Determine for which values of $a$ \begin{pmatrix} 4 & 0 & 0 \\ 4 & 4 & a \\ 4 & 4 & 4 \end{pmatrix}

The matrix is diagonalizable

So we first look at the characteristic polynomial:

$$ \begin{vmatrix} 4-\lambda & 0 & 0 \\ 4 & 4-\lambda & a \\ 4 & 4 & 4-\lambda \end{vmatrix}=(4-\lambda)\begin{vmatrix} 4-\lambda & a \\ 4 & 4-\lambda \end{vmatrix}=\\=(4-\lambda)[(4-\lambda)^2-4a]=(4-\lambda)[(\lambda^2-8\lambda+16-4a]=\\=4\lambda^2-32\lambda+64-16a-\lambda^3+8\lambda^2-16\lambda+4\lambda a=\lambda^3+12\lambda^2-48+64-16a+4\lambda a$$

How can I find the roots to this polynomial? usually with 3rd power polynomial I use to find the factor of the free element and test to find when the polynomial$=0$

What Should I do in this case?

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    $\begingroup$ There is an obvious eigenvalue $\lambda=4$, independently of $a$. $\endgroup$ – Marc van Leeuwen Aug 25 '16 at 7:16
  • $\begingroup$ @MarcvanLeeuwen Yes and for $\lambda=4$ there is one eigenvector, so I still need to find a condition on $a$ $\endgroup$ – gbox Aug 25 '16 at 7:20
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    $\begingroup$ Which you can obtain from the quadratic factor in the characteristic polynomial. Certainly you know a formula for the roots of a quadratic polynomial. $\endgroup$ – Marc van Leeuwen Aug 25 '16 at 7:21
  • $\begingroup$ @MarcvanLeeuwen missed it $\endgroup$ – gbox Aug 25 '16 at 7:22

You can do this without much computation.

First off, for $a=0$ the matrix is triangular (but not diagonal) with equal diagonal entries, therefore not diagonalisable.

Then assume $a\neq 0$. Now the matrix is block triangular, so its characteristic polynomial is the product of those of the diagonal blocks $D_1=(4)$ and $D_2=(\begin{smallmatrix}4&a\\4&4\end{smallmatrix})$. The former characteristic polynomial is $X-4$ with root $4$, while the characteristic polynomial of $D_2$ is a quadratic polynomial with the sum of its roots being (the trace) $8$. Now $4$ is clearly not a root of the latter characteristic polynomial (as $D_2-4I$ is invertible since $a\neq0$), an consequently $D_2$ cannot have multiple roots over $\Bbb C$, so the matrix is diagonalisable over $\Bbb C$.

Finally, since the question was about being diagonalisable over$~\Bbb R$, one does have to know the discriminant of the characteristic polynomial of $D_2$. Here I do have to admit some computation: the discriminant is $64-4\times\det D_2 =16 a$, so the eigenvalues of $D_2$ are real (and the original matrix diagonalisable over$~\Bbb R$) if and only if $a>0$.

  • $\begingroup$ As for "the matrix is triangular (but not diagonal) with equal diagonal entries, therefore not diagonalisable" I looked at previous questions and read your answers there (Thanks for that too), the above is true because "A matrix is diagonalizable if and only if its minimum polynomial has only single roots" and in this case it has multiple roots of the same value? $\endgroup$ – gbox Aug 25 '16 at 8:05
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    $\begingroup$ Well, if a triangular matrix is diagonalisable, then the diagonal form must have the same diagonal entries as the original (possibly permuted), because the characteristic polynomial of any triangular matrix has those entries as multiset of roots. Now if all diagonal entries are equal, then the diagonal form must be a scalar multiple of the identity matrix. But such a matrix is not similar to any other matrix at all, so the only hope for such a triangular matrix to be diagonalisable is that is is already equal to this scalar multiple of the identity. $\endgroup$ – Marc van Leeuwen Aug 25 '16 at 8:11

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