Exercises: Linear Transformations II - Invertible Maps

Problem 1

The linear maps \(L\) and \(T\) below are described in standard coordinates.

\[ \begin{array}{rl} L\from \RNrSpc{4} \longrightarrow \RNrSpc{3} &\quad L(x,y,z,t)= (-t+x,y+x,z-x) \\ T\from \RNrSpc{3} \longrightarrow \RNrSpc{4} &\quad T(x,y,z) = (x+z,y,-x,-z+y) \end{array} \]

Answer the subsequent questions

Show that \((L\Comp T)\) is invertible.
- Solution
Find \((L\Comp T)^{-1}(x,y,z)\) for an arbitrary vector \((x,y,z)\) of \(\RNrSpc{3}\).
- Solution Answer
Determine if \((T\Comp L)\) is invertible. Explain your answer.
- Solution Answer

Problem 2

Let \(\EuScript{B}=(\Vect{b}_1,\Vect{b}_2,\Vect{b}_3)\) and \(\EuScript{D}=(\Vect{d}_1,\Vect{d}_2,\Vect{d}_3)\) be the ordered bases of \(\RNrSpc{3}\) with

\[ \begin{array}{rclrclrcl} \Vect{b}_1 & = & (1,0,0), &\quad \Vect{b}_2 & = & (-1,-1,1), &\quad \Vect{b}_3 & = & (0,0,1) \\ \Vect{d}_1 & = & (1,0,1), &\quad \Vect{d}_2 & = & (-1,-1,0), &\quad \Vect{d}_3 & = & (0,-1,-1) \end{array} \]

Further, let \(L\from \RNrSpc{3} \to \RNrSpc{3}\) be the linear map which is represented by the matrix

\[ \Mtrx{A}_{\EuScript{B}\EuScript{D}} = \left[ \begin{array}{rrr} 0 & -2 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{array} \right] \]

Find the matrix \(\Mtrx{A}_{\EuScript{D}\EuScript{B}}\) representing \(L^{-1}\) with respect to the bases \(\EuScript{B}\) for the domain and \(\EuScript{D}\) for the target.
- Answer
Find the \(\EuScript{D}\)-coordinates of \(L^{-1}(a,b,c)_{\EuScript{B}}\), where \((a,b,c)_{\EuScript{B}}\) is arbitrary in \(\RNrSpc{3}\).
- Solution
Find the matrix \(\Mtrx{A}\) representing \(L^{-1}\) with respect to the standard basis of \(\RNrSpc{3}\).
- Solution Answer
If \(\Vect{x}_{\EuScript{S}}=(x,y,z)\), express \(L^{-1}(\Vect{x})\) in standard coordinates.
- Answer

Problem 3

Suppose \(L\from \RNrSpc{n} \to \RNrSpc{n}\) is a linear map such that, whenever \(\Vect{u}_1,\dots ,\Vect{u}_r\) are linearly independent vectors in \(\RNrSpc{n}\), the vectors \(L(\Vect{u}_1),\dots ,L(\Vect{u}_r)\) are also linearly independent. Show that \(L\) is invertible.

Hint Solution

Problem 4

For each of the statements below determine if it is true or false.

Given an isomorphism \(L\from V\to W\), there are always vectors \(\Vect{x}\neq \Vect{0}\) in \(V\) with \(L(\Vect{x})=\Vect{0}\).
- Hint Hint Solution Answer
Given an isomorphism \(L\from V\to W\), the vector equation \(L(\Vect{x}) = \Vect{0}\) has exactly one solution, namely \(\Vect{x} = \Vect{0}\).
- Hint Solution Answer
Given an isomorphism \(L\from V\to W\) and a vector \(\Vect{y}\) in \(W\), there is a unique \(\Vect{x}\) in \(V\) with \(L(\Vect{x}) = \Vect{y}\).
- Hint Solution Answer
Given an isomorphism \(L\from V\to W\) and a vector \(\Vect{y}\) in \(W\), there are always infinitely many \(\Vect{x}\) in \(V\) with \(L(\Vect{x}) = \Vect{y}\).
- Hint Answer

