Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$1)$ For each distinct real eigenvalue $\lambda$ of a $3 \times 3$ 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 $3\times 3$ 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??

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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.

share|cite|improve this answer

This is wrong. The cross product of two rows is orthogonal to those two rows, and can thus only be an eigenvector if its corresponding two components are zero, and they need not be.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.