0
$\begingroup$

For all $c \in \mathbb{C}$, the matrix $A = \begin{bmatrix} c & 1 \\ 0 & c \\ \end{bmatrix}$

is not diagonalizable because the rank of $A - cI = 1.$
Change one entry to make $A$ diagonalizable. Which entries could you change?

Official Solution: Changing any entry except $A(1,2) = 1$ makes $A$ diagonalizable ($A$ will have two different eigenvalues).

Solution from P4 on http://www.math.montana.edu/~geyer/2010/fall/221.html:

${\Large{\color{red}{[}}} \;$ Changing the 1 to a 0 obviously diagonalizes the matrix, and thus diagonalizable. $\; {\Large{{\color{red}{]}}}}$
Changing any of the $c$'s to a different number diagonalizes the matrix because it will have 2 different eigenvalues.
Changing the 0 to something else will also lead to two different eigenvalues and diagonalize the matrix.

$\Large{{1.}}$ The first solution states that $A(1, 2) = 1$ shouldn't be changed, while the second solution says that it can be (as I signalised with the red brackets). Which is right and which is wrong?

$\Large{{2.}}$ Are there more efficient solutions, superior to inspecting each entry of the matrix?

$\endgroup$
  • $\begingroup$ I do not understand your "The two solutions discord with A(1,2)=1. Which is right and which is wrong?" what actually are asking for? $\endgroup$ – user87543 Jan 12 '14 at 6:55
  • $\begingroup$ @PraphullaKoushik: Thanks. I just rewrote it. Better? $\endgroup$ – Greek - Area 51 Proposal Jan 12 '14 at 6:57
  • $\begingroup$ Yes it is better now... the point is they are excepting diganonalizable matrices (which are not already diagonal)... so the point of making $1$ to $0$ does not seem to be interesting... $\endgroup$ – user87543 Jan 12 '14 at 7:01
  • $\begingroup$ I would interpret the exercise as follows: Change one entry by an amount with absolute value $<\varepsilon$, where $\varepsilon$ can be any positive constant (arbitrarily small). Which entry can you pick so that you are guaranteed to get a diagonalizable matrix? $\endgroup$ – Jyrki Lahtonen Jan 12 '14 at 7:02
2
$\begingroup$
  • If you change one of the entries $A(1,1)$ or $A(2,2)$ the matrix $A$ becomes with two distinct eigenvalues so it's diagonalizable.

  • If you change the entry $A(2,1)$ then we verify easily that the characteristic polynomial of $A$ has two distinct complex eigenvalues so it's also diagonalizable.

  • Finaly if we change the entry $A(1,2)$ the matrix $A$ stil has the only eigenvalue $c$ but since it isn't a scalar matrix (except if $A(1,2)=0$) hence it isn't diagonalizable.

$\endgroup$
  • $\begingroup$ Thanks for your response. As regards your last bullet, the solution from P4 says "Changing the 1 to a 0 obviously makes the matrix diagonal, and thus diagonalizable." Thus are they wrong? $\endgroup$ – Greek - Area 51 Proposal Jan 25 '14 at 13:45

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.