Suppose we have the next transformation:

$$f : \mathbb{K}[X]_{2} \rightarrow \mathit{M}(2, \mathbb{K})$$

$$a_{0}+a_{1}x+a_{2}x^{2} \mapsto \begin{pmatrix} p(0) & p'(1)\\ p'(-1) & p''(0) \end{pmatrix} $$

  1. Find the matrix associated to the linear application with respect to the bases: $B = \left \{ 1+x+x^2, 2x+x^2, x^2 \right \}$ and $C = \left \{ \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix} \right \}$

  2. With respect to the canonical bases, get the kernel and image of the linear transformation.

  3. Deduce whether the application is injective, superjective or bijective.

  4. Obtain an orthonormal base from which the matrix associated with the linear application is diagonal.

Attempted resolution

1. Find the matrix associated to the linear application with respect to the bases: $B = \left \{ 1+x+x^2, 2x+x^2, x^2 \right \}$ and $C = \left \{ \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 3 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}, \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix} \right \}$

First, we find the images of these polynomials:

$f(1+x+x^2) = \begin{pmatrix} 1 & 1 \\ -1 & 2 \end{pmatrix}$

$f(2x+x^2) = \begin{pmatrix} 0 & 3 \\ 0 & 2 \end{pmatrix}$

$f(x^2) = \begin{pmatrix} 0 & -1 \\ -2 & 2 \end{pmatrix}$

We, then, express these matrices as a linear combination of the matrices of base $C$

$\begin{pmatrix}1&1\\ \:-1&2\:\end{pmatrix}=a\cdot \begin{pmatrix}1&2\\ \:0&4\end{pmatrix}+b\cdot \begin{pmatrix}0&0\\ \:0&3\end{pmatrix}+c\cdot \begin{pmatrix}1&0\\ \:5&1\end{pmatrix}+d\cdot \begin{pmatrix}2&0\\ \:0&1\end{pmatrix}$

$\begin{pmatrix}0&3\\ \:0&2\:\end{pmatrix}=a\cdot \begin{pmatrix}1&2\\ \:0&4\end{pmatrix}+b\cdot \begin{pmatrix}0&0\\ \:0&3\end{pmatrix}+c\cdot \begin{pmatrix}1&0\\ \:5&1\end{pmatrix}+d\cdot \begin{pmatrix}2&0\\ \:0&1\end{pmatrix}$

$\begin{pmatrix}0&1\\ \:-2&2\:\end{pmatrix}=a\cdot \begin{pmatrix}1&2\\ \:0&4\end{pmatrix}+b\cdot \begin{pmatrix}0&0\\ \:0&3\end{pmatrix}+c\cdot \begin{pmatrix}1&0\\ \:5&1\end{pmatrix}+d\cdot \begin{pmatrix}2&0\\ \:0&1\end{pmatrix}$

After solving it, we obtain that the matrix associated to the application would be:

$$M(f)_{B,C}=\begin{pmatrix} \frac{1}{2} & \frac{3}{2} & \frac{1}{2} \\ \frac{-1}{20} & \frac{-13}{12} & \frac{3}{20}\\ \frac{-1}{5} & 0 & \frac{-2}{5} \\ \frac{7}{20} & \frac{-13}{12} & \frac{-1}{20} \end{pmatrix}$$

2. With respect to the canonical bases, get the kernel and image of the linear transformation.

With respect to the standard bases, the matrix associated with the linear transformation would be as follows:

$$M(f) = \begin{pmatrix} 1 & 0 & 0\\ 1 & 3 & 1\\ -1 & 0 & -2 \\ 2 & 2 & 2 \end{pmatrix}$$

We know that the kernel are the vectors, whose image is the null vector. Then,

$$\begin{pmatrix} 1 & 0 & 0\\ 1 & 3 & 1\\ -1 & 0 & -2 \\ 2 & 2 & 2 \end{pmatrix}\begin{pmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0\\ \end{pmatrix}$$

If we solve it, we get that the kernel of $f$ is:

$$\ker f = \left \{ \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} \right \} \Rightarrow$$ $f$ is injective

By definition, we know that the $Im(f)$ is the set or generator system formed by the columns of the matrix associated with homomorphism. Therefore,

$$Im(f) = \left \langle \begin{pmatrix} 1 \\ 1 \\ - 1\\ 2 \\ \end{pmatrix}, \begin{pmatrix} 0 \\ 3 \\ 0 \\ 2 \\ \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -2 \\ 2 \end{pmatrix} \right \rangle$$

As the dimension of the image is inferior to the dimension of the vectorial space of arrival, we know that the linear transformation is not overjective.

3. Deduce whether the application is injective, superjective or bijective.

We have said this in the previous section. As the kernel is formed only by the null vector, the application is injective. Also, since the dimension of the image is smaller than the dimension of the arrival space, we know that the application is not over-jective. This means that, trivially, application f cannot be biyective, which implies that there is no inverse of this application.

4. Obtain an orthonormal base from which the matrix associated with the linear application is diagonal.

In this section, I do not know which matrix to diagonalize if the one I obtained in the first section or the matrix with respect to the usual bases. Well, once I diagonalize the matrix, the base would be the eigenvector. And, in order to ortonormalize, I apply the Gram-Schmidt algorithm, don't I?

Would the exercise be well done? And, secondly, if I had calculated the kernel and the image with respect to the matrix obtained in section (1), would I have obtained the same results? That is to say, that f is injective and that the dimension of the image is 3.

up vote 0 down vote accepted

So far your solution is correct.

To your questions: A change of basis in the domain can be interpreted as a multiplication of your matrix by a regular matrix from the right, a change of basis in the co-domain as a multiplication by a regular matrix from the left. These operations do not change the dimension of the kernel and the image.

Ad 4.): The question is not formulated precisely: In a general case you would need two orthonormal bases: one for the domain and one for the co-domain.

Since your matrix is not quadratic, the definition of eigenvalues does not apply in this case. What you are actually looking for is the singular value decomposition.

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