# Finding eigenvalues from a matrix that has constant a.

Given the matrix $$A= \left[ \begin{array}{ccc} 2&0&1\\ 1&1&a\\ 0&0&1 \end{array} \right]$$ Find all the eigenvalues of A.

My approach: $$\det( λI-A) = 0 =$$ $$\begin{vmatrix} λ-2&0&-1\\ -1&λ-1&-a\\ 0&-a&λ-1 \end{vmatrix}$$ However, after solving it, $$λ^3 -4 λ^2+(5-a^2) λ+(2a^2-2-a)=0$$. How do I proceed to solve it further to obtain my eigenvalues?

• You have a misprint in the matrix $A$ or in the determinant Nov 27 '18 at 8:22
• The error should be in the determinant, since then the eigenvalues are independent of $a$. Nov 27 '18 at 8:26

Assuming the matrix is correct, $$det(\lambda I-A)= \begin{vmatrix} λ-2&0&-1\\ -1&λ-1&-a\\ 0&0&λ-1 \end{vmatrix} =(\lambda-1)\begin{vmatrix} λ-2&0\\ -1&λ-1\end{vmatrix} =(\lambda-1)^2(\lambda-2)$$ and so, eigenvalues are $$\lambda=2$$ simple, and $$\lambda=1$$ double.

• Thanks. Corrected. May 17 '19 at 20:06

A matrix is singular amongst other things if it has a zero row or column, or if two rows or columns are linearly dependent. Subtracting $$\lambda$$ from diagonal elements just changed the diagonal.

I can see that I can change the $$1$$ in the bottom row to a zero by a suitable choice of $$\lambda$$ and if the top left entry were $$0$$ the bottom row would be a multiple of the top one. So I can read two eigenvalues off straight away. Then the trace - the sum of the diagonal entries - is the sum of the eigenvalues.

Now either that solves the problem without doing the computations, or - if the computation is expected - it acts as a robust check on your arithmetic.

I have assumed that the matrix first given is right. One of the zeros in the bottom row will appear in any sub-product of the determinant which involves $$a$$ so you can tell that the determinant will be independent of $$a$$.

This is to encourage you to look at the matrix for clues (columns will do as well as rows) which will either act as a check on your computations or give you short-cuts.

You’ve made an error in computing the determinant: it should be $$\lambda^3-4\lambda^2+5\lambda-2$$. However, it’s not necessary to go through all of that for this particular matrix.

A matrix and its transpose have the same eigenvalues, so we can examine the rows of the matrix. From the last row, we have the left eigenvector $$(0,0,1)$$ with eigenvalue $$1$$. If you add the first and third rows, you get $$(2,0,2)$$, but left-multiplying a matrix by $$(1,0,1)$$ performs this addition, so we have another eigenvector/eigenvalue pair: $$(1,0,1)$$ and $$2$$. The trace of a matrix is equal to the sum of its eigenvalues, so we can find the last eigenvalue “for free:” it’s $$4-2-1=1$$. With a bit of practice, you’ll be able to spot things like this.

Hint: $$\left[ \begin{array}{ccc} 2&0&1\\ 1&1&a\\ 0&0&1 \end{array} \right] = P\left[ \begin{array}{ccc} 2&1&1\\ 0&1&a\\ 0&0&1 \end{array} \right] P^{-1}$$

Where $$P$$ is a permutation matrix.