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I have not seen anywhere written that it is impossible, but it seems impossible, so I want to check if I missed something.

According to a theorem, an nxn matrix is diagonalizable if it has n independent eigenvectors.

Let's say, the matrix has 1 row with only zeros(worst singular case). As it has one row with only zeros, it will zero out the corresponding row of any vector it is multiplied by. That means, the corresponding rows of its eigenvectors have to be zero. And if one of the rows is fixed, there cannot be n independent eigenvectors. Am I correct?

Did I miss anything?

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I think you're discovering that the theorem you quote only gives a sufficient condition for diagonalizability, and not a necessary and sufficient one. –  Gunnar Þór Magnússon Apr 22 at 22:40
    
Any diagonal matrix is certainly diagonalizable. –  littleO Apr 22 at 22:41
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As it has one row with only zeros, it will zero out the corresponding row of any vector it is multiplied by: this sounds correct. That means, the corresponding rows of its eigenvectors have to be zero: this doesn't need to happen for $\lambda = 0$. And if one of the rows is fixed, there cannot be n independent eigenvectors: correct conclusion from the (incorrect) previous statement. –  Omnomnomnom Apr 22 at 22:51
    
@GunnarMagnusson No, I was wrong, I think this theorem gives necessary and sufficient condition. Can you tell a nxn matrix which has less than n eigenvectors and is diagonalizable? –  khajvah Apr 22 at 23:02
    
@khajvah the condition you gave is indeed necessary and sufficient. –  Omnomnomnom Apr 22 at 23:41

2 Answers 2

up vote 5 down vote accepted

A matrix is singular if and only if $0$ is one of its eigenvalues. A singular matrix can be either diagonalizable or not diagonalizable. For example, $$ \pmatrix{ 1&0\\0&0 } $$ Is diagonalizable (since it is diagonal), whereas $$ \pmatrix{ 0&1\\0&0 } $$ is not diagonalizable.

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Yes, diagonalize the zero matrix.

THM Let $A$ be a matrix associated to a linear transformation $T:V\to V$, $V$ a $k$-vector space. Let $\lambda_1,\ldots,\lambda_l$ be it's distinct eigenvalues. Then $A$ is diagonalizable iff $$V=E(\lambda_1)\oplus\cdots \oplus E(\lambda_l)$$ where $E(\lambda_i)=\ker(A-\lambda_i1)$.

The sum is always direct, but it might not comprise all of $V$. This is equivalent to the easier claim that $V$ admits a basis of eigenvectors of $A$, also equivalent to the fact that the minimal polinomial splits over $k$ and has all simple roots. In turn, this is equivalent to the fact that $\chi_A$ splits and the multiplicity of $\lambda_i$ as a root equals the dimension of $E(\lambda_i)$. We always have $\dim E(\lambda_i)\leqslant {\rm mult}\;(\chi_A,\lambda_i)$, and this is in fact a good thing to use when checking for diagonalizability, in some cases. For example, $$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ is not diagonalizable, since $\chi=(X-1)^2$ but $\dim E(1)=1<2$.

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Any diagonal matrix with a zero on the diagonal is singular. (in response to the deleted comment..) –  mathematician Apr 22 at 22:42
    
@mathematician yeah, thanks, I realized after I wrote the comment :) –  khajvah Apr 22 at 22:43

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