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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 \begin{equation}\operatorname{rank}(A-I)\geq2\end{equation}

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?

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  • $\begingroup$ 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 $\endgroup$
    – eepperly16
    Commented Nov 25, 2017 at 8:05
  • $\begingroup$ In general, there is no way of finding the rank of a matrix that is dramatically easier than finding the RREF $\endgroup$
    – eepperly16
    Commented Nov 25, 2017 at 8:06
  • $\begingroup$ The first two rows of $A-I$ are linearly independent. $\endgroup$
    – Angina Seng
    Commented Nov 25, 2017 at 8:10
  • $\begingroup$ @eepperly16 Thanks a lot, that really helped $\endgroup$
    – Sudheesh Surendranath
    Commented Nov 25, 2017 at 8:16
  • $\begingroup$ @sudh98 Please, if you are ok, you can set as solved. Thanks! $\endgroup$
    – user
    Commented Dec 3, 2017 at 12:17

2 Answers 2

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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,

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because the first and the third row of $A-I$ become equal and the second is not a multiple of them

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