Question

Suppose V and W are two subspaces in R^n such that v . w =0, for all v in V and w in W. If S is a set of linerly independent vectors in V, and T is a set of linearly independent vectors in W, show that C, defined as the union of S and T is linearly independent.

Answer 1

need help?

Answer 2

yes please

Answer 3

ok let me think

Answer 4

thank you

Answer 5

We saw last time that we can add linear transformations and multiply them by scalars. These are just two ways to generate new linear transformations. Another obvious one is composition.
Proposition 0.1. Let T :V →W and U :W →Z be linear (with all spaces over the same field F). Then the composition U ◦ T is a linear transformation from V to Z.
Proof. Let v1,v2 ∈ V and c ∈ F. Then
(U ◦ T)(cv1 + v2) = U(T(cv1 + v2)) = U(cT(v1) + T(v2))
= cU(T(v1)) + U(T(v2)) = c(U ◦ T)(v1) + (U ◦ T)(v2) .
Recall that each T : C → C that is linear is completely determined by its value at 1. Note that {1} is a basis. This fact holds true for all linear transformations and is one of the most important theorems of the course: in the words of Conway, each linear transformation is completely determined by its values on a basis, and any values will do!
Theorem 0.2 (The slogan). Let V and W be vector spaces over F. If {v1, . . . , vn} is a basis for V and w1, . . . , wn are any vectors in W (with possible duplicates) then there is exactly one T ∈ L(V,W) such that T(vi) = wi for all i = 1,...,n.
Proof. This is an existence and uniqueness statement, so let’s first prove uniqueness. Suppose that T,U ∈ L(V,W) both map vi to wi for all i. Then write an arbitrary v ∈ V uniquely as v = a1v1 + · · · + anvn. We have
T(v)=T(a1v1 +···+anvn)=a1T(v1)+···+anT(vn)=a1w1 +···+anwn =a1U(v1)+···+anU(vn)=U(a1v1 +···+anvn)=U(v).
To prove existence we must construct one such linear map. Each v ∈ V can be written uniquely as v = a1v1 + · · · + anvn, so define T : V → W by
T(v)=a1w1+···+anwn .
The fact that T is a function (that is, for each v ∈ V there is exactly one w ∈ W such that T(v) = w) follows from uniqueness of the representation of v in terms of the basis. So we must show linearity. If v, v′ ∈ V , write v = a1v1 + · · · + anvn and v′ = b1v1 + · · · + bnvn. We have for c ∈ F,
T(cv+v′)=T((ca1 +b1)v1 +···+(can +bn)vn) =(ca1 +b1)w1 +···+(can +bn)wn
=c(a1w1 +···+anwn)+(b1w1 +···+bnwn) = cT(v) + T(v′) .

Range and nullspace
Next we define two very important subspaces that are related to a linear transformation T.
Definition 0.3. Let T : V → W be linear. The nullspace, or kernel, of T is the set N(T)⊂V definedby
N ( T ) = { v ∈ V : T ( v ) = ⃗0 } . The range, or image, of T, is the set R(T) ⊂ W defined by
R(T ) = {w ∈ W : T (v) = w for some v ∈ V } .
In the definition of N (T ) above, ⃗0 is the zero vector in the space W .
Proposition 0.4. Let T : V → W be linear. Then N(T) is a subspace of V and R(T) is a subspace of W.
Proof. First N(T) is nonempty, since each linear transformation must map ⃗0 to ⃗0: T(⃗0) = T(0⃗0)=0T(⃗0)=⃗0. Ifv1,v2 ∈N(T)andc∈F,
T (cv1 + v2) = cT (v1) + T (v2) = c⃗0 + ⃗0 = ⃗0 ,
so cv1 + v2 ∈ N (T ), showing that N (T ) is a subspace of V . For R(T ), it is also non-empty, since ⃗0 is mapped to by ⃗0. If w1,w2 ∈ R(T) and c ∈ F, choose v1,v2 ∈ V such that T(v1) = w1 and T(v2) = w2. Then
cw1 + w2 = cT(v1) + T(v2) = T(cv1 + v2) ,
so cw1 +w2 is mapped to by cv1 +v2, a vector in V and we are done.
In the finite-dimensional case, the dimensions of these spaces are so important they get their own names: the rank of T is the dimension of R(T ) and the nullity of T is the dimension of N(T). The next theorem relates these dimensions to each other.
Theorem 0.5 (Rank-nullity). Let T : V → W be linear and dim(V ) < ∞. Then rank(T ) + nullity(T ) = dim(V ) .
Proof. In a way, this theorem is best proved using quotient spaces, and you will do this in the homework. We will prove it the more standard way, by counting and using bases. Let {v1,...,vk} be a basis for the nullspace of T and extend it to a basis {v1,...,vk,vk+1,...,vn} for V. We claim that T(vk+1),...,T(vn) are distinct and form a basis for R(T); this will completetheproof. IfT(vi)=T(vj)forsomei,j∈{k+1,...,n},wethenhaveT(vi−vj)= ⃗0, implying that vi − vj ∈ N(T). But we have a basis for N(T): we can write
vi−vj =a1v1+···+akvk 2

