# Some properties of a $2\times 2$ matrix with repeated eigenvalues

I got a problem in my exam

Consider the matrix $A =\left( \begin{array}{cc} a & b \\ c & d\\ \end{array} \right)$ with real entries. Suppose it has repeated eigenvalues. Pick the correct statement:

• $bc = 0$
• $A$ is always a diagonal matrix
• $det(A)\geq 0$
• $\det(A)$ can take any real value.

I took up a example of matrix $A =\left( \begin{array}{cc} 0 & 1 \\ 0 & 0\\ \end{array} \right)$ it has repeated eigen values zero and is not a diagonal matrix. so one possibility is removed. If it has repetead eigen values then $\det(A)\geq 0$, since product of eigenvalues will be $\det(A)$. In any case product will be non negative. I am not sure about other two possibilities. Is there any better way to do this problem?

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What you did was fine as far as it went, but since $A$ is only $2\times 2$, you can also simply solve for the eigenvalues. If you do, you find yourself solving the quadratic equation $$(a-\lambda)(d-\lambda)-bc=\lambda^2-(a+d)\lambda+(ad-bc)=0\;,\tag{1}$$ so $$\lambda=\frac{a+d\pm\sqrt{(a+d)^2-4(ad-bc)}}2\;,$$ and the eigenvalue is repeated iff

\begin{align*} 0&=(a+d)^2-4(ad-bc)\\ &=a^2-2ad+d^2+4bc\\ &=(a-d)^2+4bc\;, \end{align*}

i.e., iff $4bc=-(a-d)^2$. This guarantees that $bc\le 0$, but clearly $bc$ need not be $0$, and therefore $A$ need not be a diagonal matrix. Finally, it's clear from $(1)$ that $\det A$ is the product of the eigenvalues (even if you didn't know this already), so it's clear that $\det A\ge 0$ and therefore cannot assume all real values.

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All you did was right. Of course, since you've already deduced that $\det(A)\geq 0$, you already know it is not the case that $\det(A)$ can be any real value: it can't be negative! But it can be any nonnegative real value: given $r\gt 0$, take the diagonal matrix with diagonal entries $\sqrt{r}$ to get a matrix with that determinant.

So we are down to whether the matrix must have $bc=0$.

Say the characteristic polynomial is $t^2+2t+1 = (t+1)^2$. Then we can take $$\left(\begin{array}{rr} 0 & -1\\ 1 & -2 \end{array}\right)$$ and note that the characteristic polynomial is precisely $-t(-2-t)+1 = t^2 + 2t+1$, exactly what we want. However $bc=-1$.

(How did I come up with that? It's the "companion matrix" of $t^2+2t+1$; but you can come up with such a matrix for any quadratic: if you have $t^2+at+b$, write $t^2+at+b = t(t+a)+b = -t(-t-a)+b$, so put a $0$ and a $-a$ on the diagonal, and have the other two entries multiply to $-b$ and you are done; now just pick a polynomial with a double root).

Or you can obtain an example by starting with a matrix that is not diagonal and has repeated eigenvalues different from $0$, say $$\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$$ and then conjugating by an appropriate invertible matrix, say

$$\left(\begin{array}{cr} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & -\frac{1}{2} \end{array}\right) \left(\begin{array}{cc} \vphantom{\frac{1}{2}}1 & 1\\ \vphantom{\frac{1}{2}}0 & 1 \end{array}\right)\left(\begin{array}{cr} \vphantom{\frac{1}{2}}1 & 1\\ \vphantom{\frac{1}{2}}1 & -1 \end{array}\right) = \left(\begin{array}{cr} \frac{3}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{array}\right)$$ which again has $1$ as a repeated eigenvalue, but with $bc\neq 0$.

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The matrix $$\begin{pmatrix} 2&-1 \\ 1&4 \end{pmatrix}$$ has a repeated eigenvalue 3. This excludes the answer $bc=0$ and shows that $A$ need not be diagonal.

You write that $\det(A)\geq 0$ because $\det(A)$ is the square of the repeated eigenvalue. I agree. However, you should argue why the eigenvalue is real (unless you're doing a course where complex eigenvalues are not considered). A simple argument would be that the trace of $A$ is twice the repeated eigenvalue, so the eigenvalue must indeed be real.

Remark. You may wonder how to find the matrix given above. Suppose you have a matrix $\left(\begin{smallmatrix} a&b\\c&d \end{smallmatrix}\right)$. You want it to have a repeated eigenvalue of 3. This means that $a+d=6$, so you pick for example $a=2$ and $d=4$. Moreover, you need $ad-bc=9$, so $bc=-1$. This means you can take $b=-1$ and $c=1$.

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dear sir $a+d = 6$ ? – srijan May 9 '12 at 21:05
Yes, of course, thank you. Edited. – Daan Michiels May 9 '12 at 21:16