Determining the rank of a matrix, am I missing something?

Question

I have problems determining the rank of the following matrix. Note: Most probably it is me missing something, but I am sincerely confused by this:

$$ \begin{matrix} 1 & 2 & -2 \\ 0 & 0 & 5 \\ 0 & 0 & -8 \\ \end{matrix} $$

Now, I would say that the matrix is in its reduced form because each one of the three entries under the main diagonal are zeroes. I'd say that the rank of this matrix is three, because that is the number of non-zero rows, but every calculator says that the rank is two. Why?

Thanks and sorry for taking your time!

Answer 1

It is not yet in reduced form: the last row is a (scalar) multiple of the middle row so you can further reduce it to: $$\begin{pmatrix} 1 & 2 & -2 \\ 0 & 0 & 5 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

From this form, it should be clear that the rank is $2$.

Now, I would say that the matrix is in its reduced form because each one of the three entries under the main diagonal are zeroes.

Take a look at The Row Echelon Form, I quote:

A rectangular matrix is in row echelon form if it has the following three properties:

• All nonzero rows are above any rows of all zeros
• Each leading entry of a row is in a column to the right of the leading entry of the row above it
• All entries in a column below a leading entry are zeros

Answer 2

The rank is, by definition the dimension of the image of your matrix, which is the space generated by the columns of your matrix. Now, as you can see, the second column is a multiple of the first one, and so the rank has to be at MOST 2. Finally, it should be easy to figure out that the last two columns are linearily independant !

PS: Your reduced form would rather be: $$ \begin{matrix} 1 & 2 & -2 \\ 0 & 0 & 5 \\ 0 & 0 & 0 \\ \end{matrix} $$

Answer 3

The first tow coloums are linearly dependent. So the rank is either 2 or 1. But the last tow coloums are linearly independent so the rank is greter than 2. So the rank is 2.