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?

can anybody please help?!

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- Aug 13th 2008, 01:41 PMlinearalgebraduncelinear 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?

can anybody please help?! - Aug 14th 2008, 12:31 AMflyingsquirrel
Hi,

$\displaystyle T$ is injective if $\displaystyle \ker T = \{\vec{0}\}$. Is this true ?

Quote:

b. is T surjective?

- Aug 14th 2008, 01:13 AMlinearalgebradunce
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] - Aug 14th 2008, 01:48 AMflyingsquirrel
To show that $\displaystyle T$ is injective you have to show that the only vector $\displaystyle v$ such that $\displaystyle Tv=0$ is $\displaystyle v=0$; 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 $\displaystyle v$ such that $\displaystyle Tv=0$ then $\displaystyle \ker T$ contains both 0 and $\displaystyle v$ hence $\displaystyle \ker T\neq \{0\}$ and $\displaystyle T$ can't be injective.

EDIT: by the way, for $\displaystyle v=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix},\, Tv\neq 0$ :D - Aug 14th 2008, 02:07 AMlinearalgebradunce
Oh yeah I see that that is incorrect. So T(0) = 0 so T is injective!? Is that what you're saying?

- Aug 14th 2008, 03:30 AMflyingsquirrel
No! $\displaystyle T(0)=0$ is true for

*any*linear application $\displaystyle T$, even if $\displaystyle T$ is not injective (this result comes from $\displaystyle T(0)=T(x-x)=T(x)+T(-x)=T(x)-T(x)=0$). A linear application is injective if $\displaystyle T(v)=0 \implies v=0$. You are saying that a linear application is injective if $\displaystyle v=0\implies T(v)=0$. These two sentences are very different and only the former is correct.

Now, how can one show that $\displaystyle T(v)=0\implies v=0$? Well, the only thing we have to do is solve $\displaystyle T(v)=0$ for $\displaystyle v$. Let $\displaystyle v=\begin{bmatrix}w\\x\\y\\z \end{bmatrix}$.

$\displaystyle T(v)=0\implies \begin{cases} w+z=0\\y=0\\x-y=0\\w=0\end{cases}\implies\begin{cases} z=-w=0\\y=0\\x=y=0\\w=0\end{cases}\implies v=0$hence $\displaystyle T$ is injective.