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My professor gave us this definition of a diagnolizable matrix. A matrix $A$ is diagnolizable if it's invertible and

$$(Ax)_{\mathcal{B}} = Dx_{\mathcal{B}}$$

for some diagonal matrix $D$, basis $\mathcal{B}$ and all vectors $x$.

But there are non invertible matrices that have "diagonal equivalents" in other basis. For example,

$\begin{bmatrix}2 & -1\\2 & -1\end{bmatrix}$ is diagnolizable to $\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}$ even though its not invertible.

So why the restriction on the definition?

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I have never seen anyone use invertibility as part of the definition of diagonalizable... – Tom Dec 1 '13 at 4:17
This is actually a mistake of mine; I got mixed up, I ( implicitly/subconsciously) confused eigenvector with eigenvalue, so I said below that $0$ cannot be an eigenvalue, when the real point is that $0$ vector is not an eigenvector. – user99680 Dec 1 '13 at 4:20
And I've never seen a definition of diagonizability that only wants an equation like that to hold for some vector $x$. – Henning Makholm Dec 1 '13 at 4:21
@HenningMakholm Sorry my wording was wrong, I mean't all $x$. – dfg Dec 1 '13 at 4:23
I can't believe that your professor told you this. You must have misunderstood him/her. – Doc Dec 1 '13 at 4:56
up vote 2 down vote accepted

Diagonalizable matrices need not be invertible. For instance, the matrix $$A = \begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix} = \begin{bmatrix} -1/\sqrt2 & 1/\sqrt2\\ 1/\sqrt2 & 1/\sqrt2\\ \end{bmatrix} \begin{bmatrix} 0 & 0\\ 0 & 2\\ \end{bmatrix} \begin{bmatrix} -1/\sqrt2 & 1/\sqrt2\\ 1/\sqrt2 & 1/\sqrt2\\ \end{bmatrix}$$

The converse also need not hold true, i.e., invertible matrices need not be diagonalizable. For instance, $$A = \begin{bmatrix}1 & 1\\ 0 & 1 \end{bmatrix}$$ is invertible but not diagonalizable.

Diagonalizability and Invertibility do not relate to each other, in general.

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Can you have a non-invertible matrix that is diagnalizable, and the diagnolized matrix has no-zeros in the centre diagonal? – dfg Dec 1 '13 at 4:30
@dfg A matrix is invertible if and only if all the eigenvalues are non-zero. If a matrix is diagonalized, then diagonal has the eigenvalues. Hence, the answer to your question is no. – user17762 Dec 1 '13 at 4:32

First of all, my apologies for my mixup in your other post. Hope this will help clarify the issue:, by solving :$Ax=x$ and $(A-I)x=x$ for $x$.

A non-invertible matrix can be diagonalizable. A necessary and sufficient condition for a matrix $A$ to be invertible is that the matrix must have a basis of eigenvectors, so that the dimension of the eigenspace of the matrix must be $\geq n$ , for $A$ an $n \times n$ matrix. Consider your matrix:

$\begin{bmatrix}2 & -1\\2 & -1\end{bmatrix}$

Let's compute its eigenspace. first, the eigenvalues. Set:

$Det\begin{bmatrix}2 -\lambda & -1\\2 & -1-\lambda\end{bmatrix}=0$

The characteristic polynomial is : $\lambda^2-\lambda=\lambda(\lambda -1):=0$

Then there are two eigenvalues; $\lambda=1,0$. This guarantees that you have two linearly-independent eigenvectors. Maybe you want to compute the eigenspace associated with each of the eigenvalues.

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