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We have to find the eigenvalues and eigenvectors, and the dimension of eigenspace for this matrix: $$\begin{bmatrix} 5 & -6 & -6\\ -1&4&2\\ 3 & -6 & -4\\ \end{bmatrix}$$ The eigenvalues are found to be 1, 2 and 2. The eigenvectors for $$\displaystyle \lambda=2$$ will satisfy $$A ~X = 2 X$$ Where $$\displaystyle X= (x_1, x_2, x_3)$$ We will get three simultaneous equations, and all of them can be reduced (individually) to $$x_1 = 2 (x_2 +x_3)$$ How can proceed from here? All I can think of is that Eigenspace of this eigenvalue will consist of all vectors of form $$\displaystyle t (y+z, y, z)$$, that is $$\displaystyle x_2$$ and $$\displaystyle x_3$$ are independent, thus making the dimension =2. But the answer is, eigenspace = $$\displaystyle t (2,2,-1)$$ and dimension =1.

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Adesh said: Let $$\displaystyle x_i =2, ~y_i=1$$ , that implies$$\displaystyle z= 0$$ Verification: $$\begin{bmatrix} 5 & -6 & -6\\ -1 & 4 & 2\\ 3 &-6 & -4\\ \end{bmatrix} \times \begin{bmatrix} 2\\ 1\\ 0\\ \end{bmatrix} = \begin{bmatrix} 10-6 =4\\ -2+4 = 2\\ 6-6=0\\ \end{bmatrix} = 2\begin{bmatrix} 2\\ 1\\ 0\\ \end{bmatrix}$$ Yes, it works. Click to expand... No, that is not what I meant exactly. You tried a specific example. Try it with x=2a and y=1a. Then if it works you have it!
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Adesh said: I would really like to learn what @topsquark was trying to say by constructing two orthogonal vectors. Click to expand... You have a system that has three eigenvalues, no matter that one of them is repeated. So you need to have three eigenvectors. The space with the eigenvalue of 2 is two dimensional and forms a "block diagonal" matrix: $$\displaystyle \left [ \begin{matrix} \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) & 0 \\ 0 & 1 \end{matrix} \right ]$$ The eigenvector procedure will not diagonalize the 2x2 matrix. You have to solve that separately by constructing orthogonal vectors. -Dan
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I would really like to learn what @topsquark was trying to say by constructing two orthogonal vectors.
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Jomo said: Isn’t that represent the space of all vectors which satisfy the eigenvalue 2 with respect to the given matrix? Test your answer and see if it works. Click to expand... Let $$\displaystyle x_i =2, ~y_i=1$$ , that implies$$\displaystyle z= 0$$ Verification: $$\begin{bmatrix} 5 & -6 & -6\\ -1 & 4 & 2\\ 3 &-6 & -4\\ \end{bmatrix} \times \begin{bmatrix} 2\\ 1\\ 0\\ \end{bmatrix} = \begin{bmatrix} 10-6 =4\\ -2+4 = 2\\ 6-6=0\\ \end{bmatrix} = 2\begin{bmatrix} 2\\ 1\\ 0\\ \end{bmatrix}$$ Yes, it works.
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The eigenspace corresponding to the eigenvalue $$\displaystyle \lambda =2$$ has dimension $$\displaystyle 2$$ it consists of all vectors of the form $$\displaystyle t(2y+2z,y,z)$$
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Isn’t that represent the space of all vectors which satisfy the eigenvalue 2 with respect to the given matrix? Test your answer and see if it works.
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Now, you have two degrees of freedom Click to expand... Yes. In the way you have written the equations: $$\displaystyle 2z = x-2y$$, x and y are free to take any value. So, we will get a vector looking something like $$\begin{bmatrix} x_i\\ y_i\\ x_i -2y_i \end{bmatrix}$$ That is, the third component is the difference of two times the second component from the first. I think, the form of matrix which you have written should be $$\begin{bmatrix} 2x_i\\ 2y_i\\ x - 2y\\ \end{bmatrix}$$ Where, $$\displaystyle x=2x_i$$ and $$\displaystyle y=2y_i$$, because writing the third component as $$\displaystyle x_i-2y_i$$ would mean that it takes the values $$\displaystyle x_i$$ and $$\displaystyle y_i$$ as they occur in first and secone component, and not the double of it. I want to know, why cannot we stop at $$\begin{bmatrix} x_i\\ y_i\\ x_i -2y_i \end{bmatrix}$$ Isn’t that represent the space of all vectors which satisfy the eigenvalue 2 with respect to the given matrix? I’m doing the first course in Linear Algebra, from Apostol’s Calculus Vol II.
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Adesh said: We have to find the eigenvalues and eigenvectors, and the dimension of eigenspace for this matrix: $$\begin{bmatrix} 5 & -6 & -6\\ -1&4&2\\ 3 & -6 & -4\\ \end{bmatrix}$$ The eigenvalues are found to be 1, 2 and 2. The eigenvectors for $$\displaystyle \lambda=2$$ will satisfy $$A ~X = 2 X$$ Where $$\displaystyle X= (x_1, x_2, x_3)$$ We will get three simultaneous equations, and all of them can be reduced (individually) to $$x_1 = 2 (x_2 +x_3)$$ How can proceed from here? All I can think of is that Eigenspace of this eigenvalue will consist of all vectors of form $$\displaystyle t (y+z, y, z)$$, that is $$\displaystyle x_2$$ and $$\displaystyle x_3$$ are independent, thus making the dimension =2. But the answer is, eigenspace = $$\displaystyle t (2,2,-1)$$ and dimension =1. Click to expand... I don't know how far into the Linear Algebra approach you are going for, but this is my take on the concept: $$\displaystyle \left [ \begin{matrix} 5 & -6 & -6 \\ -1&4&2\\ 3 & -6 & -4\\ \end{matrix} \right ] \left ( \begin{matrix} x \\ y \\ z \end{matrix} \right ) = 2 \left ( \begin{matrix} x \\ y \\ z \end{matrix} \right )$$ $$\displaystyle \begin{cases} 5x - 6y - 6z = 2x \\ -x + 4y + 2z = 2y \\ 3x - 6y - 4z = 2z \end{cases}$$ leading to 2z = x - 2y. Now, you have two degrees of freedom... Use this to construct two vectors of the form $$\displaystyle \left ( \begin{matrix} 2x_i \\ 2y_i \\ x_i - 2y_i \end{matrix} \right )$$ that are orthogonal, ie. $$\displaystyle \left ( \begin{matrix} 2x_1 & 2y_1 & x_1 - 2y_1 \end{matrix} \right ) \left ( \begin{matrix} 2x_2 \\ 2y_2 \\ x_2 - 2y_2 \end{matrix} \right )= 0$$ -Dan
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