What is an elegant way to find the third eigenvalue of $A=\left[\begin{smallmatrix} 51&-12 & -21\\ 60 & -40&28\\ 57&-68&1 \end{smallmatrix}\right]$?

Question

We had this in an exam today: Given the matrix $$A=\begin{bmatrix} 51&-12 & -21\\ 60 & -40&28\\ 57&-68&1 \end{bmatrix}$$ Here is the precise question as asked in our exam:

Someone tells you (accurately) that $-48$ and $24$ are eigenvalues of the matrix $A$. Without using a computer, calculator or writing anything down find the third eigenvalue of $A$.

I have used the classical method (through the characteristic polynomial of A) but this leaded to a some tremendous computation and finally got the answer.

But the question intentionally asked to do not make any use of such computation.

How can one elegantly find the third eigenvalue of $A$?

score 19 · Accepted Answer · answered Nov 23 '17 at 18:50

19

Use the following property:

Trace of a matrix is equal to the sum of the eigenvalues.

answered Nov 23 '17 at 18:50

Siong Thye Goh

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Excellent!${}{}{}$ – Dave Nov 23 '17 at 20:17
Impressive! Is there a short or intuitive proof for this property? – Zacky Dec 15 '18 at 12:23
here are some proofs. – Siong Thye Goh Dec 15 '18 at 12:39

What is an elegant way to find the third eigenvalue of $A=\left[\begin{smallmatrix} 51&-12 & -21\\ 60 & -40&28\\ 57&-68&1 \end{smallmatrix}\right]$?

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