#math/linear-algebra
---
**Problem 1:** Suppose $T \in \mathcal{L}(U,V)$ and $S \in \mathcal{L}(V,W)$ are both invertible linear maps. Prove that $ST \in \mathcal{L}(U,W)$ is invertible and that $(ST)^{-1} = T^{-1}S^{-1}$.
Observe the following,
$
\begin{align}
(T^{-1}S^{-1})(ST) & = T^{-1}(S^{-1}S)T & \text{Associativity of functions} \\
& =T^{-1}IT \\
& =T^{-1}T \\
& =I
\end{align}
$
And
$
\begin{align}
(ST)(T^{-1}S^{-1}) & =S(TT^{-1})S^{-1} & \text{Associativity of functions}
& =SIS^{-1} \\
& = SS^{-1} \\
& =I
\end{align}
$
Thus by uniqueness of the inverse of a linear map, it must be that $T^{-1}S^{-1}$ is the inverse of $(ST)$ and $(ST)^{-1}=T^{-1}S^{-1}$.
---
**Problem 2:** Suppose $V$ is finite-dimensional and $\dim V > 1$. Prove that the set of noninvertible operators on $V$ is not a subspace of $\mathcal{L}(V)$.
Let $v_{1},\ldots,v_{k}$ be a basis for $V$. Construct a $T\in \mathcal{L}(V)$ such that $Tv_{i}=v_{i}$ for $i=2,\ldots,k$ and $Tv_{1}=0$. Construct an $S\in \mathcal{L}(V)$ such that $S_{v_{1}}=v_{1}$ and $S_{v_{i}}=0$ for $i=2,\ldots,k$. Since $\text{null }T\neq{}0$ and $\text{range T}\neq{}V$, $T$ is a noninvertible operator. Similarly, $S$ is a noninvertible operator. Now suppose a $v\in V$ such that $v=a_{1}v_{1}+\ldots+a_{k}v_{k}$. Then
$
\begin{align}
(S+T)v & = Sv+Tv \\
& =(a_{1}v_{1} + 0) + (0+a_{2}v_{2}+\ldots+a_{k}v_{k}) \\
& = a_{1}v_{1}+\ldots+a_{k}v_{k}
\end{align}
$
Then $S+T$ is the identity map on $V$ and $S+T$ is surjective and thus invertible.
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**Problem 3:** Suppose $V$ is finite-dimensional, $U$ is a subspace of $V$, and $S \in \mathcal{L}(U,V)$. Prove there exists an invertible operator $T \in \mathcal{L}(V)$ such that $Tu = Su$ for every $u \in U$ if and only if $S$ is injective.
Given: $U\subset V$. Some $S\in \mathcal{L}(U,V)$ and want to find a $T\in \mathcal{L}(V)$ where $Tu=Su$ for all $u\in U$ (not $\mathcal{L}(U,V)$) and show its only possible when $S$ is injective. $S$ is not an operator. Being injective does not imply $S$ is surjective as $\text{dim }U$ could be less than $V$.
The idea is that if we're given an injective function $S$ on only a subset of $V$ we should be able to construct an invertible op $T$ on $V$ built on that $S.$
Goal 1: Suppose $S$ is injective... Then $\text{null S}=\{ 0 \}$ and $\text{range S}\subset V$. Let $u$ be a basis of $U$. and extend with $v$ be a basis of $V.$ Now we have $u,v$. Now let $r$ be a basis of $\text{range }S$, and extend it with $w$ to be a basis of $V$. Now, construct $T$ with $Tu=Su$ for all $u\in U$. And $Tv=Tw$ (they should be same length, we should be able to carry over the coefficients). $T$ is an LM. T is surjective and thus invertible.
Goal 2: Okay suppose $T$ exists... its invertible. So $T^{-1}$ exists. And we know $Tu=Su$ for all $u\in U$. Well, $T$ is injective, and since $Tu=Su$ for all $u\in U$ it must be that $S$ is also injective.
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**Problem 4:** Suppose $W$ is finite-dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$. Prove that $\operatorname{null} T_1 = \operatorname{null} T_2$ if and only if there exists an invertible operator $S \in \mathcal{L}(W)$ such that $T_1 = ST_2$.
