TITLE: How to compute Nullspace on maple?
QUESTION [5 upvotes]: I have the matrix 
$$A := \begin{bmatrix}6& 9& 15\\-5& -10& -21\\ 2& 5& 11\end{bmatrix}.$$ Can anyone please tell me how to both find the eigenspaces by hand and also by using the Nullspace command on maple? Thanks.

REPLY [6 votes]: Given the matrix 
$$A = \left(\begin{matrix}6& 9& 15\\-5& -10& -21\\ 2& 5& 11\end{matrix}\right).$$
Find the Eigensystem by hand.
First, lets find the eigenvalues by solving  $det(A - \lambda I) = 0$, so we have:
$$det(A - \lambda I) = \left|\begin{matrix}6 - \lambda & 9& 15\\-5& -10 - \lambda & -21\\ 2& 5& 11 - \lambda\end{matrix}\right| = 0.$$
This gives us the characteristic polynomial: 
$$-\lambda^3 + 7\lambda^2 - 16\lambda + 12 = 0$$
From this we get two eigenvalues (one is repeated) as: $\lambda_1 = 3, ~ \lambda_{2,3} = 2$
Next, we want to find the eigenvector for the eigenvalue $\lambda_1$, by solving the equation $(A - \lambda_1) v_1 = 0$.
$(A-\lambda_1)v_1 = (A-3)v_1 = \left(\begin{matrix}3 & 9& 15\\-5& -13 & -21\\ 2& 5& 8\end{matrix}\right)v_1 = 0.$
Using the row-reduced-echelon-form, this leads to $v_1 = (1,-2,1).$
Next, we want to find the eigenvector for the eigenvalue $\lambda_2$, by solving the equation $(A - \lambda_2) v_2 = 0.$
$(A-\lambda_2)v_2 = (A-2)v_2 = \left(\begin{matrix}4 & 9& 15\\-5& -12 & -21\\ 2& 5& 9\end{matrix}\right)v_2 = 0.$
Using the row-reduced-echelon-form, this leads to $v_2 = (3,-3,1).$
Since we have a repeated eigenvalue, care needs to be taken using algebraic and geometric multiplicities (know what those are), if the matrix is diagonalizable (you can work these details).

The algebraic multiplicity of an eigenvalue is the number of times it is a root.
The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors for the eigenvalue.

To find the eigenvector for $\lambda_3$, we will solve $(A - \lambda_3 I)v_3 = v_2$ (you must have learned why this is in class), so we have (shown in augmented form):
$\left(\begin{array}{@{}ccc|c@{}}
4 & 9& 15 & 3\\-5& -13 & -21 & -3\\ 2& 5& 8 & 1 
\end{array}
\right)v_3 = 0$
Using RREF, this results in $v_3 = (3, -1, 0)$.
Thus, we have:
$$\lambda_1 = 3, v_1 = (1, -2, 1)$$
$$\lambda_2 = 2, v_2 = (3, -3, 1)$$
$$\lambda_2 = 2, v_3 = (3,-1, 0)$$
Do you know how to use the information above to write the diagonal form, otherwise known as the Jordan Normal Form?
$$A = P J P^{-1} = \begin{bmatrix} 3 & 3 & 1 \\ -3 & -1 & -2 \\ 1 & 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} -1 & -3 & -5 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{bmatrix}$$
Babak Sorouh already told you the nullspace and Mhenni Benghorbal showed the Maple commands, so no need to repeat that.
Regards