Diagonizable matrix

Question

Got this matrix:

\begin{bmatrix} 1 & 2 \\ -2 & 5 \end{bmatrix}

I should determine if the matrix is diagonalizable or not. I found the eigenvalues ( only one) = 3. My eigenvector is then \begin{bmatrix} 1 \\ 1 \end{bmatrix} This matrix is not diagonizable (from my teachers notes) but i don't know why, can someone explain this?

Answer 1

Basically you need to find the set $E$ all the eigenvalues associated with the matrix. (In this case $E =\{3\}$). Next for each eigenvalue in $E$ you find the eigenvectors. Let $W$ be the set of all eigenvectors of the matrix. Then if $W$ spans the column space of the matrix we say that the matrix is diagonalizable. In this case the column space is $\mathbb{R}^2$ but your lone eigenvector does not span $\mathbb{R}^2$ therefore the matrix is not diagonalizable.

Answer 2

If a $2\times 2$ matrix $A$ with one double eigenvalue say $\lambda$ is diagonalisable. It means there exists an invertible matrix $P$ such that

$$A=P\cdot\lambda I_2\cdot P^{-1}=\lambda P\cdot P^{-1}=\lambda I_2$$

This means the only $2\times 2$ with a double eigenvalue that is diagonalisable is diagonal

Answer 3

To answer your comment about the eigenvectors for this matrix: $$ A = \left[ \begin{array}{cc} 1&0\\ 0&1 \end{array} \right] $$ The characteristic polynomial given by $det(A - rI)$ where I is the n by n identity matrix will have roots equal to the eigenvalues of the matrix.
(Note: $r$ is eigenvalue of A such that $Ax = rx$, where x is not the zero vector and has n components. This x is called the corresponding eigenvector to the eigenvalue $r$.)

$det(A-rI) = r^2-2r+1 = (r-1)^2$, so we have the eigenvalue $r = 1$ (multiplicity two).

Now construct the eigenvectors corresponding to $r_1 = r_2 = 1$:

Since r is a scalar such that $Ax = rx$ for a non-trival vector x, we can find this eigenvector corresponding to r by solving $A - rI$.

This means to find $Null( A - rI )$, the set of vector(s) $t$ such that $(A - rI)*t = 0$

Since $$ (A - 1I) = I - I = \left[ \begin{array}{cc} 0&0\\ 0&0 \end{array} \right] $$ That means any vector that isn't the zero vector may satisfy $(A - rI)*t = 0$. So a valid $t$ is in

$$ Span \left\{ \begin{array}{c} \begin{pmatrix} 1\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1 \end{pmatrix} \end{array} \right\} $$

The vectors in the set above are the eigenvectors associated with the eigenvalue $r = 1$.

(Any identity matrix diagonalizes to the identity matrix).