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## Main Question or Discussion Point

I am trying to perpare for my final for linear algebra and am attempting to catch up in readings- in particular about diagonlization.

So it follows, for each eigenvalue [tex]\lambda_j[/tex] ( 1 <= j <= n, where n is the number of elements in our particular basis B ), there exists corresponding eigenvectors [tex]v_j[/tex] by definition. Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors, our basis B = { lambda_1, lambda_2, ..., lambda_n } is composed of eigenvectors.

1. Can someone tell me if my proof above is correct? I feel as if I'm assuming to much when I said "Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors... "

2. Also can someone justify the remaining portion of the theorem "... Furthermore, if T is diagonalizable... is the eigenvalue corresponding to [tex]v_j[/tex] for 1 <= j <= n."

THanks again,

JL

**Theorem 5.1:**A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis B for V consisting of eigenvectors of T. Furthermore, if T is diagonalizable, B = {[tex]v_1, v_2, ..., v_n[/tex]} is an ordered basis of eigenvectors of T, and D = [tex][T]_B[/tex], then D is a diagonal matrix and [tex]D_j_j[/tex] is the eigenvalue corresponding to [tex]v_j[/tex] for 1<= j <= n.**Question:**For some reason in my textbook, they omitted the proof. I'm guessing it's fairly straightfoward, but I was wondering if someone could aid me with a proof.**Idea for a proof:**If T is diagonalizable, then for some [tex]v_j[/tex], [tex]T(v_j)[/tex] = [tex]\lambda_j[/tex][tex]v_j[/tex] (by definition- from last post).So it follows, for each eigenvalue [tex]\lambda_j[/tex] ( 1 <= j <= n, where n is the number of elements in our particular basis B ), there exists corresponding eigenvectors [tex]v_j[/tex] by definition. Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors, our basis B = { lambda_1, lambda_2, ..., lambda_n } is composed of eigenvectors.

**Questions:**1. Can someone tell me if my proof above is correct? I feel as if I'm assuming to much when I said "Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors... "

2. Also can someone justify the remaining portion of the theorem "... Furthermore, if T is diagonalizable... is the eigenvalue corresponding to [tex]v_j[/tex] for 1 <= j <= n."

THanks again,

JL

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