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I have the following matrix:

$$\left[\begin{matrix}-4 & 20 & -33 \\ 0 & 1 & 12 \\ 0 & 0 & 5\end{matrix}\right]$$

Since it is a triangular matrix, we have the eigenvalues: $-4,1,5$

Then, when I try to get the eigenvector for the eigenvalue $-4$.

A-In: $$\left[\begin{matrix}0 & -20 & 33 \\ 0 & -5 & -12 \\ 0 & 0 & -9\end{matrix}\right]$$

Then I do its RREF: $$\left[\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{matrix}\right]$$

After this step, I get lost. Wouldn't the resulting vector be $\left[\begin{matrix}1 \\ 1 \\ 0\end{matrix}\right]$?

Thanks for your help in advance!

Patrick Abraham
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idelara
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2 Answers2

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After the rref you get the system: $$\left(\begin{array}{ccc|c} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) \sim \cases{0x+y+0z = 0 \\ 0x+0y+z = 0 \\ 0x+0y+0z = 0 } \Leftrightarrow \cases{y = 0 \\ z = 0 }$$ Choose $x=t$; then $(t,0,0)$ is an eigenvector for all $t \ne 0$; e.g. pick $(1,0,0)$.

StackTD
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  • Why is x being parametrized? Sorry, this is the step where I get lost, do you know of any tutorials that show how to do this? – idelara May 12 '16 at 12:04
  • Because the last row (and the corresponding equation) disappears, you then have a linear system of two equations in three unknowns. This means you'll be able to choose one variable freely, this is usually done by introducing a parameter. Since you cannot choose $y$ and $z$ in this case (they are both 0), you choose $x$. – StackTD May 12 '16 at 12:05
  • @JackGal Do you know how to use a rref to construct a general solution to a system of equations? That is all that is being done here. When you do that, you solve for the pivot variables in terms of the free variables and then let the free variables be arbitrary parameters. – Ian May 12 '16 at 12:05
  • yep... sorry... somehow I thought that y and z were equal to 1... which is definitely wrong.. I am going to compute for lambda = 1, lets see if I can get it right – idelara May 12 '16 at 12:12
  • I am going to edit the question, if that is ok, or should I create a new question?? – idelara May 12 '16 at 12:15
  • I get for lambda = 1 {{1,-4,0},{0,0,1},{0,0,0}}. After I have that x=4y & z=0, and I get stuck here... is it {{x,y,z}} = {b{4,1,0}}? – idelara May 12 '16 at 12:17
  • Good! Since $z=0$ and $x=4y$ has to be satisfied, you can take $y=t$ and then eigenvectors are of the form $(4t,t,0)$ for every non-zero $t$; e.g. pick $(4,1,0)$. – StackTD May 12 '16 at 12:54
  • If you aren't sure, if it is an eigenvector use $A v = \lambda v$ where $v$ is your $eigenvector$ and $\lambda$ your corresponding $eigenvalue$. – Patrick Abraham May 12 '16 at 13:31
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See here for how to read a basis for the kernel of a matrix from its rref. In this case, the first column is the only one without a pivot, so the referenced procedure tells you that eigenvectors corresponding to $-4$ will be of the form $(x,0,0)^T$.

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