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I'm teaching myself linear algebra and came upon this exercise. It is probably not very hard, but I just have hard time solving it.

$$\text{Which values for constants a, b, c, d, e, f make the matrix diagonalizable?}\\ A= \begin{pmatrix} 1 & a & b & c \\ 0 & 2 & d & e \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$

I have tried wrestling this for many hours now and I need some help. What I have been able to do so far, that I know of is correct:

The characteristic polynomial is $$p(\lambda)=(\lambda-2)^3(\lambda-1)$$ and therefore the eigenvalues and eigenvectors for those are $$\lambda_1=2,\space \lambda_2=1$$

Also the two eigenvectors are $$v_1=[1,0,0,0]^T \text{ and } v_2=[a,1,0,0]^T$$

That is what I have for certain.

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Then there are some other things I've been thinking, but which I'm not sure of:

I assume that in $$D=P^{-1}AP$$ the D would be $$ D= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$

as the eigenvalues should be on the diagonal and the eigenvalue 2 is repeated.

One great point of uncertainty is that I seem to be missing $2$ vectors from the $P$ (so that it would be $4\times 4$ matrix) and I don't seem to find a way to constructing them. I know that the 2 missing vectors should be linearly independent, but only vectors I could think of were vectors $[0,0,1,0]^T$ and $[0,0,0,1]^T$ and that lead to following when $D=P^{-1}AP$ is used:

$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -a & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} 1 & a & b & c \\ 0 & 2 & d & e \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} \begin{pmatrix} 1 & a & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

$$ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} = \begin{pmatrix} 1 & 2a & b & c \\ -a & 2-2a^2 & d-ab & e-ac \\ 0 & 0 & 2 & f \\ 0 & 0 & 0 & 2 \\ \end{pmatrix} $$

And if that is correct, all constants have to be $0$, otherwise it is not diagonalizable. But I'm very skeptical that this would be the correct answer. I think I need somehow link the additional two eigenvectors to the eigenvalue $2$, but I don't know how to do that.

I'm very grateful for all help.

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For $\lambda=2$, $(A-2I)v=0$ gives the augmented matrix for the homogeneous system as

$\begin{pmatrix}-1&a&b&c:0\\0&0&d&e:0\\0&0&0&f:0\\0&0&0&0:0\\\end{pmatrix}$

What you really need that is $3$ linearly independent eigenvectors from this i.e. $n-r=3$ where $n$($=4$ here) is number of variables and $r$ is rank of the above matrix. Can you see that only $d=e=f=0$ is sufficient for $r=1$.

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  • $\begingroup$ is the solution then $d=e=f=0$, and $a=b=c$ can be freely chosen? And then the eigenvectors would be $v_1=[1,0,0,0]$, $v_2=[a,1,0,0]$, $v_3=[b,0,1,0]$ and $v_4=[c,0,0,1]$, $\endgroup$ – juahan Dec 6 '16 at 11:25
  • $\begingroup$ @juahan......yes. $\endgroup$ – Nitin Uniyal Dec 7 '16 at 3:55
  • $\begingroup$ Oh man, turned out to be much easier than I thought. Again. Thanks for the help! $\endgroup$ – juahan Dec 7 '16 at 6:23
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You only have two eigenvalues: $\;\lambda=1\;,\;\;\lambda=2\;$ , the first one with algebraic multiplicity equto to one (and thus also geometric mult. one), and the other one with alg. mult. three, and you need need also its geom. mult. to be three, too. So let us check the homogeneous linear system to ge the eigenspace for $\;\lambda=2\;$ and try to check whether it ispossible to get one with solution space of dimension three:

$$\begin{cases}&x-&ay-&bz-&\;\;\;cw=0\\ &&&-dz-&\;\;\;ew=0\\ &&&&-fw=0\end{cases}\;\;\;\implies$$

if $\;f\neq0\;$ then $\;w=0\;$ , but then the first equation is just a two dimensional plane, so we need $\;w\neq0\implies f=0\;$ . Take it from here now.

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