#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.
---
**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.
$(\Rightarrow)$ Suppose $T\in \mathcal{L}(V)$ is invertible and $S\in \mathcal{L}(U,V)$. Since $T$ is invertible $\text{null }T=\{ 0 \}$ and $\text{null S} \subseteq \text{null }T$, it must be that $\text{null }S=\{ 0 \}$. Thus $S$ is injective. $\quad \blacksquare{}$
$(\Leftarrow)$ Suppose $S\in \mathcal{L}(U,V)$ is injective. Let $u_{1}, \ldots{}, u_{n}$ be a basis of $U$. Extend it with $v_{1}, \ldots{}, v_{m}$ to be a basis of $V$. Then for any $v\in V$, $v=a_{1}u_{1}+\ldots+a_{n}u_{n}+b_{1}v_{1}+\ldots+b_{m}v_{m}$ with $a_{i},b_{j}\in \mathbf{F}$. Since $S$ is injective, $Su_{1}, \ldots{}, Su_{n}$ is linearly independent and thus a basis for $\text{range }S$. Extend this with $w_{1}, \ldots{}, w_{m}$ to be a basis of $V$. Suppose $T:V\to V$ where $Tu_{i}=Su_{i}$ for $i=1,\ldots,n$ and $Tv_{j}=w_{j}$ for for $j=1,\ldots,m$. Now $T$ is well-defined on $V$. And by inspection, $T$ is a linear map and surjective. Since $T$ is an operator and surjective it is invertible.
[Tikz diagram depicting one rectangle partitioned into two representing U and V-U. Then a second rectangle to the right of this rectangle that is also partitioned into two representing rangeS and V-rangeS. Where V-rangeS and V-U are the same size].
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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 $T_1 = ST_2$.
$(\Leftarrow)$ Suppose $S \in\mathcal{L}(W)$ such that $T_{1}=ST_{2}$. Then, suppose $v\in V$ such that $T_{1}v=0$. Then $ST_{2}v=0$ and since $S$ is invertible, and therefore injective, $T_{2}v=0$. The converse is true for each as well. Thus $v\in \text{null }T_{1} \iff v\in \text{null }T_{2}$.
$(\Rightarrow)$ Suppose $\text{null }T_{1}=\text{null }T_{2}$. Define $R : \text{range }T_{2}\to W$ where
$
R(T_{2}v)=T_{1}v
$
First, we must show that $R$ is a well-defined function. Suppose $T_{2}v_{1}=T_{2}v_{2}$, then $T_{2}v_{1}-T_{2}v_{2}=T_{2}(v_{1}-v_{2})=0\implies v_{1}-v_{2}\in \text{null }T_{2}=\text{null }T_{1}$. So, $T_{1}(v_{1}-v_{2})=0$ and $T_{1}v_{1}=T_{1}v_{2}$. Thus $R$ is well-defined. By inspection, $R$ is also linear. Now suppose $Rw=0$ for some $w\in \text{range }T_{2}$. Then there must be a $v\in V$ such that $T_{2}v=w$. Then
$
0=Rw=RT_{2}v=T_{1}v
$
Since $\text{null }T_{1}=\text{null }T_{2}$, $w=T_{2}v=0$. Thus $R$ is injective. Since $W$ is finite-dimensional, problem 3 implies that there must exist an invertible operator $S\in \mathcal{L}(W)$ such that $Sw=Rw$ for all $w\in \text{range }T_{2}$. Therefore, for every $v\in V$
$
ST_{2}v=RT_{2}v=T_{1}v
$
Thus $T_{1}=ST_{2}$. $\quad \blacksquare{}$
_(In regards to $R's$ well-definedness, the concern if we have $T_{2}v_{1}=T_{2}v_{2}$ but $T_{1}v_{1}\neq{}T_{1}v_{2}$ which would make $R$ not well-defined.)_
**Follow up questions**: why didn't we need some basis remapping? Or is that what we implicitly did with $R$ but we leveraged the function definition with a generic $v$? We used null space to prove that $T_{1}v_{1}=T_{1}v_{2}$ when $T_{2}v_{1}=T_{2}v_{2}$ in our definition of $R$ via $v_{1}-v_{2}\in \text{null }T_{2}$, how is this working?
Before invoking problem 3, basis remapping wasn't required because of the following property. If $\text{null }T_{1}=\text{null }T_{2}$ then $T_{1}v_{1}=T_{1}v_{2}\iff T_{2}v_{1}=T_{2}v_{2}$. So, rather than looking at the null space as the set of vectors that map to $0$, we can interpret it as _what pair of vectors map to the same output._ Thus, if two maps share the same null space, they share the same set of vectors which map to the same output—equivalent co-domain isn't required.
