Construct a 2x2 matrix that expresses a given transformation

[Aleoa]

Homework Statement

I have to costruct a 2x2 matrix so that :

The Attempt at a Solution

M =$\begin{bmatrix} a & b\\ c &d \end{bmatrix}$

Using the first bond i got : c+2d = 2a+4b (1)
using the second bond : d = -b (2)

And then, as a nilpotent matrix has det = 0 and tr = 0, i got

a+d-2=0 (3)
ad-a-d+1=bc (4)

The problem comes here, i get 4 equations with 4 variables, but the 4th equation is not linear, in fact it's a second order equation.
Is it possible that in a linear algebra problem i get an equation that's not linear , or i made some errors ?

 

[BvU]

i get 4 equations with 4 variables

But your equation (1) is in fact two equations:
$a+2b = 1 \ \& \ c+2d = 2 \$. So you don't worry about your eq (4) which is in fact the 5th equation (luckily it is satisfied).

Linear algebra has plenty higher order equations: think e.g. of finding eigenvalues.

 

[Aleoa]

But your equation (1) is in fact two equations:
$a+2b = 1 \ \& \ c+2d = 2 \$. So you don't worry about your eq (4) which is in fact the 5th equation (luckily it is satisfied).

Linear algebra has plenty higher order equations: think e.g. of finding eigenvalues.

But... if i try to solve the system of equations (1),(2),(3) I can't find a unique solution. Only if i add the fourth equation is can find it.

ps : It could have happened, that in an exercise like the one i posted, the intersection beetween the 3 linear equation with the second order equation gives multiple solutions ?

 

[BvU]

But... if i try to solve the system of equations (1),(2),(3) I can't find a unique solution

Of course. You have lost one equation by writing " c+2d = 2a+4b "
instead of $\ \$ " $\ \$c+2d = 2 $\ \$ and $\ \$a+2b = 1 $\ \$ "

I wrote

luckily it is satisfied

for equation 5. I leave it to you to prove that the solution of { (1a), (1b), (2), (3) } satisfies equation 5. There must be a good reason for that -- I don't believe in lucky coincidences.

 

[Aleoa]

But your equation (1) is in fact two equations:
$a+2b = 1 \ \& \ c+2d = 2 \$. So you don't worry about your eq (4) which is in fact the 5th equation (luckily it is satisfied).

Linear algebra has plenty higher order equations: think e.g. of finding eigenvalues.

Of course. You have lost one equation by writing " c+2d = 2a+4b "
instead of $\ \$ " $\ \$c+2d = 2 $\ \$ and $\ \$a+2b = 1 $\ \$ "

I wrote
for equation 5. I leave it to you to prove that the solution of { (1a), (1b), (2), (3) } satisfies equation 5. There must be a good reason for that -- I don't believe in lucky coincidences.

Perfect, now my error is clear. The only thing i don't understand is how intuitively deduce that the 5th is already satisfied ( without doing the calculations)

 

[BvU]

I have the same problem