11.27 Representing linear maps by matrices

1. Let \(T: \mathbb{C}^2 \to \mathbb{C}^2\) be the linear map given by \(T(x_1, x_2) = (x_1, 0)\).

   Let \(\mathcal{B} = \{(1, 0), (0, 1)\}\) (canonical basis) and \(\mathcal{B}' = \{(1, i), (-i, 2)\}\) be two bases for \(\mathbb{C}^2\).

   1. What is the matrix of \(T\) with respect to \(\mathcal{B}\) and \(\mathcal{B}'\)?

      We have that \(T(1, 0) = (1, 0)\) and \(T(0, 1) = (0, 0)\). Let's find the coordinates of these vectors with respect to \(\mathcal{B}'\).

      \(T(1, 0) = (1, 0) = a(1, i) + b(-i, 2)\).

      This implies: \(\begin{cases}a-ib=1,\\ ia+2b=0.\end{cases}\)

      So \(a = 2\) and \(b = -i\). Therefore: \(\begin{aligned}T(1, 0) = 2(1, i) - i(-i, 2).\end{aligned}\)

      Next, consider \(T(0, 1) = (0, 0)\). In \(\mathcal{B}'\): \(\begin{aligned}T(0, 1) = 0(1, i) + 0(-i, 2).\end{aligned}\)

      Thus, the matrix of \(T\) with respect to \(\mathcal{B}\) and \(\mathcal{B}'\) is: \(\begin{aligned}[T]_{\mathcal{B}, \mathcal{B}'} = \begin{bmatrix}2 & 0 \\ -i & 0\end{bmatrix}.\end{aligned}\) 2. What is the matrix of \(T\) with respect to \(\mathcal{B}'\) and \(\mathcal{B}\)?

      We have: \(\begin{cases}T(1,i)=(1,0)=1(1,0)+0(0,1),\\ T(-i,2)=(-i,0)=-i(1,0)+0(0,1).\end{cases}\)

      Thus, the matrix of \(T\) with respect to \(\mathcal{B}'\) and \(\mathcal{B}\) is: \(\begin{aligned}[T]_{\mathcal{B}', \mathcal{B}} = \begin{bmatrix}1 & -i \\ 0 & 0\end{bmatrix}.\end{aligned}\) 3. What is the matrix of \(T\) with respect to \(\mathcal{B}'\)?

      First, compute \(T(1, i)\): \(\begin{aligned}T(1,i)=(1,0)=a(1,i)+b(-i,2)\implies\begin{bmatrix}a\\ b\end{bmatrix}=\begin{bmatrix}2\\ -i\end{bmatrix}.\end{aligned}\)

      Next, compute \(T(-i, 2)\): \(\begin{aligned}T(-i,2)=(-i,0)=a(1,i)+b(-i,2)\implies\begin{bmatrix}a\\ b\end{bmatrix}=\begin{bmatrix}-2i\\ -1\end{bmatrix}.\end{aligned}\)

      Thus, the matrix of \(T\) with respect to \(\mathcal{B}'\) is: \(\begin{aligned}[T]_{\mathcal{B}'} = \begin{bmatrix} 2 & -2i \\ -i & -1 \end{bmatrix}.\end{aligned}\)

2. Let \(T: P_3(\mathbb{R}) \to M_2(\mathbb{R})\) be the linear map given by: \(\begin{aligned}T(ax^3 + bx^2 + cx + d) = \begin{pmatrix} a + d & b - c \\ b + c & a - d \end{pmatrix}.\end{aligned}\)

   Find a basis of \(\text{Null}(T)\) and a basis for \(\text{Range}(T)\) (using the matrix of \(T\) with respect to the canonical bases).

   Let’s find the matrix of \(T\) with respect to: \(\mathcal{B} = \{x^3, x^2, x, 1\}\) (canonical basis of \(P_3(\mathbb{R})\)) and \(\mathcal{B}' = \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}\) (canonical basis of \(M_2(\mathbb{R})\)).

   We have that:
   ​\(\begin{aligned}T(x^3) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 1 \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 0 \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + 0 \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 1 \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.\end{aligned}\)

   \(\begin{aligned}T(x^2) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = 0 \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 1 \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + 1 \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 0 \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.\end{aligned}\)

   \(\begin{aligned}T(x)=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}=0\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}+-1\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+1\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+0\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.\end{aligned}\)

   \(\begin{aligned}T(1)=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}=1\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}+0\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+0\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}+1\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.\end{aligned}\)

   Therefore, the matrix of \(T\) with respect to \(\mathcal{B}\) and \(\mathcal{B}'\) is:
   ​\(\begin{aligned}[T]_{\mathcal{BB}^{\prime}}=\begin{pmatrix}1 & 0 & 0 & 1\\ 0 & 1 & -1 & 0\\ 0 & 1 & 1 & 0\\ 1 & 0 & 0 & 1\end{pmatrix}.\end{aligned}\)

   \(\text{Null}(T)\):

   \(\begin{aligned}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}.\end{aligned}\)

   \(\begin{aligned}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}.\end{aligned}\)

   \(\implies \text{Null}(T) = \{0\}\) (polynomial).