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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 $T_1 = T_2 S$.
$(\Rightarrow)$ Suppose $\text{range }T_{1}=\text{range }T_{2}$. Let $\text{dim }V=n$. By the [[3B3 - Fundamental Theorem of Linear Maps|fundamental theorem of linear maps]], $\text{dim null }T_{1}=\text{dim null }T_{2}=k$. Let $n_{1}, \ldots{}, n_{k}$ be a basis for $\text{null }T_{1}$ and let $m_{1}, \ldots{}, m_{k}$ be a basis for $\text{null }T_{2}$. Extend $n_{1}, \ldots{}, n_{k}$ with $p_{1}, \ldots{}, p_{n-k}$ to be a basis of $V$. Then $T_{1}p_{1},\ldots,T_{1}p_{n-k}$ is a basis for $\text{range }T_{1}$. Since $\text{range }T_{1}=\text{range }T_{2}$, there must exist some $q_{1},\ldots, q_{n-k}\in V$ such that
$
T_{2}q_{j}=T_{1}p_{j}
$
for $j=1,\ldots,n-k$. Since $T_{2}q_{1},\ldots, T_{2}q_{n-k}$ are linearly independent, then $q_{1},\ldots,q_{n-k}$ must be linearly independent (suppose $0=b_{1}q_{1}+\ldots+b_{n-k}q_{n-k}$, then apply $T_{2}$ to both sides... and $0=b_{1}T_{2}q_{1}+\ldots+b_{n-k}T_{2}q_{n-k}$. Since $T_{2}q_{1},\ldots,T_{2}q_{n-k}$ is independent $b_{1}=\ldots=b_{n-k}=0$ and thus $q_{1},\ldots,q_{n-k}$ is linearly independent).
Let $U=\text{span }(q_{1}, \ldots{}, q_{n-k})$. We'll show that $U\oplus \text{null }T_{2}=V$. Suppose $z\in V$. Then $T_{2}z\in \text{range }T_{2}$ so $T_{2}z=a_{1}T_{2}q_{1}+\ldots+a_{n-k}T_{2}q_{n-k}$. Then, $T_{2}(z-a_{1}q_{1}-\ldots-a_{n-k}q_{n-k})=0$ which means $z-a_{1}q_{1}-\ldots-a_{n-k}q_{n-k}\in \text{null }T_{2}$. Thus
$
\begin{align}
z & = (a_{1}q_{1}+\ldots+a_{n-k}q_{n-k}) + (z-a_{1}q_{1}-\ldots-a_{n-k}q_{n-k})
\end{align}
$
where the first term is in $U$ and the second in $\text{null }T_{2}$. Now we'll show that $U\cap \text{null }T_{2}=\{ 0 \}$. Suppose $z\in U\cap \text{null }T_{2}$. Then $z=a_{1}q_{1}+\ldots a_{n-k}q_{n-k}$ and $T_{2}z=0$. So $T_{2}(a_{1}q_{1}+\ldots a_{n-k}q_{n-k})=0$ and $a_{1}T_{2}q_{1}+\ldots+a_{n-k}T_{2}q_{n-k}=0$ implies $a_{1}=\ldots=a_{n-k}=0$ since $T_{2}q_{1},\ldots,T_{2}q_{n-k}$ is linearly independent. Thus $z=0$ and $U\cap \text{null }T_{2}=\{ 0 \}$. Therefore, $U\oplus \text{null }T_{2}=V$.
Define $S\in \mathcal{L}(V)$ such that $Sn_{i}=m_{i}$ for $i=1,\ldots,k$ and $Sp_{j}=q_{j}$ for $j=1,\ldots,n-k$. Let $v\in V$ then $v=a_{1}n_{1}+\ldots+a_{k}n_{k}+b_{1}p_{1}+\ldots+b_{n-k}p_{n-k}$ and $Sv=S(a_{1}n_{1}+\ldots+a_{k}n_{k}+b_{1}p_{1}+\ldots+b_{n-k}p_{n-k})=a_{1}m_{1}+\ldots+a_{k}m_{k}+b_{1}q_{1}+\ldots+b_{n-k}q_{n-k}.$ So $S$ is defined for all of $V$ and by inspection, $S$ is linear. Additionally, $S$ maps basis to basis so $S$ is invertible.
