Homework 7

(30 pts) Decide whether the following linear maps are monomorphisms, epimorphisms or isomorphisms. Justify.
(a) \(T: \mathbb{C}^3 \rightarrow \mathbb{C}^3, T(x, y, z) = (x + iz, y + z, ix + y + z)\).
Since \(T \in \text{End}(\mathbb{C}^3)\). Then \(T\) is isomorphism iff \(T\) is injective.
Take \(T(x, y, z) = T(a, b, c) \Rightarrow (x + iz, y + z, ix + y + z) = (a + ic, b + c, ia + b + c)\).
\(\begin{cases} x+iz=a+ic \\ y+z=b+c \\ ix+y+z=ia+b+c \end{cases}\Rightarrow \begin{cases} x-a+i(z-c)=0 \\ y-b+z-c=0 \\ i(x-a)+(y-b)+(z-c)=0 \end{cases}\Rightarrow \begin{cases} x-a+i\left(z-c\right)=0 \\ y-b+z-c=0 \\ y-b=0 \end{cases}\Rightarrow \begin{cases} x=a \\ z=c \\ y=b \end{cases}\).
Thus \((x, y, z) = (a, b, c)\). Then \(T\) is injective. Then \(T\) is isomorphism.

(b) \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}, T(x, y, z) = (x - y + 2z, 3x + y + 4z, 5x - y + 8z)\).
Since \(T \in \text{End}(\mathbb{R}^3)\). Then \(T\) is isomorphism iff \(T\) is injective.
Take \(T(x, y, z) = T(a, b, c) \Rightarrow (x - y + 2z, 3x + y + 4z, 5x - y + 8z) = (a - b + 2c, 3a + b + 4c, 5a - b + 8c)\).
\(\begin{cases} x - y + 2z = a - b + 2c \\ 3x + y + 4z = 3a + b + 4c \\ 5x - y + 8z = 5a - b + 8c \end{cases} \Rightarrow \begin{cases} x - a - (y - b) + 2(z - c) = 0 \\ 3(x - a) + (y - b) + 4(z - c) = 0 \\ 5(x - a) - (y - b) + 8(z - c) = 0 \end{cases} \Rightarrow \begin{pmatrix} 1 & -1 & 2 \\ 3 & 1 & 4 \\ 5 & -1 & 8 \end{pmatrix}\)
\(\Rightarrow \begin{pmatrix} 1 & -1 & 2 \\ 0 & 2 & -1 \\ 0 & 0 & 0 \end{pmatrix}\Rightarrow \begin{cases} x-a+y-b+z-c=0 \\ 2(y-b)-(z-c)=0 \end{cases}\) Counter example.\(\begin{cases} x=a-3 \\ y=b+1 \\ z=c+2 \end{cases}\)
Thus \(T\) is not injective. Also not surjective.

(c) \(T: \mathbb{R}^4 \rightarrow \mathbb{R}^4, T(x, y, z, w) = (x - 2y - 5w, y + w, 3x + y + z + w, 2z)\).
Since \(T \in \text{End}(\mathbb{R}^4)\). Then \(T\) is isomorphism iff \(T\) is injective.
Take \(T(x, y, z, w) = T(a, b, c, d) \Rightarrow \begin{cases} x - a - 2(y - b) - 5(w - d) = 0 \\ y - b + w - d = 0 \\ 3(x - a) + (y - b) + z - c + w - d = 0 \\ 2(z - c) = 0 \end{cases}\)
\(\Rightarrow z=c,x=a,y=b,w=d\). Then \(T\) is injective \(\Rightarrow T\) is isomorphism.
(20 pts) Let \(T: P_3(\mathbb{R}) \rightarrow P_4(\mathbb{R})\), be the linear map defined by \(T(p(x)) = (x + 1)p(x)\).
(a) Is \(T\) a monomorphism?
(b) Is \(T\) an epimorphism?
(a) Yes, Take \(T(p(x)) = T(q(x)) \Rightarrow (x + 1)p(x) = (x + 1)q(x) \Rightarrow (x + 1)(p(x) - q(x)) = 0 \Rightarrow p(x) = q(x)\). Thus \(T\) is monomorphism.
(b) No. Counter example: Take \((x - 1)^4 \in P_4(\mathbb{R})\). \(\nexists p(x)\) s.t. \(T(p(x)) = (x + 1)p(x) = (x - 1)^4\) since \(x+1\nmid(x-1)^{4}\).
Thus \(T\) is not epimorphism.
(30 pts) Let \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^2\) be a linear map and \(U: \mathbb{R}^2 \rightarrow \mathbb{R}^3\) be another linear map. Prove that the linear map \(UT\) is not an isomorphism.
Since \(UT \in \text{End}(\mathbb{R}^3)\). Then \(UT\) is isomorphism iff \(UT\) is injective.
Since dimension theorem, then \(\dim\ker T+\dim\text{Im}T=3\) and \(\dim\text{Im}T\leq 2\), then \(\dim\ker T\geq 1\)

Then \(\exists 0\neq v\in \R^3:T(v)=0\), then \(U(T(v))=0\). Thus \(\exists 0\neq v\in \R^3:U(T(v))=0\)

Then \(0\neq v\in\ker(UT)\), then \(\ker(UT)\neq \{0\}\), thus \(UT\) is not injective. Thus \(UT\) is not isomorphism
(20 pts) Suppose \(V\) is finite-dimensional and \(S, T, U \in \mathcal{L}(V, V)\) are such that \(STU = I\). Show that \(T\) is an isomorphism and that \(T^{-1} = US\).

Since \(S(TU)=I\), then \(S\) has a right inverse, then \(S\) is invertible. Similarly, \(U\) is invertible

Since \(STU=I\) and \(S, T, U \in \mathcal{L}(V, V)\), then \(S^{-1}STU=S^{-1}\Rightarrow TU=S^{-1}\Rightarrow TUU^{-1}=S^{-1}U^{-1}\Rightarrow T=S^{-1}U^{-1}\)

Then \(S^{-1}U^{-1}\in\mathcal{L}(V,V)\) and \(US\in\mathcal{L}(V,V)\)

Then let's check \((US)\circ T=US\circ\left(S^{-1}U^{-1}\right)=US\circ\left(US\right)^{-1}=I\)

\(T\circ(US)=\left(S^{-1}U^{-1}\right)\circ\left(US\right)=\left(US\right)^{-1}\circ\left(US\right)=I\)

Thus \(T\) is invertible and \(T^{-1} = US\).