# Determining the eigenvectors from a reduced-row echelon matrix

I am given the following matrix:

$$A = P\left(\begin{matrix} -5 & -6 & 3\\3 & 4 & -3\\0 & 0 & -2\end{matrix}\right)$$

After finding the following eigenvalues by finding the characteristic polynomial I get:

$\lambda_1 = \lambda_2 = -2$ and $\lambda_3 = 1$

After finding the I now need to find the eigenvectors for $\lambda_1$ and $\lambda_2$. After putting matrix into reduced-row echelon form for $\lambda_1$:

$$\left(\begin{matrix} 1 & 2 & -1\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$$

I now want to find the eigenvector from this, but am I bit puzzled how to find it an then find the basis for the eigenspace (I know this involves putting it into vector form, but for some reason I found the steps to translating-to-vector-form really confusing and still do).

A step-by-step explanation on this point would be very helpful for a linear algebra newbie.

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You are basically looking for a basis of the null space of this matrix. There should be an example of how to do that in your textbook/lecture notes. Earlier, when null spaces were covered. – Jyrki Lahtonen Mar 15 '12 at 5:35
Yes, I understand I need to put into the form $Ax = 0$, but I guess what I still have trouble with is putting the above into vector form. – Dylan Mar 15 '12 at 5:46
The row-reduced matrix has a single non-zero row, so you have just one non-trivial equation. If you write that equation down, you see that you can always satisfy that equation by giving the first coordinate an appropriate value (simply solve for that first coordinate). Irrespective of the values of the other two coordinates! So let one of the remaining coordinates be equal to $1$, and the rest equal to $0$. Vary the position of that $1$... – Jyrki Lahtonen Mar 15 '12 at 5:50
@jyrki if i am understanding you correctly, if the rr matrix has a single non-zero matrix, i can assign the first element to one and one of the other elements to one? it doesn't matter which? can you demonstrate this by putting it into the vector form with the s and t substitutions? Thanks. – Dylan Mar 15 '12 at 5:59
If you unknown coordinates are $x,y,z$ (or whatever you want to call them), then you are left with the equation $$1\cdot x+2\cdot y-1\cdot z=0.$$ Assign $y=1,z=0$ or $y=0,z=1$ and solve for $x$. – Jyrki Lahtonen Mar 15 '12 at 6:14

Hint: If $A v_i = \lambda_i v_i$ then $(A - \lambda_i I) v_i = 0.$ You can solve fox $v_i$ by Gaussian elimination on the augmented matrix $$\left( A-\lambda_i I \quad {\bf 0} \right).$$ For $\lambda = -2,$ we have $$A - (-2) I = \begin{pmatrix} -3 & -6 & 3\\ 3 & 6 & -3 \\ 0 & 0 & -1 \end{pmatrix}$$ And the reduced echelon form of $(A-\lambda_i I )v = 0$ is $$\begin{pmatrix} 1 & 2 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix} = 0. \tag{1}$$ This gives you a solution $z = 0, y = t, x = -2.$ i.e., the set of vectors satisfying $Av = -2v$ are of the form $t-$multiple of $$\begin{pmatrix} -2 \\ 1 \\0 \end{pmatrix} t.$$ The's the basis.

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Thanks, I understand this part, and have everything done up until the point after I found the matrix for $\lambda_1$ where I need to find the eigenvector. I am having trouble finding the values from the above reduced matrix for $E_1 = [x_1, x_2, x_3]$ – Dylan Mar 15 '12 at 5:49
@Dylan I updated my answer. – user2468 Mar 15 '12 at 5:54
In (1) the third row tells you $z = 0.$ The 2nd row tells you $0.y = 0$ so $y$ is arbitrary $= t$. The first row tells you $x+2y = 0.$ So $x =-2t.$ – user2468 Mar 15 '12 at 5:56
$\left(\begin{matrix} 1 & 2 & -1\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$ gives you only one equation: $1x + 2y -1z = 0$ So 2 of there variables are arbitrary say $y = t,z= u$. This gives you $x=-2t + u.$ – user2468 Mar 15 '12 at 6:15
So every eigenspace vector is of the form $$\begin{pmatrix} -2t + u \\ t \\ u \end{pmatrix} = \begin{pmatrix} -2t \\ t \\ 0 \end{pmatrix} + \begin{pmatrix} u \\ 0 \\ u \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}u.$$ – user2468 Mar 15 '12 at 6:17

For $\lambda=-2$, let $\bf{x}$$=(x_1,x_2,x_3) be an eigenvector corresponding to -2. Then, A\bf{x} =-2\bf{x}, and multplying the matrix A by \bf{x} you obtain the following system of equations:$$-5x_1-6x_2+3x_3=-2x_13x_1+4x_2-3x_3=-2x_20x_1+0x_2-2x_3=-2x_3$$From the last equation it is clear that x_3=0. The last equation does not provide any information, so we go ahead and look at the other equations. We note that the second and first equation are the same equation since if you multiply one by -1 you get the other. Ergo you have that$$3x_1+6x_2-3x_3=0\Rightarrow x_1+2x_2-x_3=0$$and we want to write this parametrically. We have that the degree of freedom is$2$, so you have that solutions are of the form$(-2t+v,t,v)$for any$t$and$v$. Hence, you can for instance take$t=0$and$v=1$and obtain$(1,0,1)$and then you can take$t=1$and$v=0$and get$(-2,1,0)$. This two vectors form a basis for$E_{-2}$the eigenspace corresponding to the value$-2\$. For the other eigenvalue, I will let you figure it out.

Let me know if you have any questions.

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Hmm, I thought I got this right. I checked it on Wolfram as well. For the characteristic polynomial, I ended up with lambda^2(lambda+3) = -4 .... is that not correct? – Dylan Mar 15 '12 at 5:54
Oh, I am sorry. I just noticed I entered the last value in the A matrix incorrectly. Fixing it now. – Dylan Mar 15 '12 at 6:04
:/ Alright Im going to do it again then. – Daniel Montealegre Mar 15 '12 at 6:13
really sorry, no worries if you don't feel like it. my bad. – Dylan Mar 15 '12 at 6:13