Eigenvector for imaginary solution

Question

During solving an eigenvector problem for matrix $$ \begin{bmatrix} 3 & -5 \\ 2 & -3 \\ \end{bmatrix} $$ I got a solution λ = $i$ and a matrix: $$ \begin{bmatrix} i-3 & 5 \\ -2 & i+3 \\ \end{bmatrix} $$ I tried to reduce it, but did not succeed with this task, then I looked to a solution of this problem, and it states that: $$ \begin{bmatrix} 3-i & -5 \\ 2 & -i-3 \\ \end{bmatrix} \rightarrow \begin{bmatrix} 3-i & -5 \\ 0 & 0 \\ \end{bmatrix} $$ Q: Why so? Is it a property, or some row operations can perform this reduction?

And furthermore, it states, that eigenvector corresponding to λ = $i$ is: $$ \begin{bmatrix} 5 \\ 3-i \\ \end{bmatrix} $$ I would rather expect to have a pivot entry on the position of $3-i$, in the resulting matrix.

Q: How do I come up with such a solution.

Answer 1

To find eigenvectors, we put $A\mathbf{v}=\lambda \mathbf{v}$.

Since your matrix is $2\times 2$, we can let $\mathbf{v}=\begin{pmatrix} x \\ y \end{pmatrix}$

Therefore, we have:

$$\begin{pmatrix} 3 & -5 \\ 2 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=i\begin{pmatrix} x \\ y \end{pmatrix}$$

From this, we can find the values of $x$ and $y$.

$$\begin{cases} 3x-5y=ix \\ 2x-3y=iy \end{cases} \tag{1}$$

You will realise that after some manipulation that the second equation is the same as the first (This is due to the eigenvalues).

Thus, we only need to solve one equation from $(1)$.

Let us pick the most convenient one:

$$3x-5y=ix$$ $$(3-i)x-5y=0$$

Which yields the eigenvector:

$$\mathbf{v}=\begin{pmatrix} 5 \\ 3-i \end{pmatrix}$$

See why?

Answer 2

$\frac{1}{3-i}\cdot\frac{3+i}{3+i}=\frac{3+i}{10}$. What if you multiplied the first row by that to get a 1 in the top left. Can you take it from there?

Once you get there, remember you are looking for a nontrivial solution to $Ax=\lambda x$ or a nontrivial solution to $(A-\lambda I)x=0$ That vector you have is a non-trivial solution to that equation.

Answer 3

The eigenspace for the eigenvalue $\lambda=i$ is formed by the vectors $[x,y]^T$ such that: $$ \begin{bmatrix} i-3&5\\ 2&-3-1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $$

that are solutions of the system: $$ \begin{cases} (i-3)x+5y=0\\ -2x+(i+3)y=0 \end{cases} $$ multiplying the first equation for $-(i+3)/5$ and adding to the second becomes:

$$ \begin{cases} (i-3)x+5y=0\\ 0x+0y=0 \end{cases} $$

that, with a change of sign, gives the matrix in your question.

The solution of the system are the vectors of the form $[5t,(3-i)t]^T $ with $ t \in \mathbb{R}$, and this is the eigenspace of the eigenvector $\lambda=i$. A possible eigenvector is, (for $t=1$) $[5,3-i]^T$.