Lecture. Linear transformations [l2]

  • Apr 7, 2026

Definition 1. linear transformation [br94]

    A mapping $A$ of a vector space $X$ into a vector space $Y$ is said to be a linear transformation if $$ A (\mathbf{x}_{1} + \mathbf{x}_{2}) = A \mathbf{x}_{1} + A \mathbf{x}_{2}, \quad A(c \mathbf{x}) = cA \mathbf{x} $$ for all $\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x} \in X$ and all scalars $c$.

    Notation 1.1. some notations [br94A]

      We often writes $A \mathbf{x}$ insteand of $A(\mathbf{x})$ if $A$ is linear.

      Linear transformations of $X$ into $X$ are often called linear operators on $X$. If $A$ is a linear operator on $X$ which is one-to-one and onto, we say that $A$ is invertible.

      Definition 2. $L(X, Y)$, products and norms [br96]

        1. Let $L(X, Y)$ be the set of all linear transformations of the vector space $X$ into the vector space $Y$. Instead of $L(X, X)$, we shall simply write $L(X)$.

        2. If $X, Y, Z$ are vector spaces, and if $A \in L(X, Y)$ and $B \in L(Y,Z)$, we define their product $BA$ to be the composition of $A$ and $B$: $$(BA)\mathbf{x} = B(A\mathbf{x}) \qquad (\mathbf{x} \in X).$$ Then $BA \in L(X, Z)$.

        3. For $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, define the norm $\lVert A \rVert$ of $A$ to be the sup of all numbers $\lvert A\mathbf{x} \rvert$, where $\mathbf{x}$ ranges over all vectors in $\mathbb{R}^{n}$ with $\lvert \mathbf{x} \rvert \leq 1$.

          Observe that the inequality $$\lvert A\mathbf{x} \rvert \leq \lVert A \rVert \lvert \mathbf{x} \rvert $$ holds for all $\mathbf{x} \in \mathbb{R}^{n}$. Also, if $\lambda$ is such that $\lvert A\mathbf{x} \rvert \leq \lambda \lvert \mathbf{x} \rvert$ for all $\mathbf{x} \in \mathbb{R}^{n}$, then $\lVert A \rVert \leq\lambda$.

        Theorem 3. some results about norms [br97]

          1. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, then $\lVert A \rVert < \infty$ and $A$ is a uniformly continuous mapping of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$.
          2. If $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $c$ is a scalar, then$$\lVert A + B \rVert \leq \lVert A \rVert + \lVert B \rVert , \qquad \lVert cA \rVert = \lvert c \rvert \lVert A \rVert.$$With the distance between $A$ and $B$ defined as $\lVert A - B \rVert$, $L(\mathbb{R}^{n}, \mathbb{R}^{m})$ is a metric space.
          3. If $A \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$ and $B \in L(\mathbb{R}^{m}, \mathbb{R}^{n})$, then$$\lVert BA \rVert \leq \lVert B \rVert \lVert A \rVert.$$

            Proof. sketch [br97A]

              1. Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ be the standard basis in $\mathbb{R}^{n}$ and suppose $\mathbf{x} = \sum c_{i}\mathbf{e}_{i}$. Now $\lvert \mathbf{x} \rvert \leq 1 \implies \lvert c_{i} \rvert \leq 1$ for all $i$. Then $$\lvert A \mathbf{x} \rvert = \left\lvert \sum c_{i} A \mathbf{e}_{i} \right\rvert \leq \sum \lvert c_{i} \rvert \lvert A \mathbf{e}_{i} \rvert \leq \sum \lvert A \mathbf{e}_{i} \rvert$$ so that $$\lVert A \rVert \leq \sum_{i=1}^{n} \lvert A \mathbf{e}_{i} \rvert < \infty. $$

                Each $A \mathbf{e}_{i}$ is a specific vector in $\mathbb{R}^{m}$, and hence $\lvert A \mathbf{e}_{i} \rvert < \infty$.

