# Cross-product technique to find the eigenspaces of a 3x3 matrix

1) For each distinct real eigenvalue $\lambda$ of a 3x3 matrix $A$, it turns out that the cross product of the transpose of any two linearly independent rows of $A-\lambda I$ gives a corresponding eigenvector (and thus easily the corresponding eigenspace, since in this case the eigenspace is an eigenline). But why does this method work?

2) I think the above may be generalisable to any 3x3 matrix with only real eigenvalues: Substitute an eigenvalue of $A$ into $A-\lambda I$. Then take the cross products of the transpose of any two pairs of rows of $A-\lambda I$. Only two possibilities exist: (a) If only one is nonzero, that gives a corresponding eigenvector and hence easily the eigenspace. (b) If both are zero, then the eigenspace is the plane orthogonal to any row of $A-\lambda I$. Is this generalisation valid, and if so, why does the method work?

3) How about for the final case whereby 2 complex eigenvalues exist??

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Have you tried looking at the case where $\mathbf A$ is a triangular matrix? –  Ｊ. Ｍ. Aug 4 '12 at 16:46
@J.M. Yes, it works too (I just considered the diagonal matrix with entries 1,2,3). –  Ryan Aug 4 '12 at 17:00
I was hinting you to look at the case of a general triangular matrix, since a matrix with distinct real eigenvalues is similar to a triangular matrix... –  Ｊ. Ｍ. Aug 4 '12 at 17:01

If $x$ is an eigen vector with corresponding eigen value $\lambda$, then $(A - \lambda I)x = 0$, and so $x$ lies in the null space of $A - \lambda I$. Since the null space is perpendicular to the subspace spanned by any two rows of $A - \lambda I$, the cross product will give you this vector.

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Thank you! Corrections for precision: not "any two rows of A−λI", but any two linearly independent rows of A−λI. Also the eigenvalue needs to be distinct or else its eigenspace will be a plane, in which case our cross-product technique will fail. –  Ryan Aug 6 '12 at 8:55
So, to answer my own Question 1 (guided by your hint): the two essential points as to why the technique works are: (i) $λ$ has multiplicity 1, and (ii) $Nul(A−λi)$ is orthogonal to $Row(A−λi)$. –  Ryan Aug 6 '12 at 8:55

It doesn't seem correct to me:

Take $$A= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & 5 & 2 \end{bmatrix} = \begin{bmatrix} a_1^T \\ a_2^T \\ a_3^T \end{bmatrix}.$$ $A$ has real eigenvalues $\{-2,1,3\}$. $a_1 \times a_2 = (1, 0, 0)^T$, but $A (a_1 \times a_2) = (0,0,-6)^T$, which is clearly not an eigenvector.

It is not a left eigenvector either, $(a_1 \times a_2)^T A = (0,1,0)^T$.

Here is the answer to the modified question:

Let $B = A-\lambda I = \begin{bmatrix} b_1^T \\ b_2^T \\ b_3^T \end{bmatrix}$, where $\lambda$ is an eigenvalue of $A$ and suppose $b_1,b_2$ are linearly independent. Since $\det B =0$, we have $b_3 \in \mathbb{sp} \{b_1,b_2\}$. The vector $b_1 \times b_2$ is orthogonal to $b_1,b_2$, and hence $b_3$ since it is in $\mathbb{sp} \{b_1,b_2\}$. Consequently $B (b_1 \times b_2) = 0$, or equivalently, $(A-\lambda I)(b_1 \times b_2) = 0$, from which the answer follows.

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Sorry I made a silly mistake. I meant to say to take the two LI rows from $A-\lambda I$ instead. Please refer to the edited question. –  Ryan Aug 4 '12 at 16:39
Hey thanks for your answer. I'm almost getting it-- can you elaborate on the "consequently" bit in your answer? I mean, how does B(b1×b2)=0 follow from b1×b2 being orthogonal to b3? Sorry for being daft. –  Ryan Aug 4 '12 at 17:41
$B (b_1 \times b_2) = \begin{bmatrix} b_1^T (b_1 \times b_2) \\ b_2^T (b_1 \times b_2) \\ b_3^T (b_1 \times b_2)\end{bmatrix}$. The properties of the cross product gives $b_1^T (b_1 \times b_2) = b_2^T (b_1 \times b_2) = 0$, and since $b_3 = \mathbb{sp} \{b_1, b_2 \}$, you can write $b_3 = \alpha b_1 + \beta b_2$, so $b_3^T (b_1 \times b_2) = (\alpha b_1 + \beta b_2)^T (b_1 \times b_2) = \alpha b_1^T (b_1 \times b_2) + \beta b_2^T (b_1 \times b_2) = 0$. –  copper.hat Aug 4 '12 at 17:44
Cheers! Btw, is cross-product meaningful when there are complex entries in the column vectors? –  Ryan Aug 4 '12 at 17:57
Yes, it is still meaningful. –  copper.hat Aug 4 '12 at 18:59
Now that you have corrected your question, it is a special case of the standard relation for the adjoint matrix, sometimes called the adjunct matrix or other names. Anyway, beginning with some $n$ by $n$ matrix $B,$ we calculate certain $n-1$ by $n-1$ subdeterminants called cofactors and throw in a transpose and some judicious $\pm 1$ factors to create a matrix $\mbox{adj} \; B$ with the property $$B \;\cdot \mbox{adj} \; B = \mbox{adj} \; B \cdot B = (\det B) \cdot I.$$ If we insert your $$B = A - \lambda I$$ where $\lambda$ is an eigenvalue of $A,$ we have $\det B = 0.$ Your construction with the cross product amounts to taking any column of $$\mbox{adj} \, (A - \lambda I),$$ since $$(A - \lambda I) \cdot \mbox{adj} \, (A - \lambda I) = 0.$$
EDIT: note that the field containing the entries of $A$ and/or the eigenvalues does not matter much. Furthermore the relation to cross product is the well-known description of the cross product as cofactors, that is three 2 by 2 determinants.
Sorry I made a silly mistake. I meant to say to take the two LI rows from $A−λI$ instead. Please refer to the edited question. –  Ryan Aug 4 '12 at 16:40