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I having a few issues finding the eigenvectors for the following matrix:

$$ \begin{bmatrix} -1 & -1\\ 0 & -2 \\ \end{bmatrix}$$

I calculated the eigenvalues to be $\lambda = -1, -2 $

And I try to solve for the eigenvector corresponding to $-1$ as follows:

$$\begin{bmatrix} -1--1 & -1\\ 0 & -2--1 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

Reducing the $2 \times 2$ matrix gives $\begin{bmatrix}0 & 1\\0 & 0\end{bmatrix}$ and my instinct says that the eigenvector should be $\begin{bmatrix}0\\1\end{bmatrix}$ however other online calculators give the answer to be $\begin{bmatrix}1\\0\end{bmatrix}$

Why is this?

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  • $\begingroup$ What is the null space for the matrix $\begin{bmatrix}0&1\\0&0\end{bmatrix}$? (hint: augment the matrix with a column of zeros) $\endgroup$ – Morgan Rodgers Aug 30 '15 at 6:54
  • $\begingroup$ The bottom right entry should be $(-2) - (-1) = -1$. The matrix should be $\begin{pmatrix} 0 & -1 \\ 0 & -1 \end{pmatrix}$. $\endgroup$ – Empiricist Aug 30 '15 at 7:11
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You started with a system of linear equations. In matrix form this can be written as: $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & -1 & 0 \end{array}\right)$$ from which you have gotten to $$\left(\begin{array}{cc|c} 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right).$$ As we have a complete row of $0$'s, there might be infinitely many solutions (which is a good thing because otherwise you'd have made a mistake somewhere). The second row now reads written as an equation $0=0$ which is obviously true but doesn't give us any information about $x$ or $y$, so we remove it from our matrix.

You now have to get to the (strict) row echelon form. To get there we need the leading coefficient in the first row to be $1$: $$\left(\begin{array}{cc|c} 1 & 0 & t \\ 0 & -1 & 0 \end{array}\right)$$ where $t$ is an arbitrary real number. We can multiply the second row with $(-1$ which yields $$\left(\begin{array}{cc|c} 1 & 0 & t \\ 0 & 1 & 0 \end{array}\right).$$ We now have the strict row echelon form; the left part is the identity matrix which means that the right part is the finished solution. Every vector of the form $\displaystyle \begin{pmatrix} t \\ 0 \end{pmatrix}$ is a solution to your system of linear equations. Thus $\langle\begin{pmatrix} 1 \\ 0 \end{pmatrix}\rangle=E_{A}(-1)$ or in other words: $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is a basis for the eigenspace of $A$ associated to $\lambda_1=-1$.


One example for the strict echelon form (using more rows for better visualization): assume that somehow we've arrived at $$\left(\begin{array}{cccc|c}{-2}&4&2&3 &0 \\ 2& 2&3&1&0 \\ 0&0& 2&5&0 \\ 0&0&2 & 5&0\end{array}\right).$$ Via row elimination this can be further reduced to $$\left(\begin{array}{cccc|c}{-2}&4&2&3 &0 \\ 0& 6&5&4&0 \\ 0&0& 2&5&0 \\ 0&0&0 & 0&0\end{array}\right).$$ This means we already have steps in the first, second and third row but are missing one step in the last row: $$\left(\begin{array}{cccc|c}\underline{-2}&4&2&3 &0 \\ 0&| \underline{6}&5&4&0 \\ 0&0&|\underline{2}&5&0 \\ 0&0&0 & 0&0\end{array}\right).$$ To this last step, we again remove the row with only $0$'s and instead write $\left(\begin{array}{cccc|c}0 & 0 & 0 & 1 & t\end{array}\right)$. This will give us the missing step in the last row: $$\left(\begin{array}{cccc|c}\underline{-2}&4&2&3 &0 \\ 0&| \underline{6}&5&4&0 \\ 0&0&|\underline{2}&5&0 \\ 0&0&0 & |\underline{1}&t\end{array}\right).$$ Now we can work our way back up until we get the identity matrix on the left side, e.g. my next move would be to get $$\left(\begin{array}{cccc|c}\underline{-2}&4&2&0 &-3t \\ 0&| \underline{6}&5&0&-4t \\ 0&0&|\underline{2}&0&-5t \\ 0&0&0 & |\underline{1}&t\end{array}\right).$$

If you're missing more than one step, we do the exact same thing twice, e.g. in $$\left(\begin{array}{cccc|c} 2 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0\end{array}\right)$$ there are two steps missing in the second and third row. This will give us $$\left(\begin{array}{cccc|c} 2 & 0 & 1 & 2 & 0 \\ 0 & 1 & 0 & 0 & t \\ 0 & 0 & 1 & 0 & s \\ 0 & 0 & 0 & 1 & 0\end{array}\right)$$ and from here we can work up again: $$\left(\begin{array}{cccc|c} 2 & 0 & 0 & 0 & -s \\ 0 & 1 & 0 & 0 & t \\ 0 & 0 & 1 & 0 & s \\ 0 & 0 & 0 & 1 & 0\end{array}\right).$$ If we now multiply the first row with $\dfrac{1}{2}$ we get the identity matrix on the left side: $$\left(\begin{array}{cccc|c} 1 & 0 & 0 & 0 & -\frac{s}{2} \\ 0 & 1 & 0 & 0 & t \\ 0 & 0 & 1 & 0 & s \\ 0 & 0 & 0 & 1 & 0\end{array}\right).$$

Assuming that this matrix originated from an eigenvalue problem we'd get two linearly independent eigenvectors from $$\begin{pmatrix}-\frac{s}{2} \\ t \\ s \\ 0 \end{pmatrix}=\begin{pmatrix}-\frac{s}{2} \\ 0 \\ s \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ t \\ 0 \\ 0 \end{pmatrix} = s\cdot \begin{pmatrix} -\frac{1}{2} \\ 0 \\ 1 \\ 0 \end{pmatrix} + t\cdot \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}$$ and therefor we'd get for the eigenspace: $E_A=\langle \begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\0 \\ 0\end{pmatrix}\rangle$ (notice that I just multiplied every entry in the first vector with $2$ to get nicer values).

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  • $\begingroup$ What steps did you take to get to the "strict" row-echelon form? Did you just plop $1$ into the first row and $t$ into the third row, or is there a formulaic method to get to it? $\endgroup$ – PythonNewb Aug 30 '15 at 7:30
  • $\begingroup$ Basically that's what I did, yes. I like to think of it in a way of "stairs" (might come from the german term "strikte Zeilenstufenform" which roughly translates to "stric row-step-form"); every row resembles one step and as with real stairs, I don't like missing a step. The maths behind this is that we have more variables than equations, thus we're missing information. I'll edit a matrix with the "steps" into my answer to highlight the process. $\endgroup$ – Hirshy Aug 30 '15 at 8:20
  • $\begingroup$ @PythonNewb see my edit in my answer. I hope this clears things up. If you have more questions, feel free to ask! $\endgroup$ – Hirshy Aug 30 '15 at 8:43
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From your equations, you have $y=0$ and $x$ does not feature in the equations, so can be any value.

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  • $\begingroup$ May you please elaborate further? $\endgroup$ – PythonNewb Aug 30 '15 at 6:59
  • $\begingroup$ Both lines say $0x-1y=0$ $\endgroup$ – David Quinn Aug 30 '15 at 7:11

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