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I've got a matrix $A = \begin{bmatrix}1&-1\\ 2&-1\end{bmatrix}$ and wish to diagonalise it. I find the eigenvalues as below.

$$\det(A - xI) = 0 = \det\begin{bmatrix}1-x&-1\\ 2&-1-x\end{bmatrix}$$

$$\det(A - xI) = (1-x)(-1-x) - (-2) = 1 + x^2$$

Then $x =i, -i$. So, now I need to find the eigenvectors, which is where I'm a little confused. In the case that $x = i$:

$$E_1 = ker\begin{bmatrix}1-i&-1\\ 2&-1-i\end{bmatrix}$$

$$\begin{bmatrix}1-i&-1\\ 2&-1-i\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$

At this point we usually row reduce and end up with a row of all zeros. However, in this case that isn't possible. So, I get

$2a + (-1-i)b \implies a = (b+ i)/2$ from the second row.

I want to write the kernel in terms of a spanning set, so get $E_1 = \{((1+i)/2, 1)\}$. Would this be a correct spanning set for $E_1$? Is the choice between which rows to use arbitrary? I'm just seeking clarification because all problems I've done before row reduce to only a single row.


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up vote 1 down vote accepted

Solving your linear system regarding $E_1$, $$\begin{bmatrix}1-i&-1\\2&-1-i\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix},$$ You can perform for operations as normal $R_2\leftarrow R_2-\frac{2}{1-i}R_1$, which will give the second row as all zeros (like 'normal').

Then you have $(1-i)a-b=0\Rightarrow a=\frac{1+i}{2}b$, as you had.

Then, $$E_1=\operatorname{span}\left\{\begin{bmatrix}1+i\\2\end{bmatrix}\right\}.$$

The same process can then be taked for the other eigenpair.

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Thanks! I don't know why I thought the matrix wouldn't work heh – user1520427 Oct 1 '12 at 3:47

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