Consider a linear transformation given by the matrix \begin{align*} A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 2 & 5 & 2 & 1 \\ -1 & 1 & 0 & 3 \end{bmatrix} \end{align*} Find a basis for $R^4$ with respect to which the matrix representation for the linear transformation above is diagonal.
Why the basis is the linear independent eigenvectors of matrix $A$? How does it come?
On
A $4\times 4$ matrix has at most $4$ eigenvalues. A whole subspace of eigenvectors must correspond to each eigenvalue (of course, there are a lot of things that I did not cover as is the multiplicity of each, the dimension of each eigenspace and all of that).
Ideally, you could choose four eigenvectors from all of them that form a basis for the whole space. And how does this help us?
If $\beta = \{ v_1,v_2,v_3,v_4 \}$ is the desired basis, each vector $v_i$ with eigenvalue $\lambda_i$, then note the following $$Av_1=\lambda_1 v_1=\lambda_1 v_1+0v_2+0v_3+0v_4$$ $$Av_2=\lambda_2 v_2=0v_1+\lambda_2 v_2+0v_3+0v_4$$ $$Av_3=\lambda_3 v_3=0v_1+0v_2+\lambda_3 v_3+0v_4$$ $$Av_4=\lambda_4 v_4=0v_1+0v_2+0v_3+\lambda_4 v_4$$ How is the matrix representation of $A$ with respect to that basis $\beta$? You see?
It comes from the fact that, for an eigenvector $v$, we have $Av=\lambda v$ for the corresponding eigenvalue value $\lambda$.
Let $B$ be the matrix of eigenvectors. Then $B^{-1}AB=D$ is equivalent to $AB=BD$. Now use $Av=\lambda v$ to see that $D$ will consist precisely of the eigenvalues on the diagonal. Both sides have $i$th column equal to $\lambda_iv_i$, where $\lambda_i$ is the $i$th eigenvalue and $v_i$ the $i$th eigenvector.