Chapter 3, Section B

Give an example of a linear map T with dimnull T = 3 and dimrange T = 2.

Define a linear map T : ℝ⁵ → ℝ² such that:

Then null T is the set of all elements of ℝ³ such that x₁ = -1∕2x₂, x₃ = x₄ - 3x₅:

Thus, dimrange T = 2. Then, by the fundamental theorem of linear maps, it makes sense that the dimension of the domain, ℝ⁵, is 2 + 3 = 5.

Proof. Suppose S,T ∈(V ) are such that range $S ⊆ null T$ . Prove that $2 (ST ) = 0$ .

Suppose that $range S ⊆ ker T$ . Then, for all $v ∈ V$ , $T (Sv) = 0$ (*). We see that:

(ST)²v	= ST(STv)
	= S(TS)(Tv) (by associativity of linear maps)
	= S0 (by (*))
	= 0 (by definition of linear map)

Thus $2 (ST)$ is equivalent to the zero map, as we wanted to show. __

What property of $T$ corresponds to $v1,...,vm$ spanning $V$ ?

For v₁,…,v_m to span V , we want range T = V . This is precisely the definition of T being surjective.

What property of $T$ corresponds to the list $v1,...,vm$ being linearly independent?

For v₁,…,v_m to be linearly independent, we want for a linear combination of v₁,…,v_m, with coefficients a₁,…,a_m ∈ F:

where the zeroes in the first and second expressions are the list of length m whose entries are all 0, and the zeroes in the linear combinations are simply scalars. In other words, we require for null T = {0}

. But this is precisely the definition of T being injective. Thus, the property of T that corresponds to the list v₁,…,v_m being linearly independent is injectivity.

Show that

is not a subspace of

(ℝ⁵, ℝ⁴).

For dimnull T > 2 to hold, we look to the fundamental theorem of linear maps:

We know that dimV = 5, and that dimrange T = 4. So, by the fundamental theorem, we know that dimnull T = dimV - dimrange T = 5 - 4 = 1. So, the set defined as { }
T ∈ L (ℝ5,ℝ4 ) | dim null T > 2

is actually the empty set, ∅.

Give an example of T ∈

(ℝ⁴) such that range T = null T.

Define a linear transformation T ∈ (ℝ⁴):

Then, range T = Span

, and dimnull T = Span

. To put it more clearly, any vector of the form (a₁,a₂,0,0), when put through the linear transformation T, will become (0,0,0,0) (since the first two entries are “discarded” by T). But, the range of T is precisely those vectors of the form (b₁,b₂,0,0).

Also see that dimrange T = dimnull T = 2, and that dimrange T +dimnull T = 2+2 = 4, which is what we expect (since the domain is ℝ⁴).

It’s worth mentioning that I didn’t get this immediately, nor on my first try. I started by playing around with 2 × 2 examples of matrices (cheating a bit, since in Axler’s book, we haven’t quite gotten around to connecting linear transformations to matrices). My math professor constantly says that creating small examples is very important in gaining intuition to solve a more general problem, a piece of advice that (to my chagrin, sometimes) holds true in basically any scenario.

Prove that there does not exist T ∈

(ℝ⁵) such that range T = null T.

This actually holds true for any linear transformation that can be encoded as a square matrix with odd dimensions. We can look at it that way (which is slightly more concrete and familiar to me), or, we can look at it through the lens of the fundamental theorem of linear maps, which is probably what Axler wants us to do here…

So we just plug in appropriate values into the fundamental theorem. This is relevant because for range T to equal null T, their dimensions surely must be equal. So we have:

where dimnull T = dimrange T. Do you see the issue here? 5 is an odd number, and yet according to our equation above, we need a positive integer (let’s call it x) such that 2x = 5. But this is impossible.

Thus, there does not exist T ∈

(ℝ⁵) such that range T = null T.

Suppose V and W are finite-dimensional with 2 ≤ dimV ≤ dimW. Show that {T ∈ L (V,W ) | T is not injective}

is not a subspace of

(V,W).

If T ∈

(V,W) and T is not injective, then T must map to a space of strictly lesser dimension; i.e., dimV > dimW. Suppose that V,W are finite-dimensional and dimV ≤ dimW. So the set {T ∈ L (V,W ) | T is not injective}

is in fact empty and therefore is not a subspace of

(V,W).

Suppose V and W are finite-dimensional with dimV ≥ dimW ≥ 2. Show that {T ∈ L (V,W ) | T is not surjective}

is not a subspace of

(V,W).