Now suppose a $z\in V$. Then
$
\begin{align}
T_{2}Sz & = T_{2}S(a_{1}n_{1}+\ldots+a_{k}n_{k}+b_{1}p_{1}+\ldots+b_{n-k}p_{n-k}) \\
& = T_{2}(a_{1}m_{1}+\ldots+a_{k}m_{k}+b_{1}q_{1}+\ldots+b_{n-k}q_{n-k}) \\
& = a_{1}T_{2}m_{1}+\ldots+a_{k}T_{2}m_{k} + b_{1}T_{2}q_{1}+\ldots+b_{n-k}T_{2}q_{n-k}\\
& = b_{1}T_{1}p_{1}+\ldots+b_{n-k}T_{1}p_{n-k}\\
\end{align}
$
And
$
\begin{align}
T_{1}z & =T_{1}(a_{1}n_{1}+\ldots+a_{k}n_{k}+b_{1}p_{1}+\ldots+b_{n-k}p_{n-k}) \\
& =b_{1}T_{1}p_{1}+\ldots+b_{n-k}T_{1}p_{n-k}
\end{align}
$
Therefore, $T_{2}S=T_{1}$.
$(\Leftarrow)$ Let $S\in \mathcal{L}(V)$ be an invertible operator such that $T_{1}=T_{2}S$. Since $S$ is invertible, it surjective, so $S(V)=V$, ($S$ defined on all of $v$ and maps onto all of $V$), and then
$
\text{range }T_{1}=T_{1}(V)=(T_{2}S)(V)=T_{2}(S(V))=T_{2}(V)=\text{range }T_{2}.
$
$\quad \blacksquare{}$
_In the forward pass, I originally tried to define $Sv$ to be some vector $q∈V$ such that $T_{2}q=T_{1}v$, so that $T_{2}(Sv)=T_{2}q=T_{1}v$. While this $q$ exists because $\text{range }T1=\text{range }T2$, it is not necessarily unique. There may be distinct $q1,q2∈V$ such that
$
T_{2}q_{1}=T_{1}v=T_{2}q_{2}
$
Thus, without making a consistent choice, the proposed rule could assign two different outputs to the same input, ($v\to q_{1}$ and $v\to q_{2}$), and would not be well-defined._
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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 $\dim \operatorname{null} T_1 = \dim \operatorname{null} T_2$.
$(\Rightarrow)$ Suppose $R\in \mathcal{L}(V)$ and $S\in\mathcal{L}(W)$ such that $T_{1}=ST_{2}R$. Then $\text{range }T_{1}=\text{range }ST_{2}R$. And by the [[3B3 - Fundamental Theorem of Linear Maps|fundamental theorem of linear maps]], $\text{dim null }T_{1}=\text{dim null }ST_{2}R$. Since $R$ is invertible on $V$, it is surjective on $V$. So $R(V)=V$. Thus the dimension of $\text{null }T_{2}$ equal to $\text{dim null }T_{2}R$. Since $S$ is invertible, it is injective and has $\text{null }S=\{ 0 \}$. Thus, $ST_{2}Rv=0\iff T_{2}Rv=0\iff T_{2}(u)=0$. Thus $\text{dim null }ST_{2}R=\text{dim null }T_{2}$. And $\text{dim null }T_{1}=\text{dim null }T_{2}$.
$(\Leftarrow)$ Suppose $\text{dim null }T_{1}=\text{dim null }T_{2}.$ Let $\text{dim }V=n$. Let $x_{1}, \ldots{}, x_{k}$ be a basis for $\text{null }T_{1}$ and extend it with $v_{1},\ldots,v_{n-k}$ to a basis of $V$. Let $y_{1}, \ldots{}, y_{k}$ be a basis for $\text{null }T_{2}$ and extend it with $u_{1}, \ldots{}, u_{n-k}$ to a basis of $V.$ Define $R\in \mathcal{L}(V)$ where $Rx_{i}=y_{i}$ for $i=1,\ldots,k$ and $Rv_{j}=u_{j}$ for $j=1,\ldots,n-k$. Since $x_{1}, \ldots{}, x_{k},v_{1}, \ldots{}, v_{n-k}$ form a basis of $V$, $R$ is defined on all of $V$. By inspection $R$ is a well-defined function and linear. Suppose $v\in \text{null }T_{1}$ then $v=a_{1}x_{1}+\ldots+a_{k}x_{k}$ and
$
\begin{align}
T_{2}Rv & =T_{2}R(a_{1}x_{1}+\ldots+a_{k}x_{k})\\
& =T_{2}(a_{1}y_{1}+\ldots a_{k}y_{k}).\\
\end{align}
$
So $v\in \text{null }T_{2}$. Suppose $u\in \text{null }T_{2}R$ then $T_{2}Ru=0\implies Ru=a_{1}y_{1}+\ldots +a_{k}y_{k}\implies$ $u=a_{1}x_{1}+\ldots a_{k}x_{k}$. So $u\in \text{null }T_{1}$. Thus by problem $3$, since $W$ is finite dimensional and $\text{null }T_{1}=\text{null }T_{2}R$ there must exist an invertible operator $S\in \mathcal{L}(W)$ such that $T_{1}=ST_{2}R$.
