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

[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]

 

[fauboca]

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.

 

[trap101]

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

 

[fauboca]

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

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

 

[trap101]

\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?

 

[fauboca]

\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?

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.

 

[trap101]

\begin{bmatrix} 0 & a & 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?

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

 

[fauboca]

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

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

 

[trap101]

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.