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I'm having trouble with the second direction of the proof, ie. Assuming $A$ is similarto a diagonal matrix, then I want to show that $A$ is diagonalizable. I know part (b) of my answer is incomplete, but its because I was attempting to generate an answer with no success. Here's what I have so far:


question: Show that $A$ is diagonalizable if and only if $A$ is similar to a diagonal matrix.

answer: I want to show that $A$ is diagonalizable if and only if $A$ is similar to a diagonal matrix. Note, if $A$ is similar to $B$, then $B = M^{-1}AM$ and $A = MBM^{-1}$ for any invertible matrix $M$ (from Section 6.6 of the textbook [Introduction to Linear Algebra, Strang, 4e]).

Let's break this into two cases, representing the two directions of the 'iff' statement.

(a) Suppose $A$ is diagonalizable. I want to show that $A$ is similar to a diagonal matrix. Since $A$ is diagonalizable, we know that $A$ has $n$ distinct eigenvalues; let $\lambda_1, ..., \lambda_n$ be those values and $x_1, ..., x_n$ be the corresponding eigenvectors. Then $A$ is similar to the diagonal matrix $\Lambda$ which has the eigenvalues on it's diagonal and 0 in other spots, since $A = [x_1~\cdots~x_n]\Lambda [x_1~\cdots~x_n]^{-1} = S\Lambda S^{-1}$. Then, $A$ is similar to a diagonal matrix.

(b) Suppose $A$ is similar to a diagonal matrix. I want to show that if $A$ is diagonalizable. We cannot assume there exists eigenvalues to form a $\Lambda$ matrix or the matrix $S$ as in case (a). So let $D$ be the diagonal matrix to which $A$ is similar to, and $M$ be an invertible matrix. Then $A = MDM^{-1}$ and $D = M^{-1}AM$. ...


Then, my question is, how do I show part (b) is true? For reference, the class defines a square matrix $A$ as diagonalizable if we can write it as $S\Lambda S^{-1}$, where $S$'s columns are it's eigenvectors and $\Lambda$ is that matrix of all 0's, except the eigenvalues on it's diagonal.

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    $\begingroup$ What's your definition of $A$ is diagonalizable? You said that $A$ has $n$ distinct eigenvalues and this is false. $\endgroup$ – user296113 Mar 31 '16 at 18:10
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    $\begingroup$ I thought the definition of diagonalizable is "similar to a diagonal matrix". $\endgroup$ – Gabriel Nivasch Mar 31 '16 at 18:12
  • $\begingroup$ Sorry, forgot to mention: We assume that we've already shown that if $A$ is diagonalizable, then it has $n$ distinct eigenvalues. and I added to the question our definition of diagonalizable. $\endgroup$ – Account Killed Mar 31 '16 at 18:13
  • $\begingroup$ But that is not true, the square identity matrix of order n is diagonalizable and has not n distinct eigenvalues $\endgroup$ – Z. L. Mar 31 '16 at 18:14
  • $\begingroup$ @DylanBrown Then you must be using a very nonstandard definition of diagonalizable. Certainly we would want the identity matrix to be diagonalizable, and yet all of its eigenvalues are 1. $\endgroup$ – Alex S Mar 31 '16 at 18:14
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By definition, a matrix is said to be diagonalizable if it is similar to a diagonal matrix. Therefore, there is nothing to prove.

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