and subtracting vi − vj to the other side, we have a linear combination of elements of a basis equal to zero with some nonzero coefficients, a contradiction.
Now we show B = {T (vk+1), . . . , T (vn)} is a basis for R(T ). They are clearly contained in the range, so Span(B) ⊂ R(T). Conversely, if w ∈ R(T) we can write w = T(v) for some v ∈ V and using the basis, find coefficients such that bi such that
w = T (v) = T (b1v1 + . . . + bnvn) .
Expanding the inside, we get b1T (v1) + · · · + bnT (vn). The first k vectors are zero, since
v1,...,vk ∈N(T),so
proving w ∈ Span(B) and therefore B spans R(T).
For linear independence, let bk+1T (vk+1) + · · · + bnT (vn) = ⃗0. Then ⃗0 = T (bk+1vk+1 + · · · + bnvn) ,
so bk+1vk+1 + ··· + bnvn ∈ N(T). As before, we can then write these vectors in terms of v1,...,vk, use linear independence of {v1,...,vn} to get bi = 0 for all i.
One reason the range and nullspace are important is that they tell us when a transfor- mation is one-to-one (injective) or onto (surjective). Recall these definitions:
Definition0.6.IfXandY aresetsandf:X→Y isafunctionthenwesaythatfis one-to-one (injective) if f maps distinct points to distinct points; that is, if x1, x2 ∈ X with x1 ̸= x2 then f(x1) ̸= f(x2). We say that f is onto (surjective) if each point of Y is mapped tobysomex;thatis,foreachy∈Y thereexistsx∈X suchthatf(x)=y.
Proposition 0.7. Let T : V → W be linear. Then 1. T is injective if and only if N(T) = {⃗0}.
2. T is surjective if and only if R(T) = W.
Proof. The second is just the definition of surjective, so we prove the first. Suppose that T is injective and let v ∈ N(T). Then T(v) = ⃗0 = T(⃗0), but because T injective, v = ⃗0, proving that N(T) ⊂ {⃗0}. As N(T) is a subspace, we have {⃗0} ⊂ N(T), giving equality.
Conversely suppose that N(T) = {⃗0}; we will prove that T is injective. So assume that T(v1) = T(v2). By linearity, T(v1 −v2) =⃗0, so v1 −v2 ∈ N(T). But he only vector in N(T) is the zero vector, so v1 − v2 = ⃗0, giving v1 = v2 and T is injective.
In the previous proposition, the second part holds for all functions T, regardless of whether they are linear. The first, however, need not be true if T is not linear. (Think of an example!)
We can give an alternative characterization of one-to-one and onto: Proposition 0.8. Let T : V → W be linear.
w = bk+1T (vk+1) + · · · + bnT (vn) ,
3
1. T is injective if and only if it maps linearly independent sets of V to linearly indepen- dent sets of W.
2. T is surjective if and only if it maps spanning sets of V to spanning sets of W.
3. T is bijective if and only if it maps bases of V to bases of W.
Proof. The third part follows from the first two. For the first, assume that T is injective and let S ⊂ V be linearly independent. We will show that T(S) = {T(v) : v ∈ S} is linearly independent. So let
a 1 T ( v 1 ) + · · · + a n T ( v n ) = ⃗0 .
ThisimpliesthatT(a1v1+···+anvn)=⃗0,implyingthata1v1+···+anvn =⃗0byinjectivity. But this is a linear combination of vectors in S, a linearly independent set, giving ai = 0 for all i. Thus T(S) is linearly independent.
Conversely suppose that T maps linearly independent sets to linearly independent sets and let v ∈ N (T ). If v ̸= ⃗0 then {v} is linearly independent, so {T (v)} is linearly indepen- dent. But if T (v) = ⃗0 this is impossible, since {⃗0} is linearly dependent. Thus v ̸= ⃗0 and N(T) = {⃗0}, implying T is injective.
For item two, suppose that T is surjective and let S be a spanning set for V . Then if w∈W wecanfindv∈V suchthatT(v)=wandalinearcombinationofvectorsofSequal to v: v = a1v1 + · · · + anvn for vi ∈ S. Therefore
w = T (v) = a1T (v1) + · · · + anT (vn) ,
meaning that we have w ∈ Span(T (S)), so T (S) spans W . Conversely if T maps spanning sets to spanning sets, then T(V ) = R(T) must span W. But since R(T) is a subspace of W, this means R(T) = W and T is onto.


Answer 6

hope this helps :)

Answer 7

i dont get it... could you explain it step by step?

Answer 8

Why don't we suppose not and try to arrive at a contradiction. Are you still here?

Answer 9

yep im still here. can you help?

Answer 10

Suppose \(S \cup T\) is not linearly independent and therefore we have that
for \(a_1s_1 + a_2s_2 + ... a_ns_n + b_1t_1 + b_2t_2 + ... + b_mt_m = 0\) where some \(a_i\neq 0\) or \(b_i \neq 0\). Do you agree so far?

