I am trying to find the eigenvectors of a $2\times2$ singular matrix, $A= \bigl( \begin{smallmatrix}0 & 1\\ 0 & -3\end{smallmatrix}\bigr)$.
My problem is that I can't.
I know the answer is, $Q= \bigl( \begin{smallmatrix}0 & 1\\ 0 & -3\end{smallmatrix}\bigr)$$($by using Matlab$)$, but I don't understand why this is the right answer.
I know how to do this for non singular $2\times2$ matrices, but I can't seem to find any info on how to do this on singular matrices.
The same goes for diagonalization on these matrices.
If you know how to, please show me the process in detail, thank you!
Best Answer
the process of finding the eigenvalues of a matrix is an algebraic operation which does not involve the singularity or non-singularity of the matrix. this latter fact will, however, impact on the nature of the set of eigenvalues and their corresponding eigenspaces.
to take an extreme case, what are the eigenvalues of the zero $n \times n$ matrix?
it should be easy to see that the characteristic equation, in this case, is: $$ (-\lambda)^n =0 $$ in this very special case the kernel of the transformation is the entire domain, and every vector is an eigenvector, albeit in a somewhat 'trivial' sense.
in the case of your matrix the equation is $$ -\lambda(-3-\lambda) = 0 $$ which has the solution set $\{0,-3\}$
now the eigenspace corresponding to a zero eigenvalue is the kernel of the linear transformation represented by the matrix. in symbols, for the column 2-vector $v$ $$ Mv = 0v \Leftarrow \Rightarrow Mv = 0 $$ if you substitute $v=(x,y)^T$ then the conditions obtained are $y=0$ and $-3y=0$, i.e. $y=0$, corresponding to the eigenvector $(1,0)^T$ the value $1$ for $x$ could be replaced with any other non-zero number, as any scalar multiple of an eigenvector is also an eigenvector (by linearity).
for the value $\lambda=-3$ the conditions are: $$ y = -3x \\ -3y = -3y $$ choosing $x=1$ (again any non-zero value is as good) we have $y=-3$ giving the eigenvector $(1,-3)^T$