Similarly to Exercise 7, we use the fact that if T ∈

(V,W) and T is not surjective, then T must map to a space of strictly higher dimension; i.e., dimV < dimW. But we suppose that V,W are finite-dimensional and that dimV ≥ dimW. So, {T ∈ L (V,W ) | T is not surjective}

is empty and therefore not a subspace of

(V,W).

Suppose T ∈

(V,W) is injective and v₁,…,v_n is linearly independent in V . Prove that Tv₁,…,Tv_n is linearly independent in W.

If T ∈

(V,W) and T is injective, then Tv = Tw implies v = w for v,w ∈ V . Suppose that the list of vectors v₁,…,v_n is linearly independent in V . We wish to show that Tv₁,…,Tv_n is linearly independent in W, so we set up a linear combination:

and we wish to show that for the above equality to hold, a₁ = ⋅⋅⋅

= a_n, all elements of F, must all be 0.

But we know that v₁,…,v_n is linearly independent in V :

and all b₁,…,b_n ∈ F must equal 0. Set these two linear combinations equal to each other:

Since v₁,…,v_n is a linearly independent list, we know that v₁ must be some non-zero element of V , so for this equality to hold, it is clear that b₁ = 0. Further, since we know that T is injective, the only vector that is mapped to the zero vector is the zero vector itself. Seeing that v₁ is a non-zero vector, we can assume that Tv₁ is also non-zero, which forces a₁ to equal 0. We repeat this process for a_iTv_i and b_iv_i, for i = 2,…,n, showing that at each step, a_i must equal 0.

We have shown for the linear combination

to equal 0, a₁ = ⋅⋅⋅

= a_n must all be 0. And since Tv₁,…,Tv_n ∈ W, we have shown that Tv₁,…,Tv_n is linearly independent in W.

Suppose v₁,…,v_n spans V and T ∈

(V,W). Show that Tv₁,…,Tv_n spans range T.

The range of a linear transformation is the subset of W consisting of those vectors that are equal to Tv for some v ∈ V . But note that any arbitrary vector v ∈ V can be expressed as a linear combination of v₁,…,v_n. So let v = c₁v₁ + ⋅⋅⋅

+ c_nv_n for some scalars c₁, ⋅⋅⋅

,c_n ∈ F. Then Tv = T(c₁v₁ + ⋅⋅⋅

+ c_nv_n. By linearity, this results in:

If any element of range T can be written as a linear combination of Tv₁,…,Tv_n, then it follows that Tv₁,…,Tv_n spans range T.

Suppose that V is finite-dimensional and that T ∈

(V,W). Prove that there exists a subspace U of V such that:

Written in plain English, the only vector in common to U and null T is the zero vector, and the range of T are all Tu where u is an element of U. We take a look at our reliable rank-nullity theorem:

and find that the constraints imposed on us earlier are quite possible, if we let dimnull T = 0, and dimrange T = dimV .

This can be done by letting dimnull T = {0}

(a quirk that is worth mentioning: the set which only contains the zero vector is considered to have dimension 0). This forces dimrange T = dimV , as we mentioned before. And since U is a subspace of V , we have a stronger argument: if U is a subspace of V and dimU = dimV , then U = V .

So, yes, there does exist a subspace U of V such that those conditions are met: U must be V itself.

Prove that T is surjective.

Since T is a linear map from F⁴ to F² (i.e., a linear map from a higher- to a lower-dimensional subspace), we can be certain that it can be surjective in the first place. So, to continue with proving that T is surjective, it is a matter of showing that range T = F².

First, let’s take a closer look at what the above equation is telling us: that all vectors in F⁴ of the form (5x₂,x₂,7x₄,x₄), are sent to 0 by the linear transformation T. How many vectors does it take to span this space? It takes two: a potential basis could be (5,1,0,0) and (0,0,7,1).

Note that the rank-nullity theorem tells us then that dimrange T = 2. This is what we wanted to show, since this implies that range T = F².

Suppose U is a three-dimensional subspace of ℝ⁸ and that T is a linear map from ℝ⁸ to ℝ⁵ such that null T = U. Prove that T is surjective.

Suppose U ⊂ ℝ³ and dimU = 3. Suppose also that T is defined as a linear map T : ℝ⁸ → ℝ⁵ such that null T = U. We want to show that range T = ℝ⁵.

