# a question about calculating eigenvectors

It has to be the silliest question, but it’s not clear to me how to calculate eigenvectors quickly. I am just talking about a very simple 2-by-2 matrix.

When I have already calculated the eigenvalues from a characteristic polynomial, I can start to solve the equations with $A\mathbf{v}_1 = e_1\mathbf{v}_1$ and A\mathbf{v}_2 = e_2\mathbf{v}_2$, but in this case it always requires writing lines of equations and solving them. On the other hand I figured out that just by looking at the matrix you can come up with the eigenvectors very quickly. But I'm a bit confused in this part. When you have the matrix with subtracted$e_1$values like this: $$\left(\begin{array}{cc} A&B\\ C&D \end{array}\right).$$ Then for me, it always worked to use the eigenvector. $$\left(\begin{array}{r}-B\\A\end{array}\right)$$ But in some guides I find that they are using A C as an eigenvector. $$\left(\begin{array}{c}A\\C \end{array}\right).$$ And when I check it, they are indeed multiples of each other. But this other method is not clear to me, how could$A$and$C$mean anything about the eigenvector, when both of them are connected to$x$, without having to do anything with$y$. But it’s still working. Was it just a coincidence? So is the recommended method for calculating them is just to subtract the eigenvalues from the matrix and look at $$\left(\begin{array}{r} -B\\A\end{array}\right)\qquad\text{or}\qquad\left(\begin{array}{r} -D\\A \end{array}\right).$$ - add comment ## 2 Answers (Note that you are using$A$for two things in your post: it is the original matrix, and then it's an entry of the matrix; that's very bad form. and likely to lead to confusion; never use the same symbol to represent two different things). So, if I understand you: you start with a matrix$\mathscr{A}$, $$\mathscr{A} = \left(\begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22} \end{array}\right).$$ Then, if you know that$e_1$is an eigenvalue, then you look at the matrix you get when you subtract$e_1$from the diagonal: $$\left(\begin{array}{cc} a_{11}-e_1 & a_{12}\\ a_{21} & a_{22}-e_1 \end{array}\right) = \left(\begin{array}{cc}A&B\\C&D \end{array}\right).$$ Now, the key thing to remember is that, because$e_1$is an eigenvalue, that means that the matrix is singular: an eigenvector corresponding to$e_1$will necessarily map to$\mathbf{0}$. That means that the determinant of this matrix is equal to$0$, so$AD-BC=0$. Essentially: one of the rows of the matrix is a multiple of the other; one of the columns is a multiple of the other. What this means is that the vector$\left(\begin{array}{r}-B\\A\end{array}\right)$is mapped to$0$: because $$\left(\begin{array}{cc} A&B\\C&D\end{array}\right)\left(\begin{array}{r}-B\\A\end{array}\right) = \left(\begin{array}{c}-AB+AB\\-BC+AD \end{array}\right) = \left(\begin{array}{c}0\\0\end{array}\right)$$ because$AD-BC=0$. If$A$and$B$are not both zero, then this gives you an eigenvector. If both$A$and$B$are zero, though, this method does not work because it gives you the zero vector. In that case, the matrix you are looking at is $$\left(\begin{array}{cc}0&0\\C&D \end{array}\right).$$ One vector that is mapped to zero is$\left(\begin{array}{r}-D\\C\end{array}\right)$; that gives you an eigenvalue unless$C$and$D$are both zero as well, in which case any vector will do. On the other hand, what about the vector$\left(\begin{array}{r}A\\C\end{array}\right)$? That vector is mapped to a multiple of itself by the matrix: $$\left(\begin{array}{cc} A&B\\C&D\end{array}\right)\left(\begin{array}{c}A\\C\end{array}\right) = \left(\begin{array}{c}A^2 + BC\\AC+DC\end{array}\right) = \left(\begin{array}{c}A^2+AD\\AC+DC\end{array}\right) = (A+D)\left(\begin{array}{c}A\\C\end{array}\right).