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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 $\dim E$?
**(a)** Suppose $S,R\in E$ then $S\in \mathcal{L}(V,W)$ where $Sv=0$ and $R\in \mathcal{L}(V,W)$ where $Rv=0$. Then, $(S+R)v=Sv+Rv=0+0=0.$ Suppose $\lambda \in \mathbf{F}$, then $(\lambda S)v=\lambda (Sv)=\lambda0=0$. And $0v=0$ where the left hand side $0$ is the zero map. So $E$ is a subspace.
**(b)** Let $\text{dim }V=n$ and $\text{dim }W=m$. Let $v,v_{2},\ldots,v_{n}$ be a basis of $V$. Then for a $T\in E$ and any $u\in V$, $Tu=Ta_{1}v+Ta_{2}v_{2}+\ldots+Ta_{n}v_{n}=Ta_{2}v_{2}+\ldots+Ta_{n}v_{n}$. Then the first column matrix of $T$ is $\mathcal{M}(T)\cdot,1=0$. And columns $j=2,\ldots,n$ are arbitrary vectors in $\mathbf{F}^{m,1}$. Thus $\text{dim }E=m(n-1)$.
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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 $u \in U$.)
Since $V$ is finite dimensional and $T:V\to W$ is surjective, $\text{dim }V\geq{}\text{dim }W$. Let $x_{1}, \ldots{}, x_{k}$ be a basis of $\text{null }T.$ Extend it with a $u_{1},\ldots,u_{n-k}$ to be a basis of $V$. Then for any $v\in V,$ $Tv=(a_{1}x_{1}+\ldots+a_{k}x_{k}+b_{1}u_{1}+\ldots+b_{n-k}u_{n-k})=b_{1}Tu_{1}+\ldots+b_{n-k}Tu_{n-k}$. So $\text{range T}=\text{span}(Tu_{1},\ldots,Tu_{n-k})$ and $Tu_{1},\ldots,Tu_{n-k}$ is linearly independent in $W$. Let $U=\text{span}(u_{1},\ldots,u_{n-k})$. Then $T|_{U}$ is surjective and injective and thus invertible. So $T|_{U}$ is an isomorphism from $U$ to $W$.
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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.
$(\Rightarrow)$ Suppose $ST$ is invertible. Then $(ST)^{-1}(ST)=I$ and $((ST)^{-1}S)T=I$. So $T$ has a left inverse. By 3.68, since $T$ is an operator then $T$ has a right inverse as well. Thus $T$ is invertible. Similarly, $ST(ST)^{-1}=I$ and $S(T(ST)^{-1})=I$, so $S$ has a right inverse and similarly a left inverse. Thus $S$ is also invertible.
$(\Leftarrow)$ Suppose $S$ and $T$ are invertible. Then consider $(T^{-1}S^{-1})(ST)=(T^{-1}(S^{-1}S)T)=T^{-1}IT=T^{-1}T=I$. Thus $(T^{-1}S^{-1})$ is the unique left inverse of $ST$. By 3.68, since $(ST)$ is an operator, then it has a right inverse as well. Thus $ST$ is invertible. $\quad \blacksquare{}$
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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 $TS = I$.
$(\Rightarrow)$ Suppose $ST=I$, the $ST$ is surjective. And since $ST$ is an operator on a finite-dimensional $V$, it is also injective and hence invertible. By problem 9, since $S,T\in \mathcal{L}(V)$ and $ST$ is invertible, then $S$ and $T$ are invertible. Then $ST=I\implies S^{-1}ST=S^{-1}I\implies T=S^{-1}$. Thus $I=S^{-1}S=TS$.
$(\Leftarrow)$ By symmetry the converse is true.
_This is the proof of 3.68._
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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 $T^{-1} = US$.
By problem 9, since $V$ is finite dimensional, $S,T,U\in \mathcal{L}(V)$, and by association of functions $S,T,U,ST,TU$ are invertible. Since $U$ is invertible and $(ST)U=I$ then $U