The rank-nullity theorem tells us that:

so we know that dimrange T = 5. This suffices to show that dimrange T = F⁵. Thus, T is surjective.

Prove that there does not exist a linear map from F⁵ to F² whose null space equals {(x1,x2,x3,x4,x5) ∈ F5 | x1 = 3x2 and x3 = x4 = x5}

.

Suppose such a linear map exists, and let it be defined T : F⁵ → F². A closer look at the null space of T as described shows that all vectors in F⁵ of the form (3x₂,x₂,x₃,x₃,x₃) are sent to 0 by T. Since there are only two “free” variables in (3x₂,x₂,x₃,x₃,x₃), we know that null T can be spanned by two linearly independent vectors; that is, dimnull T = 2.

By the rank-nullity theorem:

(This is just another way to write the same thing we already have been writing; Ker referring to null and Im referring to range.) Plugging in our known values, we get the equation:

Suppose there exists a linear map on V whose null space and range are both finite-dimensional. Prove that V is finite-dimensional.

This immediately follows from the rank-nullity theorem. If the null space and range of a linear map over V are finite-dimensional, then the dimension of V is the sum of the dimension of the null space and range; the sum of two finite numbers will always be another finite number.

Suppose V and W are finite-dimensional. Prove that there exists an injective linear map from V to W iff dimV ≤ dimW.

: Suppose T ∈

(V,W) and that T is injective. We know that for this to be the case, dimV ≤ dimW. (3.22)

: Suppose dimV ≤ dimW. Suppose T is a linear map from V to W, but note that we aren’t making any assumptions about whether it is injective or otherwise (yet!). By the rank-nullity theorem:

For T to be injective, we know that null T must be {0}

, i.e., dimnull T = 0. Note that range T ⊂ W, so dimrange T ≤ dimW. We express this inequality in the theorem above, along with our claim that dimnull T should be zero:

This is precisely the condition we assumed at the beginning. Thus, we have shown that if dimV ≤ dimW, there exists an injective linear map from V to W. (At least, I hope we did. I easily get confused when writing iff proofs.)

Suppose W is finite-dimensional and T ∈

(V,W). Prove that T is injective iff there exists S ∈

(W,V ) such that ST is the identity operator on V .

= ⇒

: Suppose T ∈

(V,W) is injective. We wish to show that there exists S ∈

(W,V ) such that ST is the identity operator on V . If T is injective, then null T = {0}

; i.e., the only vector that is sent to 0 by T is the zero vector itself. Further, T maps distinct inputs to distinct outputs. If v,w ∈ V and v≠w, then Tv≠Tw.

So, it follows that there exists an S ∈

(W,V ) that takes distinct Tv and Tw, etc, back to v,w, etc. If this is the case, then ST essentialy maps v to v and w to w, which is precisely the definition of the identity operator.

: Suppose there exists S ∈

(V,W) such that ST is the identity operator on V . We wish to show that T ∈

(V,W) is injective. If the composition of two linear maps, S and T, results in the identity operator, this means that S is the inverse mapping of T. Therefore, T is invertible. Therefore, T is injective.

This is too verbose and hand-wavy, even for my taste. I probably will edit this later.

Suppose V is finite-dimensional, T ∈

(V,W), and U is a subspace of W. Prove that {v ∈ V | Tv ∈ U}

is a subspace of V and:

Intuitively, the set {v ∈ V | Tv ∈ U }

is talking about the subset of V such that it contains all the elements in V that are being mapped to U by T. So, we should just check the usual subspace conditions:

Let p,q ∈

. So, looking at p,q from the point of view of the subspace U, we note that p + q would be of the form Tp + Tq, and by linearity, could be written T(p + q). So p + q is clearly an element of {u ∈ V | Tv ∈ U}

Similarly, take cx ∈ {v ∈ V | Tv ∈ U }

, and rewrite it as T(cx). See that it could be written (by linearity) as cTx, so cx is clearly an element of {v ∈ V | T v ∈ U}

.

For the second part of this proof, we have to show something related to dimensions, so let’s consult our trusty rank-nullity theorem, which says the dimension of the domain of a linear map is the sum of the dimension of the null space and the dimension of the range of the linear map. But the set {v ∈ V | T v ∈ U}

is only considering a subset of the domain of T.

range T is defined to be all those elements in W that are “hit” by the linear map T. This set may be larger than U, since U is a subspace of W. But U is closely connected to the subspace of V , {v ∈ V | T v ∈ U}

, because this set describes those elements in v that get mapped to elements in U by T. Further, U ∩ range T is describing the very set {Tv ∈ U}

!

