# Find eigenvalues without having the matrix.

Let $$\{v_1,v_2,v_3,v_4\}$$ be a vector basis of $$\mathbb{R}^4$$ and $$A$$ a constant matrix of $$\mathbb{R}^{4\times 4}$$ so that: $$A v_1=-2v_1,\hspace{0.5cm} Av_2=-v_1,\hspace{0.5cm} Av_3=3v_4,\hspace{0.5cm}Av_4=-3v_3$$ Can I find the eigenvalues of the matrix A? I know that $$\lambda_1=-2$$ is a trivial eigenvalue but I don't know how to calculate the others.

You actually know the representation of your matrix related to your basis. How do you represent linear application as matrix? Then you can calculate your characteristic polynomial and work from there if you want a standard way to solve this.

• Are you saying that I consider the application $f(v_1,v_2,v_3,v_4)=(-2v_1,-v_1,3v_4,-3v_3)$? So that I end up having the matrix \begin{pmatrix} -2 & 0 & 0& 0\\ -1 & 0 & 0&0\\ 0&0&0&3\\ 0&0&-3&0 \end{pmatrix} Commented Jan 2, 2021 at 11:54
• You are considering the transpose of your actual linear application if I'm not mistaken. It should do your job because you are looking for eigenvalues and $Spec(A)=Spec(A^{T})$ but i would suggest to write the "right matrix" and work from there for clarity. Commented Jan 2, 2021 at 12:46

Notice that this linear map/matrix is basically two $$R^2\to R^2$$ maps joined together: One map consists of a linear map from span$$\{v_1,v_2\}$$ to itself, the other a linear map from span$$\{v_3,v_4\}$$. Respectively, these have matrix representations $$\begin{bmatrix}{-2 \: -1 \\\quad 0 \quad 0}\end{bmatrix}$$ and  $$\begin{bmatrix}{\quad 0 \: +3 \\ -3 \quad\: 0}\end{bmatrix}$$  It is pretty easy to find eigenvalues of these one.