Question

Determine the kernel and range. HELP???

Determine the kernel and range of the transformation defined by the matrix

6 4

3 2
.
(Enter your answers as a comma-separated list. Enter each vector in the form
(x1, x2, ...).
Use r for any arbitrary scalar.)

Show that dim ker(T) + dim range(T) = dim domain(T).
dim ker(T) + dim range(T) = dim domain(T) right double arrow implies

Answer 1

`color(red)("range"(T) = {((2a),(a))| a in RR}),qquad dim"range"(T)=1`
`color(red)("ker"(T) = {((2t),(-3t))| t in RR}),qquad dim"ker"(T)=1`

Explanation:

The transformation takes the vector `((x),(y))` to the vector `((x^'),(y^'))` where

`((x^'),(y^')) = ((6,4),(3,2))((x),(y)) implies`

`x^' = 6x+4y`
`y^' = 3x+2y`

The range
It is easy to see that `x^'=2y^'`, so that vectors in the range of the map must be of the form `((2a),(a)), a in RR`. Conversely, if a vector is of the form `((2a),(a))` with `a in RR`, then any of the infinite vectors `((x),(y))` with `3x+2y=a` will map into it. Thus

`color(red)("range"(T) = {((2a),(a))| a in RR})`

It is easy to see that all the elements of `"range"(T)` are multiples of the single vector `((2),(1))`. The single element set `{((2),(1))}` is thus a spanning set for `"range"(T)`. Since it is obviously linearly independent, this set provides a basis for `"range"(T)`. Hence we have

`color(red)("dim"\ "range"(T) = 1)`

The Kernel

The kernel of the transformation is defined by

`"ker"(T) equiv {v in RR^2|Tv=0}`

So, if `v=((x),(y)) in "ker"(T)`, we have

`6x+4y = 0`
`3x+2y = 0`

Which will be satisfied by `x = 2t,\ y=-3t` for any `t in RR`. So

`color(red)("ker"(T) = {((2t),(-3t))| t in RR})`

It is obvious that `{((2),(-3))}` is a basis for `"ker"(T)`. Hence

`color(red)("dim"\ "ket"(T) = 1)`

Rank-nullity theorem

Since `"domain"(T)=RR^2`, we have `dim "domain"(T) = 2`

Thus the rank-nullity theorem

`dim"range"(T)+dim"ker"(T) = dim"domain"(T)`

is verified by `1+1 = 2`