my teacher gave me this exercise:
Determine if this matrix is diagonalizable $ \begin{pmatrix} 1 & 1&1&1\\ 1&2&3&4\\ 1&-1&2&-2\\ 0&0&1&-2 \end{pmatrix} $

I have tried to calculate the characteristic polynomial, that is $-13 + 10 x + x^2 - 3 x^3 + x^4$, but I don't know how to go on. I tried to look at the roots of the characteristic polynomial with Wolfram Alpha and they are horrible! So I think that must exist an alternative way to do it! Have you got any ideas? Thank you!

  • 1
    $\begingroup$ You should have seen at least one theorem of the form: A real square matrix is diagonalisable if and only if . . . Do you remember any such theorems? $\endgroup$ – Michael Albanese Jun 15 '14 at 13:41
  • $\begingroup$ Are you sure that you have jotted down the matrix correctly? It seems to me that the third row should be $(1,-2,-2,-2)$ instead of $(1,-1,2,-2)$. $\endgroup$ – user1551 Jun 15 '14 at 14:59

Hint: The matrix is diagonalizable (over $\mathbb C$) if, and only if, it has 'four' eigenvalues. In particular, if the matrix has four distinct eigenvalues, then it is diagonalizable. Using the characteristic polynomial, can you prove that the matrix has four distinct eigenvalues?

Further hints: Recall that given a polynomial function $P$ whose coefficients are all real, it holds that $\forall z\in \mathbb C\left(P(z)=0\implies P(\overline z)=0\right)$. Since the characteristic polynomial is a real fourth (even!) degree polynomial, either all roots are non-real, there exactly 'two' (beware of multiplicities) real roots or all roots are real. Use the intermediate value theorem theorem to prove that there are at least two real roots and check the derivative to prove that there are no more than two real roots.

  • $\begingroup$ I know this theorem, but the roots of this polynomial are $\textbf{horrible}$. So either my teacher is a sadist either there is a better solution. $\endgroup$ – Nisba Jun 15 '14 at 14:05
  • 2
    $\begingroup$ @user3343783 My point is that you don't need to find the roots, it suffices to prove that four distinct roots exist. $\endgroup$ – Git Gud Jun 15 '14 at 14:08
  • $\begingroup$ I am not able, can you show me how you do it please? $\endgroup$ – Nisba Jun 15 '14 at 14:41
  • $\begingroup$ @user3343783 See the edit. $\endgroup$ – Git Gud Jun 15 '14 at 14:49
  • $\begingroup$ @GitGud The polynomial $x^3(x-1)$ has two real roots, namely $x=0$ and $x=1$ and so satisfies the condition of having at least two real roots. Yet not all roots are distinct $\endgroup$ – Fly by Night Jun 15 '14 at 15:08

If the roots of the characteristic polynomial are distinct, the matrix is diagonalizable (the Jordan normal form is composed of $1\times1$ Jordan blocks). The roots of the polynomial $P$ are distinct if $\deg(\gcd(P,P'))=0$ (if $(x-a)^n$ divides $P$, then $(x-a)^{n-1}$ divides $P'$).

Using the Polynomial Extended Euclidean Algorithm on $\color{#C00000}{P}$ and $\color{#00A000}{P'}$, we get $$ \begin{align} &(1822x^3-1987x^2-2613x+10903)(\color{#00A000}{4x^3-9x^2+2x+10})\\ &-(7288x^2-2482x-3659)(\color{#C00000}{x^4-3x^3+x^2+10x-13})\\ &=61463 \end{align} $$ Thus, $\deg(\gcd(P,P'))=0$, and therefore, the matrix is diagonalizable.

  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Aug 6 '14 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.