Then, it is clear that the dimension of {v ∈ V | T v ∈ U}

is the sum of the vectors that are “lost”, i.e., sent to zero after T is applied, plus the dimension of the space defined in the last paragraph (U ∩ range T).

Suppose U and V are finite-dimensional vector spaces and S ∈

(V,W) and T ∈

(U,V ). Prove that:

We see that ST describes a linear map from V to itself, i.e., ST ∈

(V ).

By rank-nullity theorem, define dimrange S ≤ dimV . By similar logic, dimrange ST ≤ dimV , so it follows that dimrange ST ≤ dimrange S.

Now we wish to show that dimrange ST ≤ dimrange T. Note that null T ⊂ null ST. So, dimnull T ≤ dimnull ST. By the rank-nullity theorem, this shows that dimrange ST ≤ range T.

Thus, dimrange ST ≤ min {dim range S,dimrange T}

Suppose that W is finite-dimensional and S,T ∈

(V,W). Prove that null S ⊆ null T iff there exists E ∈

(W) such that T = ES.

: Suppose that for linear maps S and T, both from V to W (with W being finite-dimensional), there exists a E ∈

(W) such that T = ES. We wish to show that null S ⊆ null T.

Suppose that v ∈ V and v ∈ null S. Then, note that if Sv = 0 for some v ∈ V , then (ES)v = E(Sv) = E(0) = 0, so v ∈ null T.

: Suppose that null S ⊆ null T. We wish to show that there exists E ∈

(W) such that T = ES.

Suppose that v ∈ null S, and that for all such v, v ∈ null T. We wish to show that there exists E ∈

(W) such that T = ES. Does this not follow from the fact that if w ∈ null S, then Sw = 0, and by extension Tw = 0? Then, define a mapping E from W to W, so that T = ES, and use similar reasoning from above to show that null S ⊂ null E. I’m pretty confused with this part of the proof, if it wasn’t obvious. I’ll have to come back to this.

Also very verbose—my apologies. Although this one is somewhat more focused/does a better job of explaining my thought process. Plus, it actually arrives at a concrete answer.

Suppose P ∈

(V ) and P² = P. Prove that V = null P ⊕ range P.

Whenever you see ⊕, it denotes a “direct sum”. The definition of a direct sum is something I always have to look up and reconfirm for myself what it is, because it’s simple enough for you to fool yourself. But it turns out it isn’t always so easy, especially because a direct sum means something slightly different depending on which mathematical object you’re summing over. Wow, there’s nuance.

Anyways, a direct sum of two vector spaces, V ⊕ W, isn’t so bad. Suppose for any x ∈ X, it can be uniquely expressed as x = x₁ + x₂ for x₁ ∈ V and x₂ ∈ W. Then X = V ⊕ W. I suppose you could also look at it like this:

Where v₁,…,v_n is a basis of V and w₁,…,w_m is a basis of W. Okay, what else do we know? Well, P is a linear transformation mapping elements of V to itself. Interestingly enough, P² = P, so applying P twice to v ∈ V produces the same result as if you’d just applied it once. I guess this tells us something crucial: that null P = {0}

.

Intuitively, I know that this must be the case because the equality P² = P tells me that we don’t “lose” any vectors when we apply P. That is, at least when applied the first and second time, P does not force any other vectors to go to 0 other than the zero vector itself. If P did have a non-trivial null space, then P² could never equal P; the vectors that got sent to 0 on the first pass of P would then be lost forever.

Well, if null P = {0}

, then by the rank-nullity theorem, range P must span the entire space of V , i.e., we know that range P ⊂ V , and that dimV = dimrange P, so range P = V .

Then it follows that the direct sum of null P and range P is the uninteresting direct sum of V and the zero vector, which is just…V .

Suppose p ∈

(ℝ). Prove that there exists a polynomial q ∈

(ℝ) such that 5q′′ + 3q′ = p.

Axler does note here that this can be proved without linear algebra, but that “it is more fun with linear algebra”. When I saw this question, my first instinct was to use the fact that the differentiation map is almost surjective. Then I thought, what would the integration map look like? (This question was probably irrelevant, however)

Define the degree of p to be n, and consider the linear map:

Now, all we have to do is show that S is surjective, because then that would mean for all p ∈

_n(ℝ), there is a mapping from q ∈

_n+1(ℝ) to p via S.

