consider the linear transformation T:R4--> R4 defined by
[w] [w+z]
T [x] = [y]
[y] [x-y]
[z] [w]

for each i have to justify the answer
a. is T injective?
$T$ is injective if $\ker T = \{\vec{0}\}$. Is this true ?
b. is T surjective?
$T$ is surjective if, given any vector $v\in\mathbb{R}^4$, one can find $u\in\mathbb{R}^4$ such that $T(u)=v$. In other words, given $v=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}\in\mathbb {R}^4$, can you find at least a 4-tuple $(w,x,y,z)$ such that $\begin{cases} w+z=a\\y=b\\x-y=c\\w=d\end{cases}$ ?

3. Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.
eg. if I choose the vector
[1]
[0]
[0]
[-1]

Ok I understand the definitions but am still struggling to decide if T is injective or not, for the kernel(T) obviously we know this equals the null space(T) so if I find one vector in R4 such that Tv=the zero vector does this mean it is injective.
eg. if I choose the vector
[1]
[0]
[0]
[-1]
To show that $T$ is injective you have to show that the only vector $v$ such that $Tv=0$ is $v=0$; that's not what you're doing. Anyway, this idea is often used to show that a linear application is not injective. For if you find a non-zero vector $v$ such that $Tv=0$ then $\ker T$ contains both 0 and $v$ hence $\ker T\neq \{0\}$ and $T$ can't be injective.

EDIT: by the way, for $v=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\, Tv\neq 0$

5. Oh yeah I see that that is incorrect. So T(0) = 0 so T is injective!? Is that what you're saying?

No! $T(0)=0$ is true for any linear application $T$, even if $T$ is not injective (this result comes from $T(0)=T(x-x)=T(x)+T(-x)=T(x)-T(x)=0$). A linear application is injective if $T(v)=0 \implies v=0$. You are saying that a linear application is injective if $v=0\implies T(v)=0$. These two sentences are very different and only the former is correct.
Now, how can one show that $T(v)=0\implies v=0$? Well, the only thing we have to do is solve $T(v)=0$ for $v$. Let $v=\begin{bmatrix}w\\x\\y\\z \end{bmatrix}$.
$T(v)=0\implies \begin{cases} w+z=0\\y=0\\x-y=0\\w=0\end{cases}\implies\begin{cases} z=-w=0\\y=0\\x=y=0\\w=0\end{cases}\implies v=0$
hence $T$ is injective.