Constructing A Matrix From Know Qualities

Question

I am studying for my linear algebra final, and a question that has appeared on a few past exams is in the general form:

Create an example of a n by n matrix, with one of its eigenvalues equal to x, that is (not) diagonal, is (not) diagonalizable, and is (not) invertible.

The (not)s meaning a specific question may ask for the quality or state is must not have the quality. I'm unsure exactly how to approach this type of question. A specific example of this question type can be seen below:

e.g. Construct an example of a 2×2 matrix, with one of its eigenvalues equal to −1, that is not diagonal or diagonalizable, but is invertible.

if we take: $$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$ then we can deduce both $$ad-bc \neq 0$$ and $$(a+1)(d+1)-bc = 0$$

On how to make it not diagonalizable I can see how one could make -1 the only eigenvalue and somehow use an equation to ensure that it's multiplicity is 1. I'm not sure how to do this or how to move on from here. Any help would be appreciated on how to solve the general set of these questions or on how to solve this particular problem.

Thank you for the help.

Answer 1

Hint:

If a matrix, $A \in \mathbb{R}^{2 \times 2}$ has $2$ distinct eigenvalues, then it must be diagonalizable. Hence both of the eigenvalues must be $-1$.

Verify that $$\begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$$ satisfy the condition by checking that it is not diagonalizable.

Edit to response to OP's further question:

For $A\in \mathbb{R}^{4 \times 4}$, we can consider the matrix

$$\begin{bmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1 \end{bmatrix}$$

If we consider the matrix $A-\lambda I$, we get

$$A-\lambda I =\begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

We can see the $$nullity(A-\lambda I)=n-1 < n,$$ hence the geometric mulitplicity corresponding to $-1$ is less than $n$.

Answer 2

A general way to approach such problems is to construct a matrix out of Jordan blocks. The size and content of these blocks will be determined by the conditions of the problem: invertibility affects whether or not zero is an eigenvalue and each known eigenvalue contributes at least one Jordan block. If the matrix is diagonalizable, but not diagonal, you could first construct a diagonal matrix out of the eigenvalues then apply a change of basis to it.

To take your example, since the matrix is not diagonalizable, it must have a Jordan block of size $\ge2$. The matrix itself is only $2\times2$, so it only has room for a single such Jordan block and nothing else. You have a known eigenvalue $-1$ (and in fact it is the only eigenvalue) so the Jordan matrix is $$\left[\begin{array}{rr}-1&1\\0&-1\end{array}\right].$$ This is in a sense the most basic matrix that satisfies the conditions of your example, but any matrix similar to this one will also fit the bill.