Solution for Problem 1 a

We answer this problem via the following steps:

Find the matrix \(\Mtrx{A}\) representing \(L\).
Find the matrix \(\Mtrx{B}\) representing \(T\).
The matrix representing \((L\Comp T)\) is then the matrix product \(\Mtrx{C}\DefEq \Mtrx{A}\cdot \Mtrx{B}\)
We then know that \((L\Comp T)\) is invertible exactly when \(\Mtrx{C}\) is invertible. So, we may test \(\Mtrx{C}\) for invertibility by checking if its determinant is nonzero.

Step 1 The matrix \(\Mtrx{A}\) representing \(L\) has the following vectors as its columns:

\(L(\StdBssVec{1})\)	\(=\)	\((1,1,-1)\)
\(L(\StdBssVec{2})\)	\(=\)	\((0,1,0)\)
\(L(\StdBssVec{3})\)	\(=\)	\((0,0,1)\)
\(L(\StdBssVec{4})\)	\(=\)	\((-1,0,0)\)

Therefore,

\[ \Mtrx{A}\ =\ \left[ \begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \end{array} \right] \]

Step 2 Similarly, the matrix \(\Mtrx{B}\) representing \(T\) has the following vectors as its columns:

\(T(\StdBssVec{1})\)	\(=\)	\((1,0,-1,0)\)
\(T(\StdBssVec{2})\)	\(=\)	\((0,1,0,1)\)
\(T(\StdBssVec{3})\)	\(=\)	\((1,0,0,-1)\)

Therefore,

\[ \Mtrx{B}\ =\ \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 1 & -1 \end{array} \right] \]

Step 3 The matrix representing \((L\Comp T)\) is

\[ \Mtrx{C} \DefEq \left[ \begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 1 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 \end{array} \right] \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 1 & -1 \end{array} \right] = \left[ \begin{array}{rrr} 1 & -1 & 1 \\ 1 & 1 & 0 \\ -2 & 0 & 0 \end{array} \right] \]

Step 4 testing \(\Mtrx{C}\) for invertibility. We find that \(\det(\Mtrx{C})=2\). Therefore, \((\Mtrx{A} \Mtrx{B})\) is invertible, and so \((L\Comp T)\) is invertible.

Solution for Problem 4 c

Put \(M\DefEq L^{-1}\). Then we know that \(L\Comp M = \Id[W]\). This means, in particular, that

\(L(M(\Vect{y}))\)

\(=\)

\(\Vect{y}\)

So set \(\Vect{x} \DefEq L(\Vect{y})\), and we have that \(L(\Vect{x}) = \Vect{y}\). Therefore the equation

\[L(\Vect{x}) = \Vect{y}\]

has at least one solution. So we wonder if it could have more solutions. Suppose \(\Vect{s}\) in \(V\) also satisfies \(L(\Vect{s}) = \Vect{y}\). Then we know

\[L(\Vect{x} - \Vect{s}) = L(\Vect{x}) - L(\Vect{s}) = \Vect{y} - \Vect{y} = \Vect{0}\]

Now we learned in the previous parts of the exercise that \(\Vect{0}\) is the only vector which gets sent to \(\Vect{0}\) by the isomorphism \(L\); in other words \((\Vect{x} - \Vect{s}) = \Vect{0}\) and, therefore \(\Vect{x} = \Vect{0}\). – The statement in question is true.

\(\Vect{x}\)	\(=\)	\(M\Comp L(\Vect{x})\)
\(\)	\(= \)	\(M(\Vect{0})\)
\(\)	\(= \)	\(\Vect{0}\)

\(\Vect{x}\)	\(=\)	\(M\Comp L(\Vect{x})\)
\(\)	\(= \)	\(M(\Vect{0})\)
\(\)	\(= \)	\(\Vect{0}\)