# Show that $W$ is a 2 dimensional subspace or $\mathbb{R}^{3}$ and find a basis for it.

Question:

Let $$A = \begin{bmatrix} 0 & 1 & 4\\ 0 & 1 & -8\\ -9 & 3 & 15 \end{bmatrix}, W = \left \{ (x,y,z)\in \mathbb{R}^{3}:A\begin{bmatrix} x & y & z \end{bmatrix}^{T}=3\begin{bmatrix} x & y & z \end{bmatrix}^{T} \right \}$$

Show that $W$ is a 2 dimensional subspace or $\mathbb{R}^{3}$ and find a basis for it.

I don't really know what exactly the question is asking.

Attempt:

I reduced $A$ to $A = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$

$$\therefore \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=3\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

I don't know what to do after this.

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You're given a set $W$. First of all can you prove $W$ is a vector space? – Git Gud May 1 '14 at 11:57

Hint

Notice that $$W=\ker (A-3I_3)$$ so $W$ is a subspace of $\Bbb R^3$ and to find the dimension there are two ways:

• by solving the system of equations given by $$(A-3I_3)(x,y,z)^T=0$$
• or by calculating the characteristic polynomial of $A$ and seeing the eventual multiplicity of $3$.
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Nice hint and direction! – amWhy May 3 '14 at 11:41
Thanks for the edit! – user63181 May 3 '14 at 11:43
It was just a tiny "tweak" to a already fine answer! ;-) – amWhy May 3 '14 at 11:45

The condition defining $W$ indicates that $3$ is an eigenvalue of $A$, so the question is actually asking you to find the eigenspace associated with this eigenvalue. Solve the homogenous system: $A-3I=0$ to find the eigenvectors associated with $\lambda=3$...

ok, I checked and it seems $A$ has eigenvalues $3,4$ and $9$...so I cannot see how $W$ can have dimension 2.

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Eigenvectors for $9$ are not in $W$. – Santiago Canez May 1 '14 at 12:58
@SantiagoCanez Excuse me. you are right... – Christiaan Hattingh May 1 '14 at 13:23
but then $W$ does not have dimension 2, so seems as if there is an error in the question. – Christiaan Hattingh May 1 '14 at 13:26