# Which entries could you change to make this 2 by 2 matrix diagonalizable ? [Strang P309 6.2.14]

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?

• 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? – user87543 Jan 12 '14 at 6:55
• @PraphullaKoushik: Thanks. I just rewrote it. Better? – Greek - Area 51 Proposal Jan 12 '14 at 6:57
• 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... – user87543 Jan 12 '14 at 7:01
• 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? – Jyrki Lahtonen Jan 12 '14 at 7:02

• 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.