$$ However, you are looking for a vector that is mapped to$\left(\begin{array}{c}0\\0\end{array}\right)$by$\left(\begin{array}{cc}A&B\\C&D\end{array}\right)$, not for a vector that is mapped to a multiple of itself by this matrix. Now, if your original matrix has zero determinant already, and you haven't subtracted the eigenvalue from the diagonal, then here's why this will work: the sum of the two eigenvalues of the matrix equals the trace, and the product of the two eigenvalues equals the determinant. Since the determinant is$0$under this extra assumption, then one of the eigenvalues is$0$, so the other eigenvalue equals the trace,$a_{11}+a_{22}$. In this case, the vector$\left(\begin{array}{c}A\\C\end{array}\right)$is an eigenvector of$\mathscr{A}$corresponding to$a_{11}+a_{22}$, unless$A=C=0$(in which case$\left(\begin{array}{r}a_{22}\\a_{11}\end{array}\right)$is an eigenvector unless$\mathscr{A}$is the zero matrix). Added. As Robert Israel points out, though, there is another point here. Remember that$e_1+e_2 = a_{11}+a_{22}$,$A=a_{11}-e_1 = e_2-a_{22}$; and that$a_{11}a_{22}-a_{12}a_{21} = e_1e_2$; if we take the vector$\left(\begin{array}{c}A\\C\end{array}\right)$with the original matrix$\mathscr{A}, we have: \begin{align*} \left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)\left(\begin{array}{c}A\\C\end{array}\right) &= \left(\begin{array}{c} a_{11}A + a_{12}C\\a_{21}A + a_{22}C.\end{array}\right)\\ &= \left(\begin{array}{c} a_{11}(e_2-a_{22}) + a_{12}a_{21}\\ a_{21}(e_2-a_{22}) + a_{22}a_{21} \end{array}\right) = \left(\begin{array}{c} e_2a_{11} + (a_{12}a_{21}-a_{11}a_{22})\\ e_2a_{21} + (a_{22}a_{21} - a_{22}a_{21} \end{array}\right)\\ &= \left(\begin{array}{c} e_2a_{11} - e_1e_2\\ e_2a_{21} \end{array}\right) = \left(\begin{array}{c} e_2(a_{11}-e_1)\\e_2a_{21} \end{array}\right)\\ &= e_2\left(\begin{array}{c} a_{11}-e_1\\a_{21} \end{array}\right) = e_2\left(\begin{array}{c}A\\C \end{array}\right). \end{align*} So ifA$and$C$are not both zero, then$\left(\begin{array}{c}A\\C\end{array}\right)$is an eigenvector for the other eigenvalue of$\mathscr{A}$. To summarize: 1. If you subtract the eigenvalue from the diagonal, and in the resulting matrix the first row is not equal to$0$, then your method will produce an eigenvector corresponding to the eigenvalue you subtracted. 2. If you subtract the eigenvalue from the diagonal, and in the resulting matrix the first column is not equal to$0$, then taking that column will produce an eigenvector corresponding to the other eigenvalue of$\mathscr{A}$(not the one you subtracted). - The vector$\left( \matrix{A\cr C\cr} \right)$is an eigenvector (of the original matrix) for the other eigenvalue (assuming$A$and$C$are not both 0). – Robert Israel May 13 '11 at 19:52 @Robert: Aha. That makes sense. Thanks. – Arturo Magidin May 13 '11 at 20:01 Thank you for the very clear and detailed answer! – zsero May 13 '11 at 22:41 add comment Please goes through this list one by one. 1. Your method will work as long as either$A$or$B$is not zero. 2. When doesn't it work? Try calculate the eigenvector of$\left(\begin{smallmatrix} 1 & 0 \\ 1 & 1 \end{smallmatrix} \right )$. 3. Does the method of taking$\left(\begin{smallmatrix} A\\C \end{smallmatrix} \right )$work? Try calculate the eigenvector of$\left(\begin{smallmatrix} 3 & 3 \\ 4 & 7 \end{smallmatrix} \right )$and see it for yourself. 4. What is an obvious solution to the equation$Ax+By=0$? Hint: it has something to do with your method. 5. Say when either$A$or$B$is nonzero, what will happen to the second row when you do row reduction? Why? Remember, the determinant of$\left(\begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \right )\$ is zero.
6. Can you see why your method works by now?
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