Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

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 $bc\neq 0$.

share|cite|improve this answer

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 $c=1$.

share|cite|improve this answer
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

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.