Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation. Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that R is linear and invertible.
Solution
(a)
Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x):
T(x) = T(T(v))
= T 2 (v)
= 0
Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have im(T) contained in ker(T).
(b)
R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of two linear transformations, it must itself be linear as well. If you want to write the proof out, it would look like this:
Let v,w be in V, and k be a scalar.
R(k*v+w) = (k*v+w) + T(k*v+w)
= k*v + w + k*T(v) + T(w) [since T is linear]
= k*[v+T(v)] + [w+T(w)]
= k*R(v)+R(w)
Therefore R is linear.
Suppose R(v) = 0
0 = R(v)
= v + T(v)
T(v) = -v
We know that T(T(v)) = 0
From the above, we have that T(v)=-v. Therefore we have:
T(-v) = 0
-T(v) = 0 [T is linear]
Once again we know that T(v)=-v. Therefore we have:
-(-v) = 0
v = 0
Therefore R(v)=0 --> v=0
Since R is linear, this means R is invertible.
.
Let T - V rightarrow V be a linear transformation sulci that T2 -.docx
1. Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation.
Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that
R is linear and invertible.
Solution
(a)
Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x):
T(x) = T(T(v))
= T 2
(v)
= 0
Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have
im(T) contained in ker(T).
(b)
R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of
two linear transformations, it must itself be linear as well. If you want to write the proof out, it
would look like this:
Let v,w be in V, and k be a scalar.
R(k*v+w) = (k*v+w) + T(k*v+w)
= k*v + w + k*T(v) + T(w) [since T is linear]
= k*[v+T(v)] + [w+T(w)]
= k*R(v)+R(w)
Therefore R is linear.
Suppose R(v) = 0
0 = R(v)
= v + T(v)
T(v) = -v
We know that T(T(v)) = 0
From the above, we have that T(v)=-v. Therefore we have:
2. T(-v) = 0
-T(v) = 0 [T is linear]
Once again we know that T(v)=-v. Therefore we have:
-(-v) = 0
v = 0
Therefore R(v)=0 --> v=0
Since R is linear, this means R is invertible.
![Let T : V rightarrow V be a linear transformation sulci that T2 = 0 is the zero transformation.
Show that itn T ker T If R: V rightarrow V is defined by R(v) = v + T(v) for all v in V. show that
R is linear and invertible.
Solution
(a)
Let x be in im(T). Then x = T(v) for some v in V. Now consider T(x):
T(x) = T(T(v))
= T 2
(v)
= 0
Since T(x) = 0, x must be in ker(T). This was true for a generic x in im(T), so we must have
im(T) contained in ker(T).
(b)
R(v) = I(v) + T(v), where I is the identity transformation (which is linear). Since R is the sum of
two linear transformations, it must itself be linear as well. If you want to write the proof out, it
would look like this:
Let v,w be in V, and k be a scalar.
R(k*v+w) = (k*v+w) + T(k*v+w)
= k*v + w + k*T(v) + T(w) [since T is linear]
= k*[v+T(v)] + [w+T(w)]
= k*R(v)+R(w)
Therefore R is linear.
Suppose R(v) = 0
0 = R(v)
= v + T(v)
T(v) = -v
We know that T(T(v)) = 0
From the above, we have that T(v)=-v. Therefore we have:](https://image.slidesharecdn.com/lett-vrightarrowvbealineartransformationsulcithatt2-230204030508-ece099ca/85/Let-T-V-rightarrow-V-be-a-linear-transformation-sulci-that-T2-docx-1-320.jpg)
![T(-v) = 0
-T(v) = 0 [T is linear]
Once again we know that T(v)=-v. Therefore we have:
-(-v) = 0
v = 0
Therefore R(v)=0 --> v=0
Since R is linear, this means R is invertible.](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)