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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.

Martin Sleziak
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Bubblez775
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3 Answers3

6

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

Amzoti
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4

Here are the maple commands

with(LinearAlgebra):

A := <<6,9,15>|<-5,-10,-21>|<2,5,11>>;

NS := NullSpace(A);

ES := Eigenvectors(A);

Mhenni Benghorbal
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3

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 $\mathbb R^3$ $$\langle 0^*\rangle$$

Mikasa
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