Let be $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ a linear application between two vectors, such that $f(\overrightarrow{u})= A_{f}\overrightarrow{u}$. Where $A_{f}$ is the next square matrix:

$$A_{f} = \begin{pmatrix} 5 & 4 & 0 \\ 3 & 1 & 5\\ 0 & 2 & 5 \end{pmatrix} $$

I need to calculate a vector space base containing the application kernel and a vector base such that the matrix associated with application $f$ is diagonal.

I've done the following:

1) We know, by definition, that the nucleus is the set of vectors whose image is the null vector. In mathematical language,

$ker f := \left \{ \vec{u} \in \mathbb{R}^{3} : f(\vec{u}) = \vec{0}\right \}$

Then, we need the vectors such that when multiplied by the matrix A they give the null vector

$A_{f}\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix}\ $

We solve the system of linear equations, and it gives us as a solution:

$ker f = \left \langle \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} \right \rangle $

2) I know how to diagonalize the matrix $A_{f}$, but I don't know how to get the base from which the matrix is diagonal. The exercise says that the matrix it gives us is calculated with respect to the standard base.

In the link, I command you to diagonalization. I've done it in $\LaTeX{}$ code and won't let me insert it here. Once I get the diagonalization, I don't know how to find the base.


By definition eigenvector satisfies the equation $$ A(v) = \lambda v $$

So you just have to find the kernel of the matricx $A-\lambda I$, where $\lambda$ is one eigennvalue and $I$ the identity matrix


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