Given: $T_{1},T_{2}\in \mathcal{L}(V,W)$. Only $W$ is assumed finite dimensional. There are no givens on surjective or injective on $T_{1,2}$. The null space could be infinitely large. If the nulls are the same, then the remaining finite set of vectors which map to $w\in W$ and $w\neq{}0$ are the same. Why is it not that the dimension of the not-null space just needs to be the same? Is it because the form of the inputs and coefficients of the respective bases need to be the same which forces the bases to be equal i.e. a requirement for the statement? I think the non-null space can also be infinite dimensional and thus you cannot assume a basis of non-null T. Then for an finite dimensional space getting translated to a finite space, if the finite space can translate the mappings to T1, then it must be that the null spaces are equal. Kind of like a finite space linear map can only handle this infinite space input under these requirements.
Goal: If $T_{1},T_{2}$ share the same null space, then there must be an **operator** $S$ on $W$ such that $T_{1}=ST_{2}$.
Goal 1: Suppose $\text{null }T_{1}=\text{null }T_{2}$. Suppose $w\in \text{range }T_{2}$, then by definition of $T_{2}$ then there must be a $v\in V$ where $T_{2}v=w$. Define $Sw=T_{1}v$ note this is defined for all $v\in V$.
The concern: if $T_{2}v_{1}\to w$ and $T_{2}v_{2}\to w$ can we guarantee $Sw=Tv_{1}=Tv_{2}$? Yes.
> Suppose $v_{1},v_{2}$ with $T_{2}v_{1}=T_{2}v_{2}=w\in W$. Then $T_{2}(v_{1}-v_{2})=0$ and $v_{1}-v_{2}\in \text{null }T_{2}$. Then, $v_{1}-v_{2}\in \text{null }T_{1}$ and therefore $T_{1}v_{1}=T_{1}v_{2}$. $S$ is well defined on $\text{range }T_{2}$ . Note we need this to be an operator on $W$.
??? what about the range complements. map complement t2 basis to complement t1 basis. Now the function is well defined on W. This has no bearing on the T1=ST2. Its just a move that we are required to do to make S an **operator**? Ah and specifically... an invertible one (using the basis.)
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**Problem 5:** Suppose $V$ is finite-dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$. Prove that $\operatorname{range} T_1 = \operatorname{range} T_2$ if and only if there exists an invertible operator $S \in \mathcal{L}(V)$ such that $T_1 = T_2 S$.
Given: $T_{1},T_{2}$, a finite $V$, and ? $\text{dim }W$.
- Since $V$ is finite, $\text{range, null }T_{1},T_{2}$ is finite.
- We can have n1 for null of T1, n2 for null of T2
- We can have c1 for null complement of T1, and c2 as null complement of T2
- we can have bases Tc1 and Tc2 for the ranges of T1, T2
- Forward pass: We can create a function where map c2->v, T1c1 = T2v. T2v will be independent in W because T1c1 is. Need to prove its well defined. Seems fair.
- Backward pass: S exists... such that T1=T2S; Suppose arb vinV. Then T1v=T2Sv. Similarly, T2v=T2Sq=T1q. So range is same.
Goal: Find an invertible op $S:V\to V$ such that $T_{1}=T_{2}S$.
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**Problem 6:** Suppose $V$ and $W$ are finite-dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$. Prove that there exist invertible operators $R \in \mathcal{L}(V)$ and $S \in \mathcal{L}(W)$ such that $T_1 = ST_2 R$ if and only if $\dim \operatorname{null} T_1 = \dim \operatorname{null} T_2$.
- Forward pass: Use R to map basis of n2, nc2 to n1, nc1. Then T2nc1 will be independent in W. Then use S to map T2nc1 -> T1nc1. Use fact W is finite to map the rest of the space accordingly.
- Backward pass: T1 = ST2R is exists. Suppose n1,nc1... Then (n1!=0) T1n1=0=ST2Rn1. Then ST2n1-> ST2(q) q!=0 since R is invertible and thus injective. And T1 = ST2q=Sv and . Maybe easier to go with range since non-zero.
- Can we say something like R(nc1,....,ncm) -> q1...qm LI in V? yes becuase R injecitve? And we know (Tq1...Tqm) != 0 because S injective and S(Tq1) = w != 0. sooo dim range T2 >= m = dim range T1.
- Suppose n1... and n1neq0 then T1n1=0=ST2Rn1. Then q=(Rn1) neq 0. And T1n1=0=ST2(q). S is injective so T2q must be 0. So qneq0 in null of T2. Do this for n1...nn and we get dim null T2 >= n.
- Thus dim null T2 = n.
