I have a question regarding finding eigenvectors/diagonalization.

So in my book Linear Algebra S.Friedberg, A.Insel, L.Spence (4'th edition) they deal with diagonalization, and specially the problem of finding a basis of eigenvectors for a linear operator T on an n-dimensional vector space V.

The procedure they use are like this:

  1. Let $\beta=\{\beta_1,\beta_2,...,\beta_n \}$ be an ordered basis for V and calculate $[T]_\beta=[[T(\beta_1)]_\beta,...,[T(\beta_n)]_\beta]$. The eigenvalues for T are presicely the eigenvalues for $[T]_\beta$

  2. Find the egenvalues for $[T]_\beta$ by solving the equation $det([T]_\beta-\lambda I )=0$ with respect to $\lambda$. Lets say that we get the destinct eigenvalues $\lambda_1, \lambda_2,...,\lambda_m $ as a solution.

  3. Find eigenvectors of $[T]_\beta$ corresponding to each eigenvalue found in step 2. Find the eigenvector corresponding to $\lambda$ by solving the linear system $[[T]_\beta-\lambda I \;|\;\overrightarrow0 ]$

  4. Lets say we found that $\overrightarrow v$ is an eigenvector for $[T]_\beta$ corresponding to $\lambda$. Then the vector $\phi_\beta^-1(\overrightarrow v))$ is an eigenvector for T. (where $\phi_\beta:V\rightarrow F^{dim(V)}$)

Here are my questions:

1.Is this the normal procedure for finding bases for V composed of eigenvectors of T? For example, in the special case of $V=\Bbb R^2$ and $T(x,y)=(-2x+3y,-10x+9y)$, if we let $\beta=\{(1,0),(0,1)\}$, then isn't the eigenvectors of T and the eigenvectors of $[T]_\beta$ exactly the same objects? Is this only because $\phi_\beta(x)=x$ and $V=F^{dim(V)}$ in this case? (When do we need to use step 4 at all?)

  1. Does the vectors obtained in step 4 above compose a basis for V of eigenvectors of T if and only if the sum of the dimensions of the eigenspaces of the eigenvalues found in step 2 is equal to n=dim(V)?

The procedure described above can be found at page 252 to 254 :)

  • $\begingroup$ What would you do without steps 1 and 4 if $V$ were, say, the space of polynomials of degree $\le n$? $\endgroup$ – amd May 10 '18 at 23:04
  • $\begingroup$ Well, I guess one in some cases could solve T(x)=kx directly without going the process though the matrix representation of T. I did however find out that step 4 is not necessary if V=R^n or V=C^n, because in those cases V is indistinguishable (not just isomorphic to) to F^n, where F is the field of V.(Thus making step 4 of no affect, because the inverse of the coordinate mapping is equal to the identity transformation when the standard ordered basis is chosen. ) :) $\endgroup$ – AfterMath May 11 '18 at 9:37
  • 1
    $\begingroup$ Only if you happened to choose the standard basis in step one. $\endgroup$ – amd May 11 '18 at 15:56

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