# An example of a $2\times2$ matrix $A$ without zero entries and with eigenvalues $\lambda_{1}=3,\lambda_{2}=-4$

I am trying to do an exercise that asks :

Find an example of a $2\times2$ matrix $A$ without zero entries and with eigenvalues $\lambda_{1}=3,\lambda_{2}=-4$

I am having trouble thinking of a way to find such a matrix, of course $\mathrm{diag}(3,-4)$ have $\lambda_{1}=3,\lambda_{2}=-4$ as eigenvalues but I can't figure if I can use it to get to a matrix without zero entries...

Help is appreciated!

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Hint. Start with the diagonal matrix and think of it as corresponding to the basis $\{(1,0),(0,1)\}$. Then conjugate by an invertible matrix (and think of it as a change of basis) in a way that gives a result with no zero components. For instance, what happens if you change to the basis $\{(1,1), (1,-1)\}$? Will the image of either basis vector have a $0$ coordinate?

Alternate method. Find $a$, $b$, and $c$, nonzero, such that $ac-b^2 = \lambda_1\lambda_2$ and $a+c = \lambda_1+\lambda_2$. Then $$\left(\begin{array}{cc}a&b\\b&c\end{array}\right)$$ will do. (Why?)

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First method is better, though. –  Arturo Magidin Jun 25 '12 at 21:23
I don't understand how did you choose the vectors for the base in the first method, can you please add explanation ? why can we pick two entries to be the same ? (calculation will show it by is there a more theoretical reason ?) –  Belgi Jun 25 '12 at 21:31
@Belgi: I chose those vectors in the first one because their images are $(\lambda_1,\lambda_2)$ and $(\lambda_1,-\lambda_2)$. Expressing vectors in terms of $(1,1)$ and $(1,-1)$ is also very easy: $(r,s) = \frac{r+s}{2}(1,1) + \frac{r-s}{2}(1,-1)$, so the only way that you could get a $0$ coordinate would be if $a=\pm b$, which is not the case here. As for the second, because it will make computations easier, and being a symmetric matrix we know it will be diagonalizable hence similar to the diagonal matrix provided it has the correct eigenvalues. –  Arturo Magidin Jun 25 '12 at 21:35
@Belgi: in fact, diagonalizability is not an issue here because distinct eigenvalues guarantee it; so it's really just a matter of convenience. We would want different entries for full generality, of course. –  Arturo Magidin Jun 25 '12 at 21:39
PS: Method one is the standard way of writing an exam problem: start with a diagonal matrix, conjugate by a matrix corresponding to some nice "basis", and ask students to diagonlize the resulting matrix. –  Arturo Magidin Jun 25 '12 at 21:40

Here are two ways to solve the problem:

First, if you make $P$ the matrix of eigenvectors and $D$ the diagonal matrix with the eigenvalues, we have $$A = PDP^{-1}$$

You have eigenvalues, but you can pick whatever eigenvectors you want. So, let the eigenvectors be $[1,1]$ and $[3,4]$ so that $P$ is $\begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix}$. Now see if $A=PDP^{-1}$ has any zero entries. If it does, try different eigenvectors.

A dirtier way can be had by realizing that the determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues. Thus, you want your matrix to have determinant -12 and trace -1.

If we write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we want $ad-bc=-12$ and $a+d=-1$. So, let's take $a = 1$. Then $d=-2$ from the trace formula. Plugging these into the determinant formula yields $-2-bc=-12$, so $bc=10$. Take $b=2$ and $c=5$ for one possible solution.

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