# An Eigen Value of $\tiny \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :

Let $a,b,c,d$ be distinct non zero real numbers with $a+b=c+d.$ Then, an eigen value of the matrix $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$ is :

$(i)~a+c ~~(ii)~a+b~~(iii)~a-b~~(iv)~(b-d$

Attempt:

If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :

$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$

$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$

Solving $\det ( A - \lambda I) = 0$ results in a third degree equation which is difficult to solve.

Could anyone tell me of an easier method to solve it. Thank you very much for your help in this regard.

• Not really, I believe they all should add up to $a+d$, from taking the trace. – Zach466920 Apr 9 '15 at 15:09
• Try the vector $\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}$ – eranreches Apr 9 '15 at 15:10
• @eranreches Bet me to it. You should write that as an answer. I'd upvote it. – Timbuc Apr 9 '15 at 15:12
• You don't need to compute all its eigenvalues; you just need one particular one, which @eranreches suggested. – anomaly Apr 9 '15 at 15:15
• @RossMillikan My bad. I have edited it – MathMan Apr 9 '15 at 15:32

$a+b=c+d$ implies that we should add up the first two columns of the matrix using a vector of the form $\begin{pmatrix} 1\\ 1\\ t \end{pmatrix}$.

You can see that if $A$ is the matrix, then $A\begin{pmatrix} 1\\ 1\\ t \end{pmatrix}$=$\begin{pmatrix} a+b+t\\ c+d+t\\ 0 \end{pmatrix}$.

Therefore, the suitable choice is of course $t=0$, and so the eigenvalue is $a+b$.

Thank you all for your replies. While I was just reflecting, I thought of the following solution, so, thought of sharing the following solution :

If $A= \begin{pmatrix} a&b&1 \\ c&d&1 \\ 1&-1&0\\ \end{pmatrix}$, then :

$A - \lambda I = \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$

$\det ( A- \lambda I) = \det \begin{pmatrix} a-\lambda&b&1 \\ c&d-\lambda&1 \\ 1&-1&-\lambda\\ \end{pmatrix}$

Adding the first two columns and placing it in the first results in no change in the determinant value :

Hence :

$\det ( A- \lambda I) = \det \begin{pmatrix} a+b-\lambda&b&1 \\ c+d-\lambda&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$

$= (a+b-\lambda) \det \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix} = (c+d-\lambda) \det \begin{pmatrix} 1&b&1 \\ 1&d-\lambda&1 \\ 0&-1&-\lambda\\ \end{pmatrix}$

Hence, $\det ( A- \lambda I) =0 \implies a+b = \lambda = c+d$

The characteristic polynomial is not difficult to solve. Setting $d=a+b-c$ it is $$\chi (A)=t^3 + t^2( - 2a - b + c) + t(a^2 + ab - ac - bc)=t(a + b - t)(a - c - t),$$ so that $0$, $a+b$ and $d-b$ are the eigenvalues. This solution has the advantage that you obtain all eigenvalues.

• That last factoring doesn't look pretty straightforward to me. Is there any trick you used, or else you factored out $\;t\;$ and then solved the quadratic? – Timbuc Apr 9 '15 at 15:38
• @Timbuc Yes, it is only a quadratic polynomial left to factor. This is not difficult, one could use Vieta. – Dietrich Burde Apr 9 '15 at 17:43

If you write out the cubic you will see that you can group the quantity $-(c+d)$ together. If you replace that quantity with $-(a+b)$ you can cancel it with the $(a+b)$ that appears elsewhere in the cubic. Additionally, you will be able to cancel a $\lambda$ term with a $-\lambda$. When all is said and done you should find that $$\det(A-\lambda I) = -\lambda(a-\lambda)(d-\lambda)+bc\lambda$$ and you should have a much easier time solving $$-\lambda(a-\lambda)(d-\lambda)+bc\lambda=0$$ Can you proceed from here?