Question

Let T : P1 → P2 be defined by
T(a+bx) = b+ax+(a−b)x^2.
Check that T is a linear transformation. Find the matrix of the transformation with
respect to the ordered bases B1 = {1,x} and B2 = {x^2,x^2+x,x^2+x+1}. Find the
kernel of T . Further check that the range of T is
{a+bx+cx^2 ∈ P2 | a+c = b}.

Answer 1

Answer on Question #61909 – Math – Linear Algebra

Question

Let $T: P1 \to P2$ be defined by

$$T(a + bx) = b + ax + (a - b)x^2.$$

Check that $T$ is a linear transformation. Find the matrix of the transformation with respect to the ordered bases $B1 = \{1, x\}$ and $B2 = \{x^2, x^2 + x, x^2 + x + 1\}$. Find the kernel of $T$. Further check that the range of $T$ is $\{a + bx + cx^2 \in P2 \mid a + c = b\}$.

Solution

Let $T(a + bx) = b + ax + (a - b)x^2$, then $T(ac + bdx) = bd + acx + (ac - bd)x^2$,

$$cT(a) + d(T(bx)) = c(ax + ax^2) + d(b - bx^2) = bd + acx + (ac - bd)x^2 = T(ac + bdx)$$

So, $T$ is a linear transformation.

If $B_1 = \{1, x\} = \left\{\binom{1}{0}, \binom{0}{1}\right\}$ and $B_2 = \{x^2, x^2 + x, x^2 + x + 1\} = \left\{\binom{1}{0}, \binom{0}{1}, \binom{0}{1}\right\}$, then

$$T\left(\binom{1}{0}\right) = T(1 + 0 \cdot x) = x + x^2 = 0 \cdot x^2 + 1 \cdot (x^2 + x) + 0 \cdot (x^2 + x + 1) = 1 \cdot \binom{0}{1} = \binom{0}{1};$$ $$T\left(\binom{0}{1}\right) = T(0 + 1 \cdot x) = 1 - x^2 = -x^2 - (x^2 + x) + (x^2 + x + 1) = -\binom{1}{0} - \binom{0}{1} + \binom{0}{0} = \binom{-1}{1}.$$

The matrix of the transformation $T$ is $A = \begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix}$.

The following system determines the kernel of $T$:

$$\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \to -b = 0; \, b = 0; \, a - b = 0 \to \begin{pmatrix} a = 0 \\ b = 0 \end{pmatrix}.$$

Thus, the kernel is $Ker = \binom{0}{0}$, which is the same as $0 + 0 \cdot x = 0$.

The range is $T(\alpha + \beta x) = \beta + \alpha x + (\alpha - \beta)x^2 = a + bx + cx^2$,

where $a = \beta; b = \alpha; c = \alpha - \beta$.

Thus, $a + c = \beta + \alpha - \beta = \alpha = b$ and the range is $\{a + bx + cx^2 \in P_2 \mid a + c = b\}$.

Answer: $\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 0 & 1 \end{pmatrix}$, $0, \{a + bx + cx^2 \in P_2 \mid a + c = b\}$.

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