For S to be surjective, we want its range to be

_n(ℝ), so we want dimnull S = 1. But considering that S is a differentiation map, we know that any element of null S is simply a constant (the derivative of a constant is 0, the second derivative of a constant is 0). So, dimnull S = 1.

Now using our rank-nullity theorem:

The subspace of

_n(ℝ) with dimension n + 1 can only be

_n(ℝ) itself. Thus, we’ve proved that S is surjective, and we are done.

Direct sum again…okay, so if u ∈ V is not in null ϕ, then ϕ(u)≠0 by definition. In fact, any scalar multiplied by u is also not in null ϕ. This is due to the linearity of ϕ:

So, it follows that u and any scalar multiple of u is not in ϕ, which explains the second term of the direct sum above, {au | a ∈ F}

.

Further, note that the co-domain of ϕ is F. Then, dimrange ϕ ≤ 1. Since ϕ≠0, dimϕ > 0, therefore dimrange ϕ = 1.

Let dimV = n. Then:

Let a basis of null ϕ be v₁,…,v_n-1. We want to show that extending this basis by the vector u (therefore resulting in the list of vectors v₁,…,v_n-1,u) gives a valid basis for V . This suffices to show that V is a direct sum of null ϕ and {au | a ∈ F}

.

To show that v₁,…,v_n-1,u is a basis of V, we must show that this list is linearly independent. To do so, we have our usual routine:

We claim that for the above equality to hold, all coeffs c₁,…,c_n must be 0.

We write:

Which shows that since c_nu can be written as a linear combination of the basis vectors of null ϕ, c_nu ∈ null ϕ. Since we know that u

null ϕ, c_n must be 0. But wait. We have supposed that v₁,…,v_n-1 is a basis of null ϕ, so by definition they are linearly independent. So, in order for the equality to hold, we have to have c₁ = ⋅⋅⋅

= c_n-1 = 0.

Since we have shown v₁,…,v_n-1,u to be linearly independent, we have proved that it is a valid basis for V ; we are done.

Suppose V is finite dimensional, X is a subspace of V , and Y is a finite-dimensional subspace of W. Prove that there exists T ∈

(V,W) such that null T = X and range T = Y iff dimX + dimY = dimV .

: Suppose that dimX + dimY = dimV . We wish to show that there exists T ∈

(V,W) such that null T = X and range T = Y .

I’m not sure how to solve this other than directly referencing dimX + dimY = dimW. Help…

: Suppose that there exists T ∈

(V,W) such that null T = X and range T = Y . Then, we know that by the rank-nullity theorem, dimV = dimX + dimY .

I felt so insane solving this, not only because I had no idea what I was doing, but also because I had typoed the problem description: instead of typing dimX + dimY = dimV , I wrote dimX + dimY = dimW, which doesn’t make sense in the context of this problem. I should really look back at the textbook more T_T

Suppose V is finite-dimensional with dimV > 1. Show that if ϕ :

(V ) → F is a linear map such that ϕ(ST) = ϕ(S)ϕ(T) for all S,T ∈

(V ), then ϕ = 0.

Suppose that ϕ is a linear map from

(V ) to F such that ϕ(ST) = ϕ(S)ϕ(T) for all S,T ∈

(V ). Whenever I’m not sure how to approach a proof, I will rewrite the supposition word for word while I desperately try to find purchase.

Axler says that this is a hint: A subspace ξ of

(V ) is a two-sided ideal of

(V ) if TE ∈ ξ and ET ∈ ξ for all E ∈ ξ and all T ∈

(V ). Note that the only two-sided ideals of

(V ) are

(V ) and

.

I’m not entirely sure how I am supposed to use this. Suppose that:

Tv	= T(c₁v₁ + + c_nv_n)
	= Tc₁v₁ + + Tc_nv_n
	= c₁Tv₁ + + c_nTv_n

ϕ(λu)	= λϕ(u)
	= λ(0)
	= 0

dimnull ϕ	= dimV - dimrange ϕ
	= dimV - 1
	= n - 1

ϕ(ST)	= ϕ(T)ϕ(S)
	= ϕ(S)ϕ(T) by commutativity in F
	= ϕ(TS)

dimrange ϕ = dim(V ) - dimnull ϕ	≤ 1 by rank-nullity thm.
- dimnull ϕ	≤ 1 - dim(V )
dimnull ϕ	≥ n - 1

Chapter 3, Section B

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