How to compute Nullspace on maple? 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.
 A: 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
A: Here are the maple commands
with(LinearAlgebra):
A := <<6,9,15>|<-5,-10,-21>|<2,5,11>>;
NS := NullSpace(A);
ES := Eigenvectors(A);
A: You know that, the solution sets of homogeneous linear systems $Ax=0$ provide an important source of vector spaces called Nullspace. Here, $det(A)\neq0$ so the above system has only one triple in $\mathbb R^3$ as its solution $$0^*=(0,0,0)$$ so the Nullspace is a trivial vector subspace of $\mathbb R^3$ $$\langle 0^*\rangle$$