_gemini simpler proof w/ function comps: https://gemini.google.com/app/13a4b7ea894fc399_
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**Problem 7:** Suppose $V$ and $W$ are finite-dimensional. Let $v \in V$. Let
$
E = \{T \in \mathcal{L}(V,W) : Tv = 0\}.
$
**(a)** Show that $E$ is a subspace of $\mathcal{L}(V,W)$. **(b)** Suppose $v \neq 0$. What is $\dim E$?
Given: Finite dimensional $V,W$ and a _particular_ vector $v\in V$. Then $E$ is defined as the set of all linear maps from $V\to W$ where $v$ maps to $0.$
**(a)** Goal: Show if $q,p \in E$ then q+p in E. 0 in E. $\lambda q\in E$ if $q\in E$. Seems straight forward.
**(b)** Goal: $E$ is as subspace. We can either try to span $E$ with an LI list. Okay... elements of $E$ are functions. Let $\text{dim }V=n$ and $\text{dim }W=m$.
- Create a basis $v,v_{2},\ldots,v_{n}$ of $V$. Then observe $Tv, Tv_{2},\ldots,Tv_{n}$. We know $Tv=0$. So if we are to represent ALL functions of this nature what would the matrix of this look like? The remaining $n-1$ columns are arbitrary vectors in $F^{m}$. Thus dimension is $m(n-1)$
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**Problem 8:** Suppose $V$ is finite-dimensional and $T : V \to W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $T|_U$ is an isomorphism of $U$ onto $W$. (Here $T|_U$ means the function $T$ restricted to $U$. In other words, $T|_U$ is the function whose domain is $U$, with $T|_U$ defined by $T|_U(u) = Tu$ for every $u \in U$.)
Given: Finite dim $V$ and a surjective linear map $T:V\to W$. Note $T$ is not necessarily invertible.
Goal: Find a $U\subset V$ such that $T|_{U}$ is invertible. We want to 'trim down' $V$ until we just have something thats injective and while being surjective on $W$.
- Start with a basis of null T. Extend with basis $u_{1}\ldots$ where span of $u$ is $U$. Is U to be same size as $W$? I think yes because $Tu_{1},\ldots,Tu_{n}$ should be linearly independent in $W$. Also just use rank-nullity dim theorem.
We are just looking at the representation of a linear map. It doesn't care about linear independence and what not.
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**Problem 9:** Suppose $V$ is finite-dimensional and $S, T \in \mathcal{L}(V)$. Prove that $ST$ is invertible if and only if both $S$ and $T$ are invertible.
Given: Finite dim $V$ and $S,T\in \mathcal{L}(V)$ with $ST$ being invertible.
Goal: Prove $S$ and $T$ themselves are invertible.
-> Suppose $u,v\in V$ such that $Tu=Tv$. Then $S(Tu)=S(Tv)$ and by definition of $ST$, $u=v$. Thus $T$ is injective and therefore invertible. Similarly, since $ST$ is surjective, for every $w\in V$ there must be a $u\in V$ such that $STu=w$. By function composition then $S(Tu)=w$ with $Tu\in V$. Thus for every element $w\in V$ there is an input that maps to $w$ under $S$
<- Suppose $S,T$ are invertible. Applying $S^{-1}$ and $T^{-1}$ to $ST$ we have $T^{-1}S^{-1}ST=I$. Thus $ST$ is invertible.
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**Problem 10:** Suppose $V$ is finite-dimensional and $S, T \in \mathcal{L}(V)$. Prove that $ST = I$ if and only if $TS = I$.
Given: Operators $S$ and $T$ on $V$ where $V$ is finite dimensional. And $ST=I$
Goal: Show $TS=I$.
-> Suppose $ST=I$ then $ST$ is surjective and injective making $ST$ invertible. Thus $S,T$ are each invertible. ST=I -> T=S^-1 and thus TS=S^-1S=I.
By symmetry the converse is true.
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**Problem 11:** Suppose $V$ is finite-dimensional and $S, T, U \in \mathcal{L}(V)$ and $STU = I$. Show that $T$ is invertible and that $T^{-1} = US$.
Since $I$ is invertible, $STU$ is invertible. Then by definition of function composition and problem 9, $S,TU,ST,U$ must be invertible. And since $ST$ is invertible, $T$ must be invertible by problem 9. **Note this requires V to be finite dimensional at this point.**
Since $U$ is invertible and $(ST)U=I$ then $U