Determine whether A is diagonalizable using these three steps.

Question

Here is the question:

A is the matrix

$$ \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 1 & 2\end{bmatrix} $$

(a)find eigenvalues of A

(b)For each eigenvalue λ, find the rank of the matrix λI-A

(c)Is A diagonalizable? Justify your conclusion.

I learned about geometric multiplicities and algebraic multiplicities btw maybe that will help? anyway please keep it simple. thanks!

Answer 1

You are diagonalizable if you have say $k$ eigenvalues with corresponding eigenspaces, so that the sum of the dimensions of the $k$ eigen spaces adds up to the dimension of your vector space. The matrix you gave is not diagonalizable. The idea is you are diagonalizable if you have a basis for the vector space consisting of eigenvectors, so you need $n$ eigenvectors.

The rank comes in when you are determining the dimension of the eigenspaces.

Edit: That matrix is not diagonalizable either (I can tell quickly because it is in its Jordan Normal Form which you can look up if interested).

The idea is this. Say you have a diagonal matrix, where all the entries are non-zero. Then your vector space has a basis of eigenvectors, namely the standard vectors $e_{i}$. It's easy to see they are eigenvectors, since if $A$ is your matrix $Ae_{i}=a_{ii}e_{i}$ (just write out a simple example to convince yourself.

Now things like eigenvalues and dimensions of eigenspaces are not changed by similarity. That means, if $A$ is diagonalizable, in any basis it will have a basis consisting of eigenvectors. So, what you do is you first find the eigenvalues. Above, they are 3 and 2, with 2 having multiplicity 2. Now, what is the dimension of the eigenspace? It is the dimension of the kernel of $A-\lambda I$. The dimension of the kernel is the dimension of the vector space minus the rank of the matrix.

So you compute. For $\lambda=3$ you find the rank is $2$ so the eigenspace has dimension 1. But for $\lambda=2$ you have an eigenvalue with multiplicity 2 but computing gives you rank 2, so the eigenspace corresonding to $2$ has dimension 1. So you don't have enough eigenvectors, the sum of dimensions of eigenspaces is 2 instead of 3, so you are not diagonalizable.

At least that is how I think of this.

Answer 2

Without getting too technical, recall that you are trying to decompose $A$ so that

$$A = PDP^{-1}$$

where $D$ is the matrix of eigenvalues along the diagonal and $P$ is the matrix of eigenvectors. Note that in writing this, we have assumed that $P^{-1}$ exists, that is, $P$ needs to be comprised of linearly independent eigenvectors.

If you try to find eigenvectors for an eigenvalue with a repeated root, when you try to solve the system

$$(\lambda I-A)\pmb{x} = \pmb{0}$$

you will get one or more row of zeros in its reduced row echelon form. If there are not enough row of zeros (e.g. the dimension of the eigenspace corresponding to $\lambda$ is too small) you will not have enough vectors to put into the columns of $P$ because the parametric form of the solution to $(\lambda I-A)\pmb{x} = \pmb{0}$ will not have enough parameters for you to construct enough linearly independent eigenvectors (Since your eigenvectors are the columns of $P$, they must be linearly independent or $P^{-1}$ will not exist.

Note that by design, if $\lambda$ is a distinct eigenvalue, $(\lambda I -A)\pmb{x} = \pmb{0}$ will always give you a linearly independent eigenvector (why?). So diagonalizability depends on only the repeated eigenvalues.