Injective and surjective mappings
Task number: 4529
Denote by \(P\) the space of real polynomials of degree at most \(2\).
Decide which of the following linear mappings are injective and which are surjective:
Variant
\(f:\mathbb R^{2\times2} \to \mathbb R^3\) given by \(f\left(\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\right)= (a + b + c, a + b, a)^\mathsf T\)Solution
Construct a matrix of the mapping with respect to the standard basis: \[[f]_{E,E}= \begin{pmatrix} 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ \end{pmatrix}\]
The matrix rank is 3. Because the rank of the matrix is less than the dimension of the source space (the domain), this mapping is not injective. Because the rank of the matrix is the same as the dimension of the target space, this mapping is surjective.
Variant
\(f:\mathbb R^{2\times2} \to \mathbb R^4\) given by \(f\left(\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\right) = (a + b + c + d, a + b + c, a + b, a)^\mathsf T\)Answer
The mapping is a bijection, i.e. injective and surjective.Variant
\(f:\mathbb R^{2\times2} \to P\) given by \(f\left(\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\right) = (a + b)x^2 + (c + d)x + c\)Answer
The mapping is surjective but not injective.Variant
\(f : P \to \mathbb R^4\) given by \(f(ax^2 + bx + c) = (a - b + c, b + c, a + 2c, a - c)^\mathsf T\)Answer
The mapping is injective but not surjective.Variant
\(f : P \to \mathbb R^3\) given by \(f(ax^2 + bx + c) = (a + b, 2b - c, a - b + c)^\mathsf T\)Answer
The mapping is neither injective nor surjective.Variant
\(f : P \to \mathbb R^3\) given by \(f(ax^2 + bx + c) = (a + b, 2b - c, a - b + 2c)^\mathsf T\)Answer
The mapping is a bijection, i.e. injective and surjective.