Answer 11

right... because then if they were linearly independent, they would be not equal to 0. so then a and b would just be the norm?

Answer 12

If they were linearly independent then the only linear combination yielding 0 is the linear combination that includes none of the vectors (all scalar coefficients are 0).

Answer 13

Since we are assuming that \(S \cup T\) is linearly dependent, we know that this is not the case and so we can obtain the 0 vector with some \(a_i\) and \(b_i\) not 0.

Answer 14

Furthermore, we see that to obtain the 0 vector we must have both non-zero \(a_i\) terms and non-zero \(b_i\) terms. Why is that? Why is it a contradiction for us to be able to obtain the zero vector with only non-zero \(a_i\) or only non-zero \(b_i\) terms?

Answer 15

Hint, remember that \(S\) and \(T\) are defined to be linearly independent.

Answer 16

to obtain the 0 vector, a and b both has to be non-zero so that we can find the values of s and t. so that when you dot product a with s and b with t you would get 0.
correct?

Answer 17

You are jumping ahead a bit. We are assuming \(S \cup T\) is linearly dependent to obtain a contradiction. Now we need to show that the linear combination above has both \(a_i\) and \(b_i\) terms non-zero. You see why this is the case, correct?

Answer 18

im sorry im not quite sure.. apparently i dont know as much as i thought i do.. how did you get a1s1+a2s2+...ansn+b1t1+b2t2+...+bmtm=0 where some ai≠0 or bi≠0.?

Answer 19

After that, we notice the following
Let \(s = a_1s_1 + a_2s_2 + ... + a_ns_n\)
Let \(t = b_1t_1 + b_2t_2 + ... + b_mt_m\)
We know that \(s \neq 0\) and \(t \neq 0\) as mentioned earlier. Furthermore we see that
\(s + t = 0\) and therefore \(s = -t\)
What does that tell us about the dot product between \(s\) and \(t\)

Answer 20

how do you know that s+t=0?

Answer 21

Recall at the beginning we had the following linear combination.
\(a_1s_1 + a_2s_2 + ... a_ns_n + b_1t_1 + b_2t_2 + ... + b_mt_m = 0\)
Since we are assuming that \(S \cup T\) is linearly dependent, we should be able to obtain the 0 vector with non-zero coefficients \(a_i\) and \(b_i\)

Answer 22

okay yes! i definitely understand so far!
so basically the dot product of a and s + the dot product of b and t would give us 0. since it is zero, that means that they are parallel, and therefore linearly dependent.
am i understanding right so far

Answer 23

Almost, except the dot product is non-zero that is the conclusion. Recall that it is orthogonal vectors that yield a zero dot product not parallel vectors.

Answer 24

\(s = -t\)
\(s \cdot t = s \cdot (-s) = -|s|^2\)

Answer 25

if the dot product is non-zero then how come when you wrote it, you wrote it to be equal to zero?

Answer 26

The linear combination yields zero. I don't see where you get that the dot product is zero.

Answer 27

oh okay. so how do i know that the linear combination would equal to zero? or is that just a rule that i must remember?

Answer 28

Do you know the definition of linear independence and dependence?

Answer 29

i know that the vectors have to be non-parallel to be linearly independent. if they are parallel, then they are linearly dependent.

Answer 30

If we have a set of linearly dependent vectors \(\{v_1, ..., v_n\}\) then we know that there is a solution the following equation \(a_1v_1 + ... + a_nv_n = 0\) where some \(a_i \neq 0\).

This is not true if the vectors are linearly independent. Then the only way to yield the zero vector is to take none of the vectors.

@Zarkon can you take over. Perhaps I'm not explaining this very well.

Answer 31

joanncja
do you understand this equation that Alchemista wrote

\[s \cdot t = s \cdot (-s) = -\|s\|^2\]

Answer 32

yes i understand that part, but i dont understand the start

Answer 33

what do you mean by "start"

Answer 34

a1n1 +... +anvn = 0.

what is a?

Answer 35

what do you mean by that?

Answer 36

the a's in front of the vectors are just scalars

Answer 37

okay. then why is it equal to 0? would that not be for if it is linearly independent?

Answer 38

look at this again...

\[s \cdot t = s \cdot (-s) = -\|s\|^2\]


do you see by how \(s\) and \(t\) are defined that \(s\cdot t=0\)

Answer 39

we are given ...." v . w =0, for all v in V and w in W"

Answer 40

So we notice that \(-\|s\| ^2 \neq 0 = s \cdot t = 0\) a contradiction.

Answer 41

so...

\[0=s \cdot t = s \cdot (-s) = -\|s\|^2\]

and \[\|s\|=0\Leftrightarrow s=0\]

this implies that all the scalars in s are o

Answer 42

Ah or you can argue directly instead of by contradiction.

Answer 43

also

\[0=s \cdot t = -t\cdot t = -\|t\|^2\]

Answer 44

so all the scalars for t are zero......

Answer 45

I guess we lost him

well I had fun ;)

Answer 46

the connection is lost hang on im trying to understand :(