# Diagonalizability Quest

1. Feb 22, 2012

### trap101

Show that matrix A = \begin{bmatrix} a&b \\0&a \end{bmatrix} in M2 x 2(R) is diagonalizable iff b = 0

Attempt: Now I tried to solve for the eigenvalues and eigenvectors, which gave me a matrix of this form: \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}

but this matrix won't provide me with any information to show diagonalizability. What am I missing?

Quest 2: 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?

2. Feb 22, 2012

### lanedance

b=0 ->
well this one is as the matrix is already diagonal

A is diagonalisable ->
well what do you get for the chracteristic equation

hint: i get a repeated eigenvalue - this means the eigenspace must span R^2 for the matrix to be diagonalisable - can you show this is only possible when b=0?

3. Feb 22, 2012

### fauboca

For your first question, I am assuming you obtained the characteristic polynomial of $(\lambda - a)^2 = 0$.

So we know we have a repeated eigenvalue.

$$\begin{bmatrix} 0 & b\\ 0 & 0 \end{bmatrix}$$

Nothing extreme has been done. Simply following the protocol to diagonalize.

What does this matrix tell us?

For one, $x_1$ is a free variable. If b isn't a free a variable, how many eigenvectors would we have?

4. Feb 22, 2012

### trap101

If b isn't a free variable, then I only have one eigenvector, but my multiplicity is 2 which implies that the matirx isn't diagonalizable. but if b = 0 as in the question then i obtain a diagonal matrix. As for showing it, the only way I could show it is if I obtain the 0 matrix that I talked about below.

5. Feb 22, 2012

### fauboca

The only value that allows x_2 to be a free variable is 0. With out it, you said you only have one eigenvector. You need n eigenvectors for a nxn matrix. If you only have 1 eigenvector, you have a problem. How does knowing this not help you?

6. Feb 22, 2012

### lanedance

ok so you get $\lambda = a$ with algebraic multiplicity 2

to be diagonalisable you have to have and eigenspace that spans R^2, and as such there exists two linearly independent eigenvectors

so you can represent your eigenvectors as (1,0)^T and (1,0)^T, and x(1,0)^T +y(1,0)^T =(x,y)^T is still an eignevector

then, any eigenvector (x,y)^T must then satisfy
$$\begin{bmatrix} a&b \\0&a \end{bmatrix}\begin{bmatrix} x \\y \end{bmatrix} = a \begin{bmatrix} x \\ y\end{bmatrix}$$

which gives
$$\begin{bmatrix} ax+by = ax\\ay =ay \end{bmatrix}$$

so what constraints does $b \neq 0$ give?

Last edited: Feb 22, 2012
7. Feb 22, 2012

### trap101

I don't think I follow... what is the (1,0)^T? What's the raised to the T part?

8. Feb 23, 2012

### Deveno

it means "transpose" and just means we are talking about a column vector, rather than a row vector (since we're dealing with matrices, where rows and columns are different things).

9. Feb 24, 2012

### trap101

Well simplifying I obtain y = 0, does this mean y has to equal zero in order for the expression to be satisfied? but if that happens then ay = ay becomes 0 = 0

10. Feb 25, 2012

### lanedance

ok so if y = 0, what is the dimension of the eigenspace (how many linearly independent eigenvectors do you have?)