   \(\text{Range}(T)\):

   We need to think of \(\begin{aligned}([T]_{\mathcal{B}\mathcal{B}'})^T\end{aligned}\) in the same way as \(\text{Null}(T)\).

   \(\begin{aligned}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 6 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}.\end{aligned}\)

   Therefore, \(\text{Range}(T) = \text{span}\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} = M_2(\mathbb{R})\).

Definition: A linear map \(T \in \mathcal{L}(V, W)\) is called INVERTIBLE if \(\exists S \in \mathcal{L}(W, V)\) such that \(S \circ T = \text{Id}_V\) and \(T \circ S = \text{Id}_W\) (\(S\) is called the inverse of \(T\)).

Theorem: \(T \in \mathcal{L}(V, W)\) is invertible \(\iff T\) is injective and surjective.

Theorem: Suppose \(V\) is finite-dimensional and we have \(T \in \mathcal{L}(V, V)\), the following are equivalent:

1. \(T\) is invertible.

2. \(T\) is injective.

3. \(T\) is surjective.

We can use dimension theorem to prove it. For example, \(T\) is surjective means \(\dim\)Range\(=W=V\), then \(\dim\)Null\(={0}\) which is injective

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1. Suppose \(V\) is finite-dimensional and \(S, T \in \mathcal{L}(V, V)\). Prove that \(S \circ T\) is invertible \(\iff S\) and \(T\) are invertible.

   \(\Leftarrow)\) Let’s show that \((S \circ T)^{-1} = T^{-1} \circ S^{-1}\) (these exist since \(S, T\) are invertible).

   • For \(v \in V\), \((T^{-1} \circ S^{-1})(S \circ T)(v) = T^{-1} \circ S^{-1}(S(T(v))) = T^{-1}(T(v)) = v\)
   • In the same way, \((S \circ T) \circ (T^{-1} \circ S^{-1}) = \text{Id}_V\).

   So \(S \circ T\) is invertible with inverse \(T^{-1} \circ S^{-1}\).

   \(\Rightarrow)\) Suppose \(S \circ T\) is invertible. By the last theorem, it is enough to show that \(T\) is injective and \(S\) is surjective.

   Injective: Let \(v_1, v_2 \in V\) such that \(T(v_1) = T(v_2)\). \(\implies S \circ T(v_1) = S \circ T(v_2)\), and since \(S \circ T\) is invertible, in particular \(S \circ T\) is injective \(\implies v_1 = v_2\). So \(T\) is injective.

   \(S\) surjective: Since \(S \circ T\) is invertible, in particular \(S \circ T\) is surjective. Therefore if \(v \in V\), \(\exists w \in V\) such that \((S \circ T)(w) = v\). Since \(T\) is invertible, in particular \(T^{-1}\) is surjective \(\implies\) since \(w \in V\), \(\exists u \in V\) such that \(T^{-1}(u) = w\).

   I want to write \(S(u) = v\).

   So note that \((S\circ T)(T^{-1}(u))=S\circ T(w)=v\).

   So therefore, given \(v \in V\), \(\exists u \in V\) such that \(S(u) = v \implies S\) is surjective \(\implies S\) is invertible by the last theorem.

   So, \(S\) and \(T\) are invertible.

2. Let \(T: \mathbb{R}^3 \to \mathbb{R}^3\) be the linear map given by \(T(x, y, z) = (3x + z, -2x + y, -x - 2y + 4z)\).

   1. What is the matrix of \(T\) in the canonical basis of \(\mathbb{R}^3\)?

      The canonical basis for \(\mathbb{R}^3\) is given by \(\{(1, 0, 0), (0, 1, 0), (0, 0, 1)\}\).

      We have that:

      • \(T(1, 0, 0) = (3, -2, -1) = 3(1, 0, 0) + (-2)(0, 1, 0) + (-1)(0, 0, 1)\).
      • \(T(0, 1, 0) = (0, 1, 2) = 0(1, 0, 0) + 1(0, 1, 0) + 2(0, 0, 1)\).
      • \(T(0, 0, 1) = (1, 0, 4) = 1(1, 0, 0) + 0(0, 1, 0) + 4(0, 0, 1)\).

      Therefore, \([T]_{\mathcal{B}} = \begin{pmatrix} 3 & 0 & 1 \\ -2 & 1 & 0 \\ -1 & 2 & 4 \end{pmatrix}\). 2. Prove that \(T\) is invertible and describe \(T^{-1}(x, y, z)\).

      To show that \(T\) is invertible, it is enough to show that \([T]_{\mathcal{B}}\) is invertible.

      This will also give us \(T^{-1}(x, y, z)\).

      \(\begin{pmatrix}\begin{array}{ccc|ccc}1 & 0 & 0 & 4/9 & 2/9 & -1/9\\ 0 & 1 & 0 & 8/9 & 13/9 & -2/9\\ 0 & 0 & 1 & -1/3 & -2/3 & 4/3\end{array}\end{pmatrix}\)​

      So \(T\) is invertible, and now we can write:
      \(T^{-1}(x, y, z) = \left( \frac{4}{9}x + \frac{2}{9}y - \frac{1}{9}z, \frac{8}{9}x + \frac{13}{9}y - \frac{2}{9}z, -\frac{1}{3}x - \frac{2}{3}y + \frac{2}{3}z \right)\).