# Kernel and image of linear transformation

1. Apr 6, 2006

### stunner5000pt

Find a basis for Ker T and a basis for Im T

a) T: P_{2} -> R^2 \ T(a+bx+cx^2) = (a,b)

for Ker T , both a and b must be zero, but c can be anything
so the basis is x^2

for hte image we have to find the find v in P2 st T(v) = (a,b) \in P^2
the c can be anything, right?
cant our basis be (1,0) or (0,1) ??
But wshould hte dimensions of the kernel and image add up to the dimension of the preimage?

Latex is acting funny...

Last edited: Apr 6, 2006
2. Apr 6, 2006

### Hurkyl

Staff Emeritus
I think you have the right idea, but there's a problem: (0, 0, 1) isn't a polynomial! It's not an element of P_{2}!

I'm not really sure at all what you're doing here.

But there is one thing that's clearly wrong: T is a map from P_{2} to R². Therefore, the image of T is a subset of R². So, a basis of it must consist of elements of R².

3. Apr 6, 2006

### stunner5000pt

you're right i corrected it, it should be x^2, yes?

also corrected should be (1,0), (0,1)

4. Apr 6, 2006

### stunner5000pt

a couple of more questions

b) T:R3->R3, T(x,y,z) = (x+y,x+y,0)

basis for ker T-> (1,-1,0)

basis for the imT = (1,1,0),(0,0,1)

c)
$$T: M_{22} \rightarrow M_{22$$
$$T \left[\begin{array}{cc} a&b \\ c&d \end{array}\right] = \left[\begin{array}{cc} a+b&b+c \\ c+d&d +a\end{array}\right]$$
for the ker T: a =-b and b = -c, c= -d, a = -d
basis $$\left[\begin{array}{cc} 1&-1 \\ 0&0 \end{array}\right]$$
$$\left[\begin{array}{cc} 0&0 \\ 1&-1 \end{array}\right]$$

im T basis includes
$$\left[\begin{array}{cc} 1&0 \\ 0&1 \end{array}\right]$$
$$\left[\begin{array}{cc} 0&1 \\ 1&0 \end{array}\right]$$
$$\left[\begin{array}{cc} 1&1 \\ 0&0 \end{array}\right]$$
$$\left[\begin{array}{cc} 0&0 \\ 1&1 \end{array}\right]$$

Last edited: Apr 6, 2006