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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.2012-03-15
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    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.2012-03-15
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    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 $1$...2012-03-15
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    @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.2012-03-15
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    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 $x$.2012-03-15
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    OK. Thanks. That makes sense now.2012-03-15

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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]$2012-03-15
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    @Dylan I updated my answer.2012-03-15
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    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.$2012-03-15
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    this I understand. However, I incorrectly entered the last element in the matrix $a_{33}$. It should be -2. What I am getting hung up on is when there is only one non-zero row; the vector should be [1,2,-1] and for some reason I am having trouble solving this.2012-03-15
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    $\left(\begin{matrix} 1 & 2 & -1\\0 & 0 & 0\\0 & 0 & 0\end{matrix}\right)$ 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.$2012-03-15
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    So every eigenspace vector is of the form $$ \begin{pmatrix} -2t + u \\ t \\ u \end{pmatrix} = \begin{pmatrix} -2t \\ t \\ 0 \end{pmatrix} + \begin{pmatrix} u \\ 0 \\ u \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}u. $$2012-03-15
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    In other words, The eigenspace is spanned by linear combination of $$ \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} $$ so these are the basis of eignspace corresponding to that eigenvalue.2012-03-15
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    @Dylan I added comments above.2012-03-15
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    Thank you! This is what I was stuck on - putting it in that eigenspace vector form you mentioned above.2012-03-15
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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_1$$$$3x_1+4x_2-3x_3=-2x_2$$$$0x_1+0x_2-2x_3=-2x_3$$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?2012-03-15
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    Oh, I am sorry. I just noticed I entered the last value in the A matrix incorrectly. Fixing it now.2012-03-15
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    :/ Alright Im going to do it again then.2012-03-15
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    really sorry, no worries if you don't feel like it. my bad.2012-03-15