# Thread: Linear Algebra Proof Help.

1. ## Linear Algebra Proof Help.

I'm not sure if this is the right section to post in but any help would be appreciated since I'm new here.

Suppose V is finite dimenstional and S, T, U are in L(V) where L(V) is the set of linear operators from V to itself and STU = I. Show that T is invertible and that T^-1 = US.

What I tried was saying something along the lines of ST(U) shows ST is invertible with inverse U or S(TU) shows S is invertible with inverse TU, I'm just stuck since T is in the middle and linear maps don't commute, or at least I thought they didn't. This might be different since these are all operators. Any help? Thank you!

2. ## Re: Linear Algebra Proof Help.

One way is to use the fact that if $\operatorname{rank}(A)\overset{\text{def}}{=} \dim\operatorname{Im}(A)$, then $\operatorname{rank}(AB)\le\operatorname{rank}(A)$ and $\operatorname{rank}(AB)\le \operatorname{rank}(B)$. The former is obvious since $\operatorname{Im}(AB)\subseteq \operatorname{Im}(A)$. The latter holds because $\operatorname{Im}(AB)= A(\operatorname{Im}(B))$ and if $C\in L(V_1,V_2)$, then $\operatorname{rank}(C)\le\dim V_1$.

Therefore, $\operatorname{rank}(STU)\le\operatorname{rank}(T)$, but if $T$ is not invertible, then $\operatorname{rank}(T)<\dim V=\operatorname{rank}(I)$, a contradiction. Then $STU=I$ implies that $T=S^{-1}U^{-1}$, so $T^{-1}=US$.

3. ## Re: Linear Algebra Proof Help.

I'm sorry I don't exactly follow what you did in the first part. So if rank(A) = dim Im(A) then that means that A is surjective and therefore invertible because its an operator, correct? But I don't understand what you wrote after that do you think you could elaborate a bit more?

4. ## Re: Linear Algebra Proof Help.

Originally Posted by soaps666
So if rank(A) = dim Im(A) then that means that A is surjective and therefore invertible because its an operator, correct?
The equality $\operatorname{rank}A = \dim\operatorname{Im} A$ is the definition of rank. It is not a premise that guarantees something, such as that $A$ is surjective. It's just a suggestion to use $\operatorname{rank}A$ to refer to the dimension of the image of $A$.

The idea of this approach is to show that if a composition of operators is non-degenerate (i.e., invertible, or injective), then so is every operator in the composition. One measure of degeneracy is rank: invertible operators have rank equal to the dimension of the vector space; non-invertible operators have smaller rank (being non-injective, they map some linearly independent vectors into the same vector, and so the dimension of their image is smaller). When we take a composition, the rank cannot increase: it is less than or equal to the smallest rank or the participating operators. Therefore, if $\operatorname{rank}(STU)=\dim V$, which is the maximal possible value, then $\operatorname{rank}S= \operatorname{rank}T= \operatorname{rank}U= \dim V$.

Using this approach relies on several facts.

1. $A$ is invertible iff $\operatorname{rank}A=\dim V$.
2. $\operatorname{rank}(AB)\le\operatorname{rank}(A)$ .
3. $\operatorname{rank}(AB)\le\operatorname{rank}(B)$ .

Ideally, they are proved in your textbook prior to this problem, but their proofs are not long. Briefly, $A$ is invertible $\iff$ $A$ is injective $\iff$ $\ker A=\{\vec{0}\}$ $\iff$ $\dim\ker A=0$ $\iff$ $\dim\operatorname{Im} A=\dim V$ (by the rank–nullity theorem) $\iff$ $\operatorname{rank} A=\dim V$. As I said in the previous post, fact 2 is obvious, and fact 3 follows from the fact that if $C:V_1\to V_2$ is a linear mapping where $V_1$ does not have to coincide with $V_2$, then $\dim\operatorname{Im}C\le\dim V_1$. Indeed, if $(e_1,\dots,e_n)$ is a basis of $V_1$, then $\operatorname{Im} C$ is the span of $Ce_1,\dots,Ce_n$, so it cannot have the dimension greater than $n$. When applied to fact 3, $C=A$ is a mapping from $V_1=\operatorname{Im}B$ to $V_2=V$ and $\dim V_1=\operatorname{rank}B$.

If you think that your course expects a different proof, feel free to post theorems that you have covered and that you suspect could be used here.

5. ## Re: Linear Algebra Proof Help.

One way to do this problem is to use the fact that det(AB)= det(A)det(B) together with the fact that a linear operator is invertible if and only if its determinant is non-zero,. In particular that means that if the determinant of a product of linear operators is invertible, then the determinant of every one of them is non-zero so each is invertible.

6. ## Re: Linear Algebra Proof Help.

A third way, not far from your original thought.

Since STU = I, it follows that U is invertible with inverse ST (in FINITE-DIMENSIONAL vector spaces, a matrix in L(V) is invertible if it is right-invertible or left-invertible)

By the same reasoning, it follows that S is invertible with inverse TU.

So T = S-1U-1 (multiply on the left by S-1, on the right by U-1).

The RHS is a product of invertible matrices, and thus invertible, so T is invertible.

Thus T-1 = (S-1U-1)-1 = (U-1)-1(S-1)-1 = US

Some things to always remember with nxn matrices (and thus elements of L(V) where dim(V) = n)

Invertible = injective = surjective = bijective = non-zero determinant

Injective = trivial kernel = 0 nullity

Surjective = full rank

Singular = non-trivial kernel = non-zero nullity = less than full rank = 0 determinant

Most of the above follows from the rank-nullity theorem, the facts about determinants follows from the fact that the map det:L(V) --> F is multiplicative, and sends units (invertible things) to units.

emakarov's answer uses surjectivity, HallsofIvy's answer uses determinant properties. You could have also argued that since:

ker(STU) = ker(I) = {0}, that ker(U) = {0} (since if not, then for some non-zero v we have Uv = 0, so that STU(v) ST(U(v)) = ST(0) = S(T(0)) = S(0) = 0, but STU(v) = I(v) = v, contradiction).

Thus for non-zero v, Uv is non-zero. Similarly, we must have ker(ST) = {0}, so that ker(T) = {0}, and thus T is invertible.

This then shows that S = U-1T-1 = (TU)-1, and S is invertible.

from S = U-1T-1, multiplying on the left by U gives US = T-1.