00:01
Let v via finite dimensional vector space and let t from v to b be a linear transformation such that t -square is equal to t, where t -square means t compose with t.
00:17
We want to prove first that kernel of t is exactly equal to the set of b of vectors v minus t of b for vectors b in v.
00:30
In part 2, we want to prove that v is exactly equal to the kernel, the nucleus of t plus range of t.
00:39
And in part 3, the kernel of t intersection range of t is only the set formed by the zero vector.
00:50
So let's start with part one.
00:54
So we got to prove first that kernel of t or the other way around better first.
01:05
I'm going to prove that the set of vectors v minus t of v taking v any vector in v is contained in the kernel of t that is we take an element in this set any element that say let v minus t of v a vector v or v a vector in v or v a vector in fields that is we take any vector in v and we form this operation v minus t of v which is the form of any element on this set.
01:57
So we are taking any element of the set on the left and we want to prove that is in the kernel of t.
02:05
So to prove that far better we start already doing the calculations then if we apply t to this vector we get t of this vector b minus d of b by the linearity of t, linearity of t, we get t of b minus t of t of v.
02:37
That is t of b minus t square of b.
02:45
But you know that t square is equal to t.
02:55
So t v minus t of v, that is this vector here, the transformation is, we are right here and we are going to say now that t squared being equal to t, this t squared is t squared of b is t of b.
03:23
But we are subtracting the vector minus itself is the zero vector.
03:29
And that means that this vector here is in the kernel of the transformation because the transformation applied to that vector is zero.
03:43
Then b minus t of v belongs to the kernel of t as we wanted to prove because we wanted to prove that any element we took here on this set, which was this element here, v minus t of v.
04:01
You prove that this on the set on the right kernel of t.
04:07
Correct.
04:08
So that we have proved that and now we proved the other way around that is.
04:12
We're going to prove that kernel of t is contained in the set of vectors v minus t of v or b in this vector space v so let's be or let not use the letter that u belongs to kernel of t then t of u is zero by definition of kernel of t is equal to a zero vector and now we want to see that this vector u is equal to u minus t of u.
05:08
That is because we can put it this way to see clearer.
05:15
U is the same as u minus zero vector, of course, because subtracting zero vector do not change the vector u.
05:21
But zero, as we saw here, is the image of t of u.
05:26
So that is u minus t of u.
05:30
And for u is a vector of course in vector space because it's in the general so it's vector in the vector space so we have proved we have found that u is exactly equal to a vector of the form of the vectors in the set over here then we can say that u belongs to the set of v minus t of b for b in the vector's phase okay, so we have proved the two inclusion, the two subsets, and so we have equality.
06:13
From a and b, we have that the set of v minus t of v for b in v is just the kernel of t is equal.
06:34
We prove the two inclusions and that implies equality.
06:38
So we go to the second part.
06:40
We want to prove in the second part that v vector space is exactly equal to the sum of the kernel plus the range of t.
07:00
And the sum means adding all the possible elements in the kernel of t with the elements in range of t.
07:10
The first inclusion here, i mean we had to prove equality of sets, so we got to produce two inclusions that's within part one, but there is one that is evident.
07:25
It is evident that kernel of t plus range of t is contained in space v.
07:38
That is because the kernel and the range both are formed by vectors in the vector space v.
07:46
So the sum is again in b because v is a vector space and that implies this inclusion over here so that it's evident.
07:56
So let us prove the other one.
08:01
Let's prove that v is in the sum of the kernel of t plus the range of t.
08:15
And with that we will have equality.
08:18
So let v, let's say u, v a vector in v.
08:27
I want to prove that is the sum of a vector in the kernel and a vector in the range of t.
08:35
So in this case we see that u is equal to u minus t of u plus t of u of course because we are subtracting and adding t of u to the vector u so this quality is evident we have written the vectors that way so now u minus t of u belongs to the kernel of t in why we know that because part once tell us that.
09:14
That is part one, we prove that the kernel of t is exactly equal to this set, the set of vectors of the form v minus t of v for any vector v in the vector space v.
09:26
So this we have here, u is a vector in v, and we have formed the difference, u minus, image of u through the linear transformation t.
09:40
So this element is in the kernel of t because it's in the set of vectors of the form b minus t of b for b in the vector space v and that set is equal to the kernel of t following part one.
09:57
So see in t of v using one.
10:14
So this first part here is in the kernel and this is obviously in the range because the image of a vector...