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