# Spotting the geometric multiplicity

I know that the geometric multiplicity is defined as $dim(ker(A-\lambda I))$, but when I look up how to find the kernel of a matrix it seems to be slightly long and difficult. However, in all of the problem sheets and examples in my unit's lecture notes, the geometric multiplicity is found without calculation, and I'm not sure how. Is there a way you can 'spot' what it is?

For example, what is it for the matrices:

Where $\lambda = 1$ and $A= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ 0 & 1 & 1 \end{bmatrix}$

or where $\lambda=1$ and $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$

While it is not always possible, for your matrices, we can see the number of independent rows of $A-\lambda I$. Note that number of independent row is equal to the number of independent columns.

Using the formula

$$rank(A)+nullity(A)=n$$ where $n$ is the number of columns, we can compute the nullity quickly.

$$nullity(A-\lambda I)=dim(ker(A- \lambda I))$$

For example, $\lambda=1$ and $$A= \begin{bmatrix} 1 & 0 & 0 \\ 1 & 3 & 0 \\ 0 & 1 & 1 \end{bmatrix}$$

$$A-\lambda I=\begin{bmatrix} 0 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 1 & 0 \end{bmatrix}$$

Clearly, there are $2$ independent rows, hence the geometric multiplicy$=3-2=1$.

$$A-\lambda I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$
Clearly, there is $1$ independent row, hence the geometric multiplicty$=3-1=2$.