                Since $\lvert A \mathbf{x} - A \mathbf{y} \rvert \leq \lVert A \rVert \lvert \mathbf{x} - \mathbf{y} \rvert$ if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, we see that $A$ is uniformly continuous.

              2. Omitted.

              3. Omitted.

            Theorem 4. the set of invertible linear operators [br98]

              Let $\Omega$ be the set of all invertible linear operators on $\mathbb{R}^{n}$.

              1. If $A \in \Omega, B \in L(\mathbb{R}^{n})$, and$$\lVert B - A \rVert \cdot \lVert A^{-1} \rVert < 1,$$then $B \in \Omega$.
              2. $\Omega$ is an open subset of $L(\mathbb{R}^{n})$, and the mapping $A \to A^{-1}$ is continuous on $\Omega$.

              Proof. sketch [br98A]

                1. Suppose $B$ is not invertible $\iff$ $B$ is not injective $\iff$ there exists $\mathbf{x} \in \mathbb{R}^{n}$ such that $\mathbf{x} \neq \mathbf{0}$ and $B\mathbf{x} = \mathbf{0}$. Since $A \in \Omega$, there exists $\mathbf{y} \in \mathbb{R}^{n}$ such that $\mathbf{y} \neq 0$ and $A^{-1}\mathbf{y} = \mathbf{x}$. Then,$$\begin{align*}\lvert (B - A)A^{-1} \mathbf{y} \rvert &= \lvert (B - A)\mathbf{x} \rvert = \lvert -A\mathbf{x} \rvert = \lvert \mathbf{y} \rvert \\ &\leq \lVert (B - A)A^{-1} \rVert \lvert \mathbf{y} \rvert < \lvert \mathbf{y} \rvert. \end{align*}$$This is impossible. So our assumption cannot be true. Therefore, $B \in \Omega$.
                2. Openness is trivial. Continuity can be immediately induced from the observation that$$B^{-1} - A^{-1} = B^{-1}(A - B)A^{-1}.$$

                Definition 5. matrices [br99]

                  Suppose $\{ \mathbf{x}_{1}, \dots, \mathbf{x}_{n} \}$ and $\{ \mathbf{y}_{1}, \dots, \mathbf{y}_{m} \}$ are bases of vector spaces $X$ and $Y$, respectively. Then every $A \in L(X, Y)$ determines a set of numbers $a_{ij}$ such that $$ A \mathbf{x}_{j} = \sum_{i=1}^{m} a_{ij} \mathbf{y}_{i} \qquad (1 \leq j \leq n). $$ We represent $A$ by an $m$ by $n$ matrix: $$ [A] = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{mn} \end{bmatrix} $$

                  $[A]$ depends not only on $A$ but also on the choice of bases in $X$ and $Y$.

                  If $\mathbf{x} = \sum_{j} \mathbf{x}_{j}$, then the Schwarz inequality shows that $$ \lvert A \mathbf{x} \rvert^{2} = \sum_{i} \left( \sum_{j} a_{ij} c_{j} \right)^{2} \leq \sum_{i} \left( \sum_{j} a_{ij}^{2} \cdot \sum_{j} c_{j}^{2} \right) = \sum_{i, j} a_{ij}^{2} \lvert \mathbf{x} \rvert ^{2}. $$ Thus $$ \lVert A \rVert \leq \sqrt{ \sum_{i, j} a_{ij}^{2} }. $$ Moreover, if we replace $A$ by $B - A$, where $A, B \in L(\mathbb{R}^{n}, \mathbb{R}^{m})$, and view each $a_{ij}$ as continuous functions of a single parameter, then we have the following:

                  If $S$ is a metric space, if $a_{11}, \dots, a_{mn}$ are real continuous functions on $S$, and if, for each $p \in S$, $A_{p}$ is the linear transformation of $\mathbb{R}^{n}$ into $\mathbb{R}^{m}$ whose matrix has entries $a_{ij}(p)$, then the mapping $p \mapsto A_{p}$ is a continuous mapping of $S$ into $L(\mathbb{R}^{n}, \mathbb{R}^{m})$.