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I got the following matrix:

$$ A = \begin{pmatrix} a & 0 & 0 \\ b & 0 & 0 \\ 1 & 2 & 1 \\ \end{pmatrix} $$

I need to answer when this matrix is diagonalizable. Its characteristic polynomial is $ t(t-a)(t-1) $. So its 3 eigenvalues are 0, 1 and a. Both the algebraic and geometry multiplicities of those values are 1 (for all of them).

Let's look at the matrices for those eigenvalues: $$ A - 0I = \begin{pmatrix} a & 0 & 0 \\ b & 0 & 0 \\ 1 & 2 & 1 \\ \end{pmatrix} $$ $$ A - I = \begin{pmatrix} a -1 & 0 & 0 \\ b & -1 & 0 \\ 1 & 2 & 0 \\ \end{pmatrix} $$ $$ A - aI = \begin{pmatrix} 0 & 0 & 0 \\ b & -a & 0 \\ 1 & 2 & 1 - a \\ \end{pmatrix} $$ $\rho (A - 0I) = 2 $

$\rho (A - 1I) = 2 $

$\rho (A - aI) = 2 $

It seems that for every $a$ and $b$ this matrix would be diagonalizable. But it's not. Where am I wrong?

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    $\begingroup$ $a$ must be different from $0$ or $1$. Otherwise the characteristic polynomial is not a product of distinct linear factors and hence we cannot conclude that the matrix is diagonalizable. $\endgroup$ – Spenser Mar 26 '16 at 16:51
  • $\begingroup$ @Spencer that is not a sufficient condition. The necessary and sufficient condition is that hte minimal polynomial of the matrix must be a product of different linear polynomials. $\endgroup$ – DonAntonio Mar 26 '16 at 16:52
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    $\begingroup$ @Joanpemo I know. I am just saying that we cannot conclude that $A$ is diagonalizable. I am not saying that it is not diagonalizable. $\endgroup$ – Spenser Mar 26 '16 at 16:53
  • $\begingroup$ @Joanpemo But you are right that my statement was not clear. Thanks for the precision. $\endgroup$ – Spenser Mar 26 '16 at 16:55
  • $\begingroup$ If $a=b=0$ then why can't it be diagonalizable? $ρ(A−0I)=1$ and $ρ(A−1I)=2$, $\endgroup$ – MyNick Mar 26 '16 at 17:14
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If $\;a\neq0,1\;$ the matrix has three different eigenvalues and is thus diagonalizable. Now, upon substitution in $\;\det(A-\lambda I)\;$ in the other two cases we get the homogeneous systems:

$$a=0:\;\;\begin{cases}bx=0\\x+2y-z=0\end{cases}\;\;\;\text{if}\; b\neq0\;,\;\;\text{then the solution space's}\;\;\left\{\,\begin{pmatrix}0\\y\\2y\end{pmatrix}\,\right\}$$

which is of dimension one and thus the matrix isn't diagonalizable since the algebraic multiplicity of the eigenvalue zero $\;\neq\;$ the geometric one, but if

$$b=0\implies\text{ the solution space's}\;\left\{\,\begin{pmatrix}x\\y\\x+2y\end{pmatrix}\,\right\}$$

of dimension two and thus the matrix is diagonalizable.

Now you try to do something similar with the case $\;a=1\;$ .

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1) If $a\ne 0, 1,\;$ then A is diagonalizable since it has 3 distinct eigenvalues.

2) If $a=0$, then A is diagonalizable $\iff$ $\text{nullity}(A-0I)=\text{nullity}(A)=2 \iff \text{rank}(A)=1$

$\hspace{2.3 in}\iff\text{rank}\begin{pmatrix} 0&0&0\\b&0&0\\1&2&1\end{pmatrix}=1\iff b=0$

3) If $a=1$, then A is diagonalizable $\iff$ $\text{nullity}(A-1I)=\text{nullity}(A-I)=2 \iff \text{rank}(A-I)=1$

$\hspace{2.3 in}\iff\text{rank}\begin{pmatrix} 0&0&0\\b&-1&0\\1&2&0\end{pmatrix}=1\iff b=-\frac{1}{2}$

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  • $\begingroup$ So to make sure i understood it, if !=0,1 then it diagnozable. if a= 0 then b = 0, if a =1 then b = -0.5? $\endgroup$ – MyNick Mar 26 '16 at 18:39
  • $\begingroup$ @MyNick That's right, except the first part should say $a\ne0,1$ -- I will edit my answer slightly. $\endgroup$ – user84413 Mar 26 '16 at 20:41
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Your matrix has at least $2$ eigenvalues, namely $0$ and $1$, and maybe a third, namely$~a$ it it is different from those two others. In the latter case we have $3$ simple roots of the characteristic polynomial, and $A$ is automatically diagonalisable. So the remaining interesting case is $a\in\{0,1\}$.

In that case $A$ is diagonalisable if and only if the polynomial $(X-0)(X-1)=X^2-X$ annihilates $A$, in other words if $A^2-A=0$ (the kernel of $A^2-A$ is the sum of the eigenspaces of $\lambda=0$ and $\lambda=1$). Now $$ A^2-A = \begin{pmatrix} a^2-a & 0 & 0 \\ b(a-1) & 0 & 0 \\ a+2b & 0 & 0 \\ \end{pmatrix} $$ so in the remaining cases $A$ is diagonalisable only if either $a=0$ and $b=0$, or $a=1$ and $b=-\frac12$.

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