# finding rank of matrix easily

So I was reading "Linear Algebra" by Hoffman and Kunze and I came across this, (since I don't have enough reputation to post pictures, I am quoting straight from the book)

$$A-I=\begin{bmatrix}4&-6&-6\\-1&3&2\\3&-6&-5\\\end{bmatrix}$$ We know $A-I$ is singular and obviously $\operatorname{rank}(A-I)\geq2$. Therefore, $\operatorname{rank}(A-I)=2$.

My question is, how is it obvious that $$\operatorname{rank}(A-I)\geq2$$

I know we can find the RREF and determine the rank. But is there any other way we can find it by just looking at the matrix?

• If one row of a matrix is not a multiple of the other, then matrix has at least two linearly independent rows and thus has rank at least two, which is why the author says this is obvious Commented Nov 25, 2017 at 8:05
• In general, there is no way of finding the rank of a matrix that is dramatically easier than finding the RREF Commented Nov 25, 2017 at 8:06
• The first two rows of $A-I$ are linearly independent. Commented Nov 25, 2017 at 8:10
• @eepperly16 Thanks a lot, that really helped Commented Nov 25, 2017 at 8:16
• @sudh98 Please, if you are ok, you can set as solved. Thanks!
– user
Commented Dec 3, 2017 at 12:17

Because the rank is $1$ if and only if one of the columns is non-null and all the other columns are that one times a scalar. That's obviously not the case here,
because the first and the third row of $A-I$ become equal and the second is not a multiple of them