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I am reading Linear algebra done right. I was going to a Proposition 5.21. Which goes like this.

Suppose $T \in L(V)$. Let $\lambda_1 , . . . , \lambda_n$ denote the distinct eigenvalues of $T$ . Then the following are equivalent:

(a) $T$ has a diagonal matrix with respect to some basis of $V$ ;

(b) $V$ has a basis consisting of eigenvectors of $T$ ;

My question is, Isn't all linear operators $V \rightarrow V$ is just a change of basis? In which case, isn't there always a diagonal matrix (Identity) with the right basis? Or am I wrong somewhere? What does this some basis really mean in

Suppose $T \in L(V)$ has an upper-triangular matrix with respect to some basis of $V$ . Then the eigenvalues of $T$ consist precisely of the entries on the diagonal of that upper-triangular matrix. (Proposition 5.18)

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  • $\begingroup$ Not all matrix has an eigenbasis. You may refer to Jordan canonical form. $\endgroup$ Nov 16, 2017 at 11:14
  • $\begingroup$ The linear map $\Bbb R^2\to \Bbb R^2$ that is represented in the standard basis as the matrix $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ does not have a diagonal matrix representation in any basis. $\endgroup$
    – Arthur
    Nov 16, 2017 at 11:15
  • $\begingroup$ There is no basis that makes $\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$ diagonal. $\endgroup$
    – A.Γ.
    Nov 16, 2017 at 11:15
  • $\begingroup$ @Arthur What if I chose $(1,0)$ and $(1,1)$ as the basis? $\endgroup$ Nov 16, 2017 at 11:17
  • $\begingroup$ @VinothkumarRaman The matrix representing my map in your basis is $\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ (yes, it's the same matrix). $\endgroup$
    – Arthur
    Nov 16, 2017 at 11:19

2 Answers 2

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Consider $n = 2$ and $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$. Obviously the eigenvalues of $A$ cannot be different from $0$.

If $A$ were diagonizable, then there would exist an invertible matrix $V$ such that $V^{-1} A V = D$ where $D$ is a diagonal matrix containing the eigenvalues of $A$. But this implies $D = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} $, leading to the contradiction $A = D$. Thus there is no basis such that $A$ is a diagonal matrix (in this base).

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With the examples so far you might come to think that a matrix is diagonalizable iff it is a bijection. However it might also pose a hurdle, if the underlying field is not complete:

Consider the rotation matrix $A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ on $\mathbb{R}^{2\times2}$:

It's eigenvectors are $\begin{pmatrix} 1 \\ \pm i \end{pmatrix}$, its eigenvalues $\mp i$, therefore it is not diagonalizable on $\mathbb{R}^{2\times2}$.

Maybe wikipedia can give you some more insights: https://en.wikipedia.org/wiki/Diagonalizable_matrix#Examples

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