Question

Suppose that A =
1 L 0
0 L^2 3
2 1 1

where L belongs to R

Find all the values of L in order for the Nullity A>0 (strictly positive).

Answer 1

I would use row reduction to change the matrix to
\[\left[\begin{matrix}1 & L&0 \\ 0 & L^2 & 3 \\ 0 & 1-2L &1\end{matrix}\right]\]

Answer 2

(I add -2 times the first row to the last row)
if we scale the last row by 3
\[ \left[\begin{matrix}1 & L&0 \\ 0 & L^2 & 3 \\ 0 & 3-6L &3\end{matrix}\right] \]
we see we can can a repeated row (i.e. nullity >0 ) if
\[ L^2 = 3-6L\]
solve for L

Answer 3

how is L ^2 equal to 3 - 6L @phi

Answer 4

do you mean that I need to solve for L on that equation

Answer 5

I am saying *if* L^2 equals 3-6L
then the 2nd and 3rd rows will be duplicates

Answer 6

*if* aha ok

Answer 7

in this case, there are two (ugly) values for L that "work"