Worked examples · Sheet 2

Associated matrices with respect to the standard bases

Theory used: Associated matrix Coordinates of the image Matrix with respect to the standard bases The coordinate isomorphism

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Write the matrix associated with the function $f : R^{3} \to R^{2}$

f (x, y, z) = (x - y + 2 z, 3 x + z)

with respect to the standard bases and compute $f (1, 2, - 1)$ .

A = (13 - 1 0 21)

f (1, 2, - 1) = (13 - 1 0 21) 12 - 1 = (1 - 2 - 2 3 + 0 - 1) = (- 3 2)

The domain has dimension $3$ and the codomain dimension $2$ , so the matrix has $3$ columns and $2$ rows. With respect to the standard bases the $i$ -th row contains the coefficients of $x$ , $y$ , $z$ in the $i$ -th component of $f$ ; $y$ does not appear in the second component, therefore its coefficient is $0$ . The columns are $f (e_{1}) = (1, 3)$ , $f (e_{2}) = (- 1, 0)$ and $f (e_{3}) = (2, 1)$ . By the theorem on the coordinates of the image, $f (1, 2, - 1)$ is obtained by multiplying the matrix by the column $(1, 2, - 1)$ , and the result is $f (1, 2, - 1) = (- 3, 2)$ , in agreement with direct substitution into the formula.

Exercise 2

Let $f : R^{2} \to R^{3}$ be linear with

f (e_{1}) = (1, 0, 2), f (e_{2}) = (- 1, 3, 1) .

Write $f (x, y)$ and compute $f (2, - 1)$ .

A = 102 - 1 31

f (x, y) = x f (e_{1}) + y f (e_{2}) = (x - y, 3 y, 2 x + y)

f (2, - 1) = (2 + 1, - 3, 4 - 1) = (3, - 3, 3)

The columns of the matrix with respect to the standard bases are the images of the vectors of the standard basis of the domain, so $A$ is written by placing $f (e_{1})$ and $f (e_{2})$ side by side as columns. Since $(x, y) = x e_{1} + y e_{2}$ , by linearity $f (x, y) = x f (e_{1}) + y f (e_{2})$ , and adding componentwise we obtain the explicit formula, which coincides with the product $A (x y)$ . Substituting $x = 2$ and $y = - 1$ we find $f (2, - 1) = (3, - 3, 3)$ .

Exercise 3

Let $T : R_{2} [x] \to R_{2} [x]$ be defined by $T (p) = (x + 1) p^{'}$ . Write the matrix of $T$ with respect to the basis $(1, x, x^{2})$ and use it to compute $T (3 - x + 2 x^{2})$ .

T (1) T (x) T (x^{2}) = 0 = x + 1 = 1 + x = (x + 1) \cdot 2 x = 2 x + 2 x^{2}

M (T) = 000110022

000110022 3 - 1 2 = - 1 34 ⟹ T (3 - x + 2 x^{2}) = - 1 + 3 x + 4 x^{2}

The matrix is built by computing the image of each basis polynomial and writing its coordinates as a column. The derivative of $1$ is zero, so the first column is zero; $T (x) = 1 + x$ has coordinates $(1, 1, 0)$ and $T (x^{2}) = 2 x + 2 x^{2}$ has coordinates $(0, 2, 2)$ . The polynomial $3 - x + 2 x^{2}$ has coordinates $(3, - 1, 2)$ , and the product with the matrix gives the coordinates $(- 1, 3, 4)$ of the image. The direct computation confirms the result, since $p^{'} = - 1 + 4 x$ and $(x + 1) (- 1 + 4 x) = - 1 + 3 x + 4 x^{2}$ .

Exercise 4

Write the matrix of transposition $T : M_{2} (R) \to M_{2} (R)$ , $T (X) = X^{T}$ , with respect to the standard basis $(E_{11}, E_{12}, E_{21}, E_{22})$ , and use it to compute the transpose of

X = (1324) .

T (E_{11}) = E_{11}, T (E_{12}) = E_{21}, T (E_{21}) = E_{12}, T (E_{22}) = E_{22}

M (T) = 1000001001000001

M (T) 1234 = 1324 ⟹ X^{T} = (1234)

The space $M_{2} (R)$ has dimension $4$ , so the matrix of $T$ is $4 \times 4$ . The matrix $E_{12}$ has the entry $1$ in position $(1, 2)$ , and its transpose has the entry $1$ in position $(2, 1)$ , that is $T (E_{12}) = E_{21}$ ; in the same way $T (E_{21}) = E_{12}$ , while $E_{11}$ and $E_{22}$ remain fixed. The coordinates of $E_{21}$ are $(0, 0, 1, 0)$ and form the second column, those of $E_{12}$ are $(0, 1, 0, 0)$ and form the third. The matrix $X$ has coordinates $(1, 2, 3, 4)$ by the coordinate isomorphism, and the matrix of $T$ swaps the second and third coordinates. We obtain the coordinates $(1, 3, 2, 4)$ , which correspond to the transpose $X^{T}$ .