Find the necessary and sufficient conditions on the real numbers a,b,c

1. Feb 24, 2012

trap101

Find the necessary and sufficient conditions on the real numbers a,b,c for the matrix:
\begin{bmatrix} 1 & a & b\\ 0 & 1 & c \\ 0 & 0 & 2 \end{bmatrix} to be diagonalizable.

Attempt: Now for this one I also solved for the eigenvlues which were: λ1 = 1, λ2 = 1, λ3 = 2

So the problematic eigenvalues will be the one of multiplicity 2, i.e λ = 1.

So this means I'd have to obtain two linearly independent eigenvectors for λ = 1.

I tried solving and got to this matrix: \begin{bmatrix} 0 & a & b \\ 0&0&c \\ 0&0 & 1 \end{bmatrix}

But I won't be able to find two linearly independent eigenvectors from setting any of the variables equal to anything...I don't think. What's the next step?[/QUOTE]

2. Feb 24, 2012

fauboca

Re: Diagonalizability

Take your matrix down to reduce row echelon form.

From that point, you should be able to figure what one of the variables should be.

3. Feb 24, 2012

trap101

Re: Diagonalizability

Do you mean just reduce the variables: a , b, c? i.e: divide out a by 1/a, etc and row reduce those?

4. Feb 24, 2012

fauboca

Re: Diagonalizability

Start with the 2nd and 3rd row first. Reduces those and then see what you can do to the top.

5. Feb 24, 2012

trap101

Re: Diagonalizability

\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} a row of zeros which means one free variable, or would this mean that a, c = 0 and b could be any value?

6. Feb 25, 2012

fauboca

Re: Diagonalizability

Instead of dividing out the 1st row by a, leave it as a. You will still have a = 0.

Why is c = 0? the second row means x_3 = 0.

7. Feb 25, 2012

trap101

Re: Diagonalizability

So then a = 0, x2 is a free variable, and x3 = 0. Now this leaves me with still only one free variable......actually......x1 is a free variable as well isn't it?......ahhhhhh, if that's the case I see what your getting at. But I didn't really state any conditions, all I did was solve a reduced matrix.

Thanks

8. Feb 25, 2012

fauboca

Re: Diagonalizability

What you know now is a must be 0. The question is can b and c be anything?

9. Feb 25, 2012

trap101

Re: Diagonalizability

Well if the same process would have to be applied to find any matrix that is diagonalizable, it must mean that b, c can be any real number because they will